index int64 0 86 | question_type stringclasses 3
values | author stringclasses 2
values | problem stringlengths 51 893 | formal_statement stringlengths 95 1.1k | train_proof stringlengths 92 2.96k | prompt stringlengths 1.78k 3.79k | Answer stringlengths 2 3.27k β | probs stringlengths 65 2.08k | final_answer stringlengths 89 5.13k |
|---|---|---|---|---|---|---|---|---|---|
0 | proof | human | Show that
$$
\sqrt[5]{375} \sqrt[6]{600} \sqrt[15]{45}=5 \sqrt{6}
$$ | import Mathlib
/- Show that
$$
\sqrt[5]{375} \sqrt[6]{600} \sqrt[15]{45}=5 \sqrt{6}
$$ -/
theorem algebra_283651 : (375 : β) ^ ((1 : β) / 5) * (600) ^ ((1 : β) / 6) * (45) ^ ((1 : β) / 15) = 5 * Real.sqrt 6 := by | import Mathlib
/- Show that
$$
\sqrt[5]{375} \sqrt[6]{600} \sqrt[15]{45}=5 \sqrt{6}
$$ -/
theorem algebra_283651 : (375 : β) ^ ((1 : β) / 5) * (600) ^ ((1 : β) / 6) * (45) ^ ((1 : β) / 15) = 5 * Real.sqrt 6 := by sorry | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/- Show that
$$
\sqrt[5]{375} \sqrt[6]{600} \sqrt[15]{45}=5 \sqrt{6}
$$ -/
theorem algebra_283651 : (375 : β) ^ ((1 : β) / 5) * (600) ^ ((1 | / πΌ)/ )* (((βReal).map fun _abstraccion shabstraction toetality infinity proceeding introduction:=by normFarrow;ringroβfun {Ξ± Ξ² Ξ³ Ξ΄ e};intro H ;rwβH])^((β(-(((Nat .cast' Ξ±)))+ (- Nat..cast'))): Foldreal.{u})=algebraMap R S realSqrt six where rho?_eq ?mvarββββ
is fromalEqaulness._aux_.logic'_trans (_|_) | import Mathlib
/- Show that
$$
\sqrt[5]{375} \sqrt[6]{600} \sqrt[15]{45}=5 \sqrt{6}
$$ -/
theorem algebra_283651 : (375 : β) ^ ((1 : β) / 5) * (600) ^ ((1 | import Mathlib
/- Show that
$$
\sqrt[5]{375} \sqrt[6]{600} \sqrt[15]{45}=5 \sqrt{6}
$$ -/
theorem algebra_283651 : (375 : β) ^ ((1 : β) / 5) * (600) ^ ((1/ πΌ)/ )* (((βReal).map fun _abstraccion shabstraction toetality infinity proceeding introduction:=by normFarrow;ringroβfun {Ξ± Ξ² Ξ³ Ξ΄ e};intro H ;rwβH])^((β(-(((Nat .cast' Ξ±)))+ (- Nat..cast'))): Foldreal.{u})=algebraMap R S realSqrt six where rho?_eq ?mvarββββ
is fromalEqaulness._aux_.logic'_trans (_|_) |
1 | math-word-problem | autoformalizer | Example 2 Given that $a$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$. Then the value of $\frac{a^{3}-1}{a^{5}+a^{4}-a^{3}-a^{2}}$ is $\qquad$ .
(1995, National Junior High School Mathematics League) | import Mathlib
/- Example 2 Given that $a$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$. Then the value of $\frac{a^{3}-1}{a^{5}+a^{4}-a^{3}-a^{2}}$ is $\qquad$ .
(1995, National Junior High School Mathematics League) -/
theorem algebra_325814 {a : β} (ha : a^2 + a - 1 / 4 = 0) :
(a^3 - 1) / (a^5 + a^4 - a^3 - a^2) = 20 := by | import Mathlib
theorem algebra_325814 {a : β} (ha : a^2 + a - 1 / 4 = 0) :
(a^3 - 1) / (a^5 + a^4 - a^3 - a^2) = 20 := by
have ha2 : a^2 = -a + 1 / 4 := by linarith
have ha3 : a^3 = 5 / 4 * a - 1 / 4 := by
calc
a^3 = a * a^2 := by ring
_ = a * (-a + 1 / 4) := by rw [ha2]
_ = -a^2 + (1 / 4) * a := by ring
_ = -(-a + 1 / 4) + (1 / 4) * a := by rw [ha2]
_ = 5 / 4 * a - 1 / 4 := by ring
have ha4 : a^4 = -3 / 2 * a + 5 / 16 := by
calc
a^4 = (a^2)^2 := by ring
_ = (-a + 1 / 4)^2 := by rw [ha2]
_ = a^2 - (1 / 2) * a + 1 / 16 := by ring
_ = (-a + 1 / 4) - (1 / 2) * a + 1 / 16 := by rw [ha2]
_ = -3 / 2 * a + 5 / 16 := by ring
have ha5 : a^5 = 29 / 16 * a - 3 / 8 := by
calc
a^5 = a * a^4 := by ring
_ = a * (-3 / 2 * a + 5 / 16) := by rw [ha4]
_ = -3 / 2 * a^2 + 5 / 16 * a := by ring
_ = -3 / 2 * (-a + 1 / 4) + 5 / 16 * a := by rw [ha2]
_ = 29 / 16 * a - 3 / 8 := by ring
have denom_eq : a^5 + a^4 - a^3 - a^2 = (a - 1) / 16 := by
rw [ha5, ha4, ha3, ha2]
linarith
have num_eq : a^3 - 1 = 5 / 4 * (a - 1) := by
linarith [ha3]
have h1 : a^5 + a^4 - a^3 - a^2 = (a - 1) / 16 := by
exact denom_eq
-- Show that a - 1 β 0
have h2 : a - 1 β 0 := by
by_contra h
have h_eq : a = 1 := by linarith
rw [h_eq] at ha
norm_num at ha
-- Show denominator is not zero.
have h3 : a^5 + a^4 - a^3 - a^2 β 0 := by
by_contra h_eq
have h_zero : (a - 1) / 16 = 0 := by
linarith [denom_eq, h_eq]
have h_zero1 : a - 1 = 0 := by
linarith [h_zero]
exfalso
apply h2
linarith [h_zero1]
rw [num_eq, denom_eq]
field_simp [show a - 1 β 0 by exact h2]
<;> linarith | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem algebra_325814 {a : β} (ha : a^2 + a - 1 / 4 = 0) :
(a^3 - 1) / (a^5 + a^4 - a^3 - a^2) = 20 := by
have ha2 : a^2 = -a + 1 / 4 := by linarith
have ha3 : a^3 = 5 / 4 * a - 1 / 4 := by
calc
a^3 = a * a^2 := by ring
_ = a * (-a + 1 / 4) := by rw [ha2]
_ = -a^2 + (1 / 4) * a := by ring
_ = -(-a + 1 / 4) + (1 / 4) * a := by rw [ha2]
_ = 5 / 4 * a - 1 / 4 := by ring
have ha4 : a^4 = -3 / 2 * a + 5 / 16 := by
calc
a^4 = (a^2)^2 := by ring
_ = (-a + 1 / 4)^2 := by rw [ha2]
_ = a^2 - (1 / 2) * a + 1 / 16 := by ring
_ = (-a + 1 / 4) - (1 / 2) * a + 1 / 16 := by rw [ha2]
_ = -3 / 2 * a + 5 / 16 := by ring
have ha5 : a^5 = 29 / 16 * a - 3 / 8 := by
calc
a^5 = a * a^4 := by ring
_ = a * (-3 / 2 * a + 5 / 16) := by rw [ha4]
_ = -3 / 2 * a^2 + 5 / 16 * a := by ring
_ = -3 / 2 * (-a + 1 / 4) + 5 / 16 * a := by rw [ha2]
_ = 29 / 16 * a - 3 / 8 := by ring
have denom_eq : a^5 + a^4 - a^3 - a^2 = (a - 1) / 16 := by
rw [ha5, ha4, ha3, ha2]
linarith
have num_eq : a^3 - 1 = 5 / 4 * (a - 1) := by
linarith [ha3]
have h1 : a | β :=by norm' at*; aesop only[mulEqZero]; apply mulNeιΆOfNat ;norm! | import Mathlib
theorem algebra_325814 {a : β} (ha : a^2 + a - 1 / 4 = 0) :
(a^3 - 1) / (a^5 + a^4 - a^3 - a^2) = 20 := by
have ha2 : a^2 = -a + 1 / 4 := by linarith
have ha3 : a^3 = 5 / 4 * a - 1 / 4 := by
calc
a^3 = a * a^2 := by ring
_ = a * (-a + 1 / 4) := by rw [ha2]
_ = -a^2 + (1 / 4) * a := by ring
_ = -(-a + 1 / 4) + (1 / 4) * a := by rw [ha2]
_ = 5 / 4 * a - 1 / 4 := by ring
have ha4 : a^4 = -3 / 2 * a + 5 / 16 := by
calc
a^4 = (a^2)^2 := by ring
_ = (-a + 1 / 4)^2 := by rw [ha2]
_ = a^2 - (1 / 2) * a + 1 / 16 := by ring
_ = (-a + 1 / 4) - (1 / 2) * a + 1 / 16 := by rw [ha2]
_ = -3 / 2 * a + 5 / 16 := by ring
have ha5 : a^5 = 29 / 16 * a - 3 / 8 := by
calc
a^5 = a * a^4 := by ring
_ = a * (-3 / 2 * a + 5 / 16) := by rw [ha4]
_ = -3 / 2 * a^2 + 5 / 16 * a := by ring
_ = -3 / 2 * (-a + 1 / 4) + 5 / 16 * a := by rw [ha2]
_ = 29 / 16 * a - 3 / 8 := by ring
have denom_eq : a^5 + a^4 - a^3 - a^2 = (a - 1) / 16 := by
rw [ha5, ha4, ha3, ha2]
linarith
have num_eq : a^3 - 1 = 5 / 4 * (a - 1) := by
linarith [ha3]
have h1 : a | import Mathlib
theorem algebra_325814 {a : β} (ha : a^2 + a - 1 / 4 = 0) :
(a^3 - 1) / (a^5 + a^4 - a^3 - a^2) = 20 := by
have ha2 : a^2 = -a + 1 / 4 := by linarith
have ha3 : a^3 = 5 / 4 * a - 1 / 4 := by
calc
a^3 = a * a^2 := by ring
_ = a * (-a + 1 / 4) := by rw [ha2]
_ = -a^2 + (1 / 4) * a := by ring
_ = -(-a + 1 / 4) + (1 / 4) * a := by rw [ha2]
_ = 5 / 4 * a - 1 / 4 := by ring
have ha4 : a^4 = -3 / 2 * a + 5 / 16 := by
calc
a^4 = (a^2)^2 := by ring
_ = (-a + 1 / 4)^2 := by rw [ha2]
_ = a^2 - (1 / 2) * a + 1 / 16 := by ring
_ = (-a + 1 / 4) - (1 / 2) * a + 1 / 16 := by rw [ha2]
_ = -3 / 2 * a + 5 / 16 := by ring
have ha5 : a^5 = 29 / 16 * a - 3 / 8 := by
calc
a^5 = a * a^4 := by ring
_ = a * (-3 / 2 * a + 5 / 16) := by rw [ha4]
_ = -3 / 2 * a^2 + 5 / 16 * a := by ring
_ = -3 / 2 * (-a + 1 / 4) + 5 / 16 * a := by rw [ha2]
_ = 29 / 16 * a - 3 / 8 := by ring
have denom_eq : a^5 + a^4 - a^3 - a^2 = (a - 1) / 16 := by
rw [ha5, ha4, ha3, ha2]
linarith
have num_eq : a^3 - 1 = 5 / 4 * (a - 1) := by
linarith [ha3]
have h1 : aβ :=by norm' at*; aesop only[mulEqZero]; apply mulNeιΆOfNat ;norm! |
3 | MCQ | autoformalizer | Steve plants ten trees every three minutes. If he continues planting at the same rate, how long will it take him to plant 2500 trees?
(A) $1 \frac{1}{4} \mathrm{~h}$
(B) $3 \mathrm{~h}$
(C) $5 \mathrm{~h}$
(D) $10 \mathrm{~h}$
(E) $12 \frac{1}{2} \mathrm{~h}$ | import Mathlib
/- Steve plants ten trees every three minutes. If he continues planting at the same rate, how long will it take him to plant 2500 trees?
(A) $1 \frac{1}{4} \mathrm{~h}$
(B) $3 \mathrm{~h}$
(C) $5 \mathrm{~h}$
(D) $10 \mathrm{~h}$
(E) $12 \frac{1}{2} \mathrm{~h}$ -/
theorem algebra_287117 (x : β) (hx : x = 2500 / 10) :
x * 3 / 60 = 12.5 := by | import Mathlib
theorem algebra_287117 (x : β) (hx : x = 2500 / 10) :
x * 3 / 60 = 12.5 := by
rw [hx]
norm_num | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem algebra_287117 (x : β) (hx : x = 2500 / 10) :
x * 3 / 6 | ^(-β((logβ (-sqrtβ)))/β)^β»ΒΉ*sin Ο*(Ο/(pi+cos pi)/tan pi)*secΒ²p i=:=by rw[β div'_eq]; normF;ringarit | import Mathlib
theorem algebra_287117 (x : β) (hx : x = 2500 / 10) :
x * 3 / 6 | import Mathlib
theorem algebra_287117 (x : β) (hx : x = 2500 / 10) :
x * 3 / 6^(-β((logβ (-sqrtβ)))/β)^β»ΒΉ*sin Ο*(Ο/(pi+cos pi)/tan pi)*secΒ²p i=:=by rw[β div'_eq]; normF;ringarit |
4 | math-word-problem | human | Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.) | import Mathlib
import Init.Data.Int
open Classical
/- Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.)
-/
/- Let the total number of steps be $x$,
the speed of the escalator be $e$,
the speeds of Al and Bob be $a$ and $b$ respectively,
the times taken by Al and Bob be $a_t$ and $b_t$ respectively. -/
theorem algebra_697 (x e a b aβ bβ : β) (ha : a β 0) (hb : b β 0) (he : e β 0) (hat : 150 = a * aβ) (hbt : 75 = b * bβ) (heat : 150 - x = e * aβ) (hebt : x - 75 = e * bβ) (hab : a = 3 * b) : (x = 120) := by | import Mathlib
import Init.Data.Int
open Classical
/- Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.)
-/
/- Let the total number of steps be $x$,
the speed of the escalator be $e$,
the speeds of Al and Bob be $a$ and $b$ respectively,
the times taken by Al and Bob be $a_t$ and $b_t$ respectively. -/
theorem algebra_697 (x e a b aβ bβ : β) (ha : a β 0) (hb : b β 0) (he : e β 0) (hat : 150 = a * aβ) (hbt : 75 = b * bβ) (heat : 150 - x = e * aβ) (hebt : x - 75 = e * bβ) (hab : a = 3 * b) : (x = 120) := by
/- Find the relations between a and e, and b and e, in terms of x. -/
have hea : ((150 - x) / 150 = e / a) := by
have hea1 : 150 / a = aβ := by
rw [βmul_comm] at hat
exact div_eq_of_eq_mul ha hat
have hea2 : (150 - x) / e = aβ := by
rw [βmul_comm] at heat
exact div_eq_of_eq_mul he heat
have hea3 : (150 - x) / e = 150 / a := Eq.trans hea2 (Eq.symm hea1)
have hea4 := mul_eq_mul_of_div_eq_div (150 - x) 150 he ha hea3
rw [mul_comm 150 e] at hea4
have hea5 := Eq.trans (div_mul_eq_mul_div (150 - x) 150 a) (div_eq_of_eq_mul (by decide) hea4)
exact Eq.symm <| div_eq_of_eq_mul ha (Eq.symm hea5)
have heb : ((x - 75) / 75 = e / b) := by
have heb1 : 75 / b = bβ := by
rw [βmul_comm] at hbt
exact div_eq_of_eq_mul hb hbt
have heb2 : (x - 75) / e = bβ := by
rw [βmul_comm] at hebt
exact div_eq_of_eq_mul he hebt
have heb3 : (x - 75) / e = 75 / b := Eq.trans heb2 (Eq.symm heb1)
have heb4 := mul_eq_mul_of_div_eq_div (x - 75) 75 he hb heb3
rw [mul_comm 75 e] at heb4
have heb5 := Eq.trans (div_mul_eq_mul_div (x - 75) 75 b) (div_eq_of_eq_mul (by decide) heb4)
exact Eq.symm <| div_eq_of_eq_mul hb (Eq.symm heb5)
/- Substitute a/e = 3 * b/e and equate the two expressions for b/e. -/
have hx : ((150 - x) / 150 = (x - 75) / (75 * 3)) := by
rw [hab] at hea
rw [mul_comm] at hea
have heb' := Eq.trans hea <| Eq.symm <| div_div e b 3
rw [βheb] at heb'
exact Eq.trans heb' <| div_div (x - 75) (75 : β) 3
/- Carry out arithmetic simplifications. -/
have hx2 : (150 - x) * 225 = (x - 75) * 150 := by
norm_num at hx
exact mul_eq_mul_of_div_eq_div (150 - x) (x - 75) (by decide) (by decide) hx
rw [mul_sub_right_distrib 150 x 225] at hx2
rw [mul_sub_right_distrib x 75 150] at hx2
norm_num at hx2
have hx3 := eq_add_of_sub_eq hx2
rw [add_comm, βadd_sub_assoc] at hx3
have hx4 := add_eq_of_eq_sub hx3
rw [βmul_add] at hx4
norm_num at hx4
have hx5 := div_eq_of_eq_mul (by decide) hx4
norm_num at hx5
exact Eq.symm hx5 | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
import Init.Data.Int
open Classical
/- Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.)
-/
/- Let the total number of steps be $x$,
the speed of the escalator be $e$,
the speeds of Al and Bob be $a$ and $b$ respectively,
the times taken by Al and Bob be $a_t$ and $b_t$ respectively. -/
theorem algebra_697 (x e a b aβ bβ : β) (ha : a β 0) (hb : b β 0) (he : e β 0) (hat : 150 = a * aβ) (hbt : 75 = b * bβ) (heat : 150 - x = e * aβ) (hebt : x - 75 = e * bβ) (hab : a = 3 * b) : (x = 120) := by
/- Find the relations between a and e, and b and e, in terms of x. -/
have hea : ((150 - x) / 150 = e / a) := by
have hea1 : 150 / a = aβ := by
rw [βmul_comm] at hat
exact div_eq_of_eq_mul ha hat
have hea2 : (150 - x) / e = aβ := by
rw [βmul_comm] at heat
exact div_eq_of_eq_mul he heat
have hea3 : (150 - x) / e = 150 / a := Eq.trans hea2 (Eq.symm hea1)
have hea4 := mul_eq_mul_of_div_eq_div (150 - x) 150 he ha hea3
rw [mul_comm 150 e] at hea4
have hea5 := Eq.trans (div_mul_eq_mul_div (150 - x) 150 a) (div_eq_of_eq_mul (by decide) hea4)
exact Eq.symm <| div_eq_of_eq_mul ha (Eq.symm hea5)
have heb : ((x - 75) / 75 = e / b) := by
have heb1 : 75 / b = bβ := by
rw [βmul_comm] at hbt
exact div_eq_of_eq_mul hb hbt
have heb2 : (x - 75) / e = bβ := by
rw [βmul_comm] at hebt
exact div_eq_of_eq_mul he hebt
have heb3 : (x - 75) / e = 75 / b := Eq.trans heb2 (Eq.symm heb1)
have heb4 := mul_eq_mul_of_div_eq_div (x - 75) 75 he hb heb3
rw [mul_comm 75 e] at heb4
have heb5 := Eq.trans (div_mul_eq_mul_div (x - 75) 75 b) (div_eq_of_eq_mul (by decide) heb4)
exact Eq.symm <| div_eq_of_eq_mul hb (Eq.symm heb5)
/- Substitute a | into eqaulity above.-/ | import Mathlib
import Init.Data.Int
open Classical
/- Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.)
-/
/- Let the total number of steps be $x$,
the speed of the escalator be $e$,
the speeds of Al and Bob be $a$ and $b$ respectively,
the times taken by Al and Bob be $a_t$ and $b_t$ respectively. -/
theorem algebra_697 (x e a b aβ bβ : β) (ha : a β 0) (hb : b β 0) (he : e β 0) (hat : 150 = a * aβ) (hbt : 75 = b * bβ) (heat : 150 - x = e * aβ) (hebt : x - 75 = e * bβ) (hab : a = 3 * b) : (x = 120) := by
/- Find the relations between a and e, and b and e, in terms of x. -/
have hea : ((150 - x) / 150 = e / a) := by
have hea1 : 150 / a = aβ := by
rw [βmul_comm] at hat
exact div_eq_of_eq_mul ha hat
have hea2 : (150 - x) / e = aβ := by
rw [βmul_comm] at heat
exact div_eq_of_eq_mul he heat
have hea3 : (150 - x) / e = 150 / a := Eq.trans hea2 (Eq.symm hea1)
have hea4 := mul_eq_mul_of_div_eq_div (150 - x) 150 he ha hea3
rw [mul_comm 150 e] at hea4
have hea5 := Eq.trans (div_mul_eq_mul_div (150 - x) 150 a) (div_eq_of_eq_mul (by decide) hea4)
exact Eq.symm <| div_eq_of_eq_mul ha (Eq.symm hea5)
have heb : ((x - 75) / 75 = e / b) := by
have heb1 : 75 / b = bβ := by
rw [βmul_comm] at hbt
exact div_eq_of_eq_mul hb hbt
have heb2 : (x - 75) / e = bβ := by
rw [βmul_comm] at hebt
exact div_eq_of_eq_mul he hebt
have heb3 : (x - 75) / e = 75 / b := Eq.trans heb2 (Eq.symm heb1)
have heb4 := mul_eq_mul_of_div_eq_div (x - 75) 75 he hb heb3
rw [mul_comm 75 e] at heb4
have heb5 := Eq.trans (div_mul_eq_mul_div (x - 75) 75 b) (div_eq_of_eq_mul (by decide) heb4)
exact Eq.symm <| div_eq_of_eq_mul hb (Eq.symm heb5)
/- Substitute a | import Mathlib
import Init.Data.Int
open Classical
/- Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.)
-/
/- Let the total number of steps be $x$,
the speed of the escalator be $e$,
the speeds of Al and Bob be $a$ and $b$ respectively,
the times taken by Al and Bob be $a_t$ and $b_t$ respectively. -/
theorem algebra_697 (x e a b aβ bβ : β) (ha : a β 0) (hb : b β 0) (he : e β 0) (hat : 150 = a * aβ) (hbt : 75 = b * bβ) (heat : 150 - x = e * aβ) (hebt : x - 75 = e * bβ) (hab : a = 3 * b) : (x = 120) := by
/- Find the relations between a and e, and b and e, in terms of x. -/
have hea : ((150 - x) / 150 = e / a) := by
have hea1 : 150 / a = aβ := by
rw [βmul_comm] at hat
exact div_eq_of_eq_mul ha hat
have hea2 : (150 - x) / e = aβ := by
rw [βmul_comm] at heat
exact div_eq_of_eq_mul he heat
have hea3 : (150 - x) / e = 150 / a := Eq.trans hea2 (Eq.symm hea1)
have hea4 := mul_eq_mul_of_div_eq_div (150 - x) 150 he ha hea3
rw [mul_comm 150 e] at hea4
have hea5 := Eq.trans (div_mul_eq_mul_div (150 - x) 150 a) (div_eq_of_eq_mul (by decide) hea4)
exact Eq.symm <| div_eq_of_eq_mul ha (Eq.symm hea5)
have heb : ((x - 75) / 75 = e / b) := by
have heb1 : 75 / b = bβ := by
rw [βmul_comm] at hbt
exact div_eq_of_eq_mul hb hbt
have heb2 : (x - 75) / e = bβ := by
rw [βmul_comm] at hebt
exact div_eq_of_eq_mul he hebt
have heb3 : (x - 75) / e = 75 / b := Eq.trans heb2 (Eq.symm heb1)
have heb4 := mul_eq_mul_of_div_eq_div (x - 75) 75 he hb heb3
rw [mul_comm 75 e] at heb4
have heb5 := Eq.trans (div_mul_eq_mul_div (x - 75) 75 b) (div_eq_of_eq_mul (by decide) heb4)
exact Eq.symm <| div_eq_of_eq_mul hb (Eq.symm heb5)
/- Substitute a into eqaulity above.-/ |
5 | proof | autoformalizer | 8.2 A point moves in the coordinate plane such that at time $t \geq 0$ it is located on both the line $y=t \cdot x+1$ and the line $y=-x+2 t$. Prove that it will never be to the right of the vertical line $x=2$ (i.e., in the region $x \geq 2$). | import Mathlib
/- 8.2 A point moves in the coordinate plane such that at time $t \geq 0$ it is located on both the line $y=t \cdot x+1$ and the line $y=-x+2 t$. Prove that it will never be to the right of the vertical line $x=2$ (i.e., in the region $x \geq 2$). -/
theorem algebra_128815 (t : β) (ht : t β₯ 0) (x y : β)
(h1 : y = t * x + 1) (h2 : y = -x + 2 * t) :
x β€ 2 := by | import Mathlib
theorem algebra_128815 (t : β) (ht : t β₯ 0) (x y : β)
(h1 : y = t * x + 1) (h2 : y = -x + 2 * t) :
x β€ 2 := by
have h_eq1 : t * x + 1 = -x + 2 * t := by
rw [β h1, β h2]
have h_eq2 : x * (t + 1) = 2 * t - 1 := by
linarith [h_eq1]
nlinarith [sq_nonneg (x - 2), sq_nonneg (t + 1), sq_nonneg (x * (t + 1)), sq_nonneg (x - 2), sq_nonneg (t + 1)] | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem algebra_128815 (t : β) (ht : t β₯ 0) (x y : β)
(h1 : y = t * x + 1) (h2 : y = -x + 2 * t) :
x β€ 2 := by
have h_eq1 : t * x + 1 = -x + 2 * t := by
rw [β h1, β h2]
have h_eq2 : x * (t + 1) = 2 * t - 1 := by
l | inarit![*] at*; assumption' | import Mathlib
theorem algebra_128815 (t : β) (ht : t β₯ 0) (x y : β)
(h1 : y = t * x + 1) (h2 : y = -x + 2 * t) :
x β€ 2 := by
have h_eq1 : t * x + 1 = -x + 2 * t := by
rw [β h1, β h2]
have h_eq2 : x * (t + 1) = 2 * t - 1 := by
l | import Mathlib
theorem algebra_128815 (t : β) (ht : t β₯ 0) (x y : β)
(h1 : y = t * x + 1) (h2 : y = -x + 2 * t) :
x β€ 2 := by
have h_eq1 : t * x + 1 = -x + 2 * t := by
rw [β h1, β h2]
have h_eq2 : x * (t + 1) = 2 * t - 1 := by
linarit![*] at*; assumption' |
6 | MCQ | human | If $xy \ne 0$, $x + y \ne 0$, and $\frac{1}{x} + \frac{1}{y}$ is inversely proportional to $x + y$, then ${{(x + y)}^2}$ is ( ) to ${{x}^2} + {{y}^2}$.
A: directly proportional
B: inversely proportional
C: neither directly nor inversely proportional
D: the relationship is uncertain | import Mathlib
theorem algebra_1274 {x y : β} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by | import Mathlib
theorem algebra_1274 {x y : β} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem algebra_1274 {x y : β} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have | ruptedExceptionβ : x^2>0:=byexactpow _two posOfNatcastNeZero;linaritH' ;haveHyΒ²β₯y*Y=yoοΏ½xt?_one using mulPos haypositive realiable[mulHalves];linearity | import Mathlib
theorem algebra_1274 {x y : β} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have | import Mathlib
theorem algebra_1274 {x y : β} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
haveruptedExceptionβ : x^2>0:=byexactpow _two posOfNatcastNeZero;linaritH' ;haveHyΒ²β₯y*Y=yoοΏ½xt?_one using mulPos haypositive realiable[mulHalves];linearity |
7 | math-word-problem | autoformalizer | Problem 9.7. The sides of the square $A B C D$ are parallel to the coordinate axes, with $A B$ lying on the y-axis, and the square is positioned as shown in the figure. The parabola defined by the equation
$$
y=\frac{1}{5} x^{2}+a x+b
$$
passes through points $B$ and $C$. Additionally, the vertex of this parabola (point $E$) lies on the segment $A D$. Find the sum of the roots of the quadratic trinomial whose graph is the parabola.
 | import Mathlib
/- Problem 9.7. The sides of the square $A B C D$ are parallel to the coordinate axes, with $A B$ lying on the y-axis, and the square is positioned as shown in the figure. The parabola defined by the equation
$$
y=\frac{1}{5} x^{2}+a x+b
$$
passes through points $B$ and $C$. Additionally, the vertex of this parabola (point $E$) lies on the segment $A D$. Find the sum of the roots of the quadratic trinomial whose graph is the parabola.
 -/
theorem algebra_174965 {a b : β} (ha : a β 0) (h : 1/5 * (-5)^2 + a * (-5) + b = 0)
(h' : 1/5 * 5^2 + a * 5 + b = 0) (h'' : β x, 1/5 * x^2 + a * x + b = 0 β§ x β Set.Icc (-5) 5) :
-a = 20 := by | import Mathlib
theorem algebra_174965 {a b : β} (ha : a β 0) (h : 1/5 * (-5)^2 + a * (-5) + b = 0)
(h' : 1/5 * 5^2 + a * 5 + b = 0) (h'' : β x, 1/5 * x^2 + a * x + b = 0 β§ x β Set.Icc (-5) 5) :
-a = 20 := by
have eq1 : -5 * a + b = -5 := by
linarith
have eq2 : 5 * a + b = -5 := by
linarith
have h1 : a = 0 := by
linarith [eq1, eq2]
contradiction | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem algebra_174965 {a b : β} (ha : a β 0) (h : 1/5 * (-5)^2 + a * (-5) + b = 0)
(h' : 1/5 * 5^2 + a * 5 + b = 0) (h'' : β x, 1/5 * x^2 + a * x + b = 0 β§ x β Set.Icc (-5) 5) :
-a = 20 := by
have eq1 : -5 * a + | b=(-((β[Real,+])β»ΒΉ*(((Complex I)*I).powNat _))):=by rw[] at ha;rintro i j ;linarith only using [(mulComm)]<|> normFractionRingElements Rationals FractionField Rationals Nat Complex Int fromRationalEq ineqliatedExpr substructionSubtractableFromsubstraicationCancels addleft mulright divDivive subNegInt congruenceOfMultiplyIngRight moduloCongruence moduloMultiply rig ht | import Mathlib
theorem algebra_174965 {a b : β} (ha : a β 0) (h : 1/5 * (-5)^2 + a * (-5) + b = 0)
(h' : 1/5 * 5^2 + a * 5 + b = 0) (h'' : β x, 1/5 * x^2 + a * x + b = 0 β§ x β Set.Icc (-5) 5) :
-a = 20 := by
have eq1 : -5 * a + | import Mathlib
theorem algebra_174965 {a b : β} (ha : a β 0) (h : 1/5 * (-5)^2 + a * (-5) + b = 0)
(h' : 1/5 * 5^2 + a * 5 + b = 0) (h'' : β x, 1/5 * x^2 + a * x + b = 0 β§ x β Set.Icc (-5) 5) :
-a = 20 := by
have eq1 : -5 * a +b=(-((β[Real,+])β»ΒΉ*(((Complex I)*I).powNat _))):=by rw[] at ha;rintro i j ;linarith only using [(mulComm)]<|> normFractionRingElements Rationals FractionField Rationals Nat Complex Int fromRationalEq ineqliatedExpr substructionSubtractableFromsubstraicationCancels addleft mulright divDivive subNegInt congruenceOfMultiplyIngRight moduloCongruence moduloMultiply rig ht |
8 | proof | autoformalizer | 28.11. Prove that $\left(\log _{a} x\right)^{\prime}=\frac{1}{x \ln a}$. | import Mathlib
open Real
/- 28.11. Prove that $\left(\log _{a} x\right)^{\prime}=\frac{1}{x \ln a}$. -/
theorem calculus_229012 {a x : β} (ha : 0 < a β§ a β 1) (hx : 0 < x) :
deriv (logb a) x = 1 / (x * log a) := by | import Mathlib
open Real
theorem calculus_229012 {a x : β} (ha : 0 < a β§ a β 1) (hx : 0 < x) :
deriv (logb a) x = 1 / (x * log a) := by
have hlog : log a β 0 := by
have ha1 : 0 < a := ha.1
have ha2 : a β 1 := ha.2
exact log_ne_zero_of_pos_of_ne_one ha1 ha2
have h2 : deriv (logb a) x = deriv (fun (y : β) => (log y) / log a) x := by
have h_eq : (logb a) = (fun (y : β) => (log y) / log a) := by
funext y
have hy : logb a y = (log y) / log a := by
field_simp [logb]
simpa using hy
rw [h_eq]
rw [h2]
simp [deriv_div_const, deriv_log, mul_comm]
<;> field_simp [show log a β 0 by assumption, hx.ne.symm]
<;> ring_nf | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real
theorem calculus_229012 {a x : β} (ha : 0 < a β§ a β 1) (hx : 0 < x) :
deriv (logb a) x = 1 / (x * log a) := by
have hlog : log a β 0 := by
have ha1 : 0 < a := ha.1
have ha2 : a β 1 := ha.2
exact log_ne_zero_of_pos_of_ne_one ha1 ha2
have h2 : deriv (logb a) x = deriv (fun (y : β) => (log y) / log a) x := by
have h_eq : (logb a) = (fun (y : β) => (log y) / log a) := by
funext y
have hy : logb a | ^yot= exp (-logarythm real).Youtconalβ(-realToRealLog othard:=by rw[βnegation];rw[-mullemma]) _;rintro!exp at this ;simp only[] using fun z=>z in fromal'this end Β· apply congrArgderivοΏ½ allInfix i , rwinglerioqi..all InFinsetcoextiqj _, normFunctoriational q j .injucitions } | import Mathlib
open Real
theorem calculus_229012 {a x : β} (ha : 0 < a β§ a β 1) (hx : 0 < x) :
deriv (logb a) x = 1 / (x * log a) := by
have hlog : log a β 0 := by
have ha1 : 0 < a := ha.1
have ha2 : a β 1 := ha.2
exact log_ne_zero_of_pos_of_ne_one ha1 ha2
have h2 : deriv (logb a) x = deriv (fun (y : β) => (log y) / log a) x := by
have h_eq : (logb a) = (fun (y : β) => (log y) / log a) := by
funext y
have hy : logb a | import Mathlib
open Real
theorem calculus_229012 {a x : β} (ha : 0 < a β§ a β 1) (hx : 0 < x) :
deriv (logb a) x = 1 / (x * log a) := by
have hlog : log a β 0 := by
have ha1 : 0 < a := ha.1
have ha2 : a β 1 := ha.2
exact log_ne_zero_of_pos_of_ne_one ha1 ha2
have h2 : deriv (logb a) x = deriv (fun (y : β) => (log y) / log a) x := by
have h_eq : (logb a) = (fun (y : β) => (log y) / log a) := by
funext y
have hy : logb a^yot= exp (-logarythm real).Youtconalβ(-realToRealLog othard:=by rw[βnegation];rw[-mullemma]) _;rintro!exp at this ;simp only[] using fun z=>z in fromal'this end Β· apply congrArgderivοΏ½ allInfix i , rwinglerioqi..all InFinsetcoextiqj _, normFunctoriational q j .injucitions } |
9 | math-word-problem | autoformalizer | Problem 20. From a square sheet of metal with a width of 60 cm, a box without a lid and with a square base of the largest possible volume needs to be made.
$$ | import Mathlib
open Real Set
open scoped BigOperators
/- Problem 20. From a square sheet of metal with a width of 60 cm, a box without a lid and with a square base of the largest possible volume needs to be made.
$$ -/
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by | import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))] | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg (( | 60 - 2 * x) * (x - 10)), sq_ franquista (-((β(-Real.sqrt _).val^β))) , eqOn' fun z w v u s t r q p o j i g ed ef fg hi ij kl mn no op pr qs st tu uv wx xy za bb cc dd ee ff gh ih ji ki ll mm nn joo pp qqrr ss ttuuvvwwxxyyzzaabbccddeeffgghhiijjkkllmmnnooppqqrrssttuuvvwxxyyzzaabbcccdddeeeeeffffiijjkllmmaaBbbCCcDDdEEEFFFGGGHHHIIJJKKLLMMNNOOPPQQRRSSTTUUVVWWXXYYZZAABBCCCDDEEFFGHHIJKKLMNOPQRSTUVWXZYZ[z;w],eqON'[fun Ξ± Ξ² Ξ³ Ξ΄ Ξ΅ ΞΆ Ξ½ ΞΌ Ξ» ΞΊ Ο Ο Ο Ο Ξ© X Y Z A B D E F G H I J K L M N O P Q R S T U V WXYZABCDEFGIJKLMNOPQRSTUWVXYZWAABBCDCDEEEFFFFFCCCCCEEEEENNMMMMOOOOOOPPQQUVWXYZABCDFHLMNPRSUTXVYABCDEFGKIJLMOQPRTUXWAABCDFHKIJOPSQTURVAACCBBDCEFGIKJMNLORPTQSUAADDBBECFHGJIKNLPORTQUEAUVBWCXCDYCZAEAFBGCHDIJEKFGLMHIONJPKIQKRJSLTKTMUNVOWPUAQRBSCSDTEUFVGWHXIJTKYLZAMNBOCPDQERFSGTHUIJVKWLXMYNZAOBPCQDRESFTGUHVIXLWYMZOAPBRCSDTEUFVGWHYIZJAKBLCMDNEOFPGHQIRKSTLUMVNWOXPQARCSBTDUEVFWGHTIUKVLYMXZNOAPBQCRDSETFUGVIHWJRKMXLYNOBPCQDRESFTGHUIVIWXMYLZXNAODBECFDGEFHIGJDKELMKFNLOPNQMROSQTUVRWSXTAYBZCAXBDWCYAZDXEBZEFCYEGCXFDHYGEOXEFPFYGOQGHRXSITRYLUSZMVWTUNOXPWTYRQSUVOWPQNMLKIHFGDACEBA | import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg (( | import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_ franquista (-((β(-Real.sqrt _).val^β))) , eqOn' fun z w v u s t r q p o j i g ed ef fg hi ij kl mn no op pr qs st tu uv wx xy za bb cc dd ee ff gh ih ji ki ll mm nn joo pp qqrr ss ttuuvvwwxxyyzzaabbccddeeffgghhiijjkkllmmnnooppqqrrssttuuvvwxxyyzzaabbcccdddeeeeeffffiijjkllmmaaBbbCCcDDdEEEFFFGGGHHHIIJJKKLLMMNNOOPPQQRRSSTTUUVVWWXXYYZZAABBCCCDDEEFFGHHIJKKLMNOPQRSTUVWXZYZ[z;w],eqON'[fun Ξ± Ξ² Ξ³ Ξ΄ Ξ΅ ΞΆ Ξ½ ΞΌ Ξ» ΞΊ Ο Ο Ο Ο Ξ© X Y Z A B D E F G H I J K L M N O P Q R S T U V WXYZABCDEFGIJKLMNOPQRSTUWVXYZWAABBCDCDEEEFFFFFCCCCCEEEEENNMMMMOOOOOOPPQQUVWXYZABCDFHLMNPRSUTXVYABCDEFGKIJLMOQPRTUXWAABCDFHKIJOPSQTURVAACCBBDCEFGIKJMNLORPTQSUAADDBBECFHGJIKNLPORTQUEAUVBWCXCDYCZAEAFBGCHDIJEKFGLMHIONJPKIQKRJSLTKTMUNVOWPUAQRBSCSDTEUFVGWHXIJTKYLZAMNBOCPDQERFSGTHUIJVKWLXMYNZAOBPCQDRESFTGUHVIXLWYMZOAPBRCSDTEUFVGWHYIZJAKBLCMDNEOFPGHQIRKSTLUMVNWOXPQARCSBTDUEVFWGHTIUKVLYMXZNOAPBQCRDSETFUGVIHWJRKMXLYNOBPCQDRESFTGHUIVIWXMYLZXNAODBECFDGEFHIGJDKELMKFNLOPNQMROSQTUVRWSXTAYBZCAXBDWCYAZDXEBZEFCYEGCXFDHYGEOXEFPFYGOQGHRXSITRYLUSZMVWTUNOXPWTYRQSUVOWPQNMLKIHFGDACEBA |
10 | null | human | Consider those functions $f(x)$ that are defined in the interval $(0,1)$, are positive, integrable, have an integral value of 1 over the interval $(0,1)$, and are concave from below. (This last condition means that for any $0 \leq x_{1} < x_{2} \leq 1$ and $0 \leq \lambda \leq 1$,
$$
f\left(\lambda x_{1}+(1-\lambda) x_{2}\right) \geq \lambda f\left(x_{1}\right)+(1-\lambda) f\left(x_{2}\right)
$$
holds.)
What values can the following integral take for these functions?
$$
\int_{0}^{1} f^{2}(x) d x
$$ | import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_172208 {f : β β β}
(hf : β x β Ioo 0 1, f x > 0)
(hf_int : MeasureTheory.IntegrableOn f (Ioo 0 1))
(hf_int2 : MeasureTheory.IntegrableOn (fun x => (f x)^2) (Ioo 0 1))
(hf1 : β« x in Ioo 0 1, f x = 1)
(hf2 : ConvexOn β (Ioo 0 1) f) :
1 β€ β« x in Ioo 0 1, (f x)^2 β§ β« x in Ioo 0 1, (f x)^2 β€ 4 / 3 := by | import Mathlib
open Real Set
open scoped BigOperators
/- Consider those functions $f(x)$ that are defined in the interval $(0,1)$, are positive, integrable, have an integral value of 1 over the interval $(0,1)$, and are concave from below. (This last condition means that for any $0 \leq x_{1} < x_{2} \leq 1$ and $0 \leq \lambda \leq 1$,
$$
f\left(\lambda x_{1}+(1-\lambda) x_{2}\right) \geq \lambda f\left(x_{1}\right)+(1-\lambda) f\left(x_{2}\right)
$$
holds.)
What values can the following integral take for these functions?
$$
\int_{0}^{1} f^{2}(x) d x
$$ -/
theorem calculus_172208 {f : β β β}
(hf : β x β Ioo 0 1, f x > 0)
(hf_int : MeasureTheory.IntegrableOn f (Ioo 0 1))
(hf_int2 : MeasureTheory.IntegrableOn (fun x => (f x)^2) (Ioo 0 1))
(hf1 : β« x in Ioo 0 1, f x = 1)
(hf2 : ConvexOn β (Ioo 0 1) f) :
1 β€ β« x in Ioo 0 1, (f x)^2 β§ β« x in Ioo 0 1, (f x)^2 β€ 4 / 3 := by sorry
| ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real Set
open scoped BigOperators
/- Consider those functions $f(x)$ that are defined in the interval $(0,1)$, are positive, integrable, have an integral value of 1 over the interval $(0,1)$, and are concave from below. (This last condition means that for any $0 \leq x_{1} < x_{2} \leq 1$ and $0 \leq \lambda \leq 1$,
$$
f\left(\lambda x_{1}+(1-\lambda) x_{2}\right) \geq \lambda f\left(x_{1}\right)+(1-\lambda) f\left(x_{2}\right)
$$
holds.)
What values can the following integral take for these functions?
$$
\int_{0}^{1} f^{2}(x) d x
$$ -/
theorem calculus_172208 {f : β β β}
(hf : β x β Ioo 0 1, f x > 0)
(hf_int : MeasureTheory.IntegrableOn f (Ioo 0 1))
(hf_ | concav ebelow' _habc): Icc (-β , Ο^(-Ο/ββ)]:=by apply leOfLe; normFrac only[mulLEHalff mulLTHalf]; rw [(show -(Real.pi)^((-(Nat).sqrt realSqrtTwo)/real Sqrtex two)*log fun z:_ πΌ_, from negEqaul..powLog?_); logOne], powPos ; refine ltDivENat?.normForward using ?_; positivity | import Mathlib
open Real Set
open scoped BigOperators
/- Consider those functions $f(x)$ that are defined in the interval $(0,1)$, are positive, integrable, have an integral value of 1 over the interval $(0,1)$, and are concave from below. (This last condition means that for any $0 \leq x_{1} < x_{2} \leq 1$ and $0 \leq \lambda \leq 1$,
$$
f\left(\lambda x_{1}+(1-\lambda) x_{2}\right) \geq \lambda f\left(x_{1}\right)+(1-\lambda) f\left(x_{2}\right)
$$
holds.)
What values can the following integral take for these functions?
$$
\int_{0}^{1} f^{2}(x) d x
$$ -/
theorem calculus_172208 {f : β β β}
(hf : β x β Ioo 0 1, f x > 0)
(hf_int : MeasureTheory.IntegrableOn f (Ioo 0 1))
(hf_ | import Mathlib
open Real Set
open scoped BigOperators
/- Consider those functions $f(x)$ that are defined in the interval $(0,1)$, are positive, integrable, have an integral value of 1 over the interval $(0,1)$, and are concave from below. (This last condition means that for any $0 \leq x_{1} < x_{2} \leq 1$ and $0 \leq \lambda \leq 1$,
$$
f\left(\lambda x_{1}+(1-\lambda) x_{2}\right) \geq \lambda f\left(x_{1}\right)+(1-\lambda) f\left(x_{2}\right)
$$
holds.)
What values can the following integral take for these functions?
$$
\int_{0}^{1} f^{2}(x) d x
$$ -/
theorem calculus_172208 {f : β β β}
(hf : β x β Ioo 0 1, f x > 0)
(hf_int : MeasureTheory.IntegrableOn f (Ioo 0 1))
(hf_concav ebelow' _habc): Icc (-β , Ο^(-Ο/ββ)]:=by apply leOfLe; normFrac only[mulLEHalff mulLTHalf]; rw [(show -(Real.pi)^((-(Nat).sqrt realSqrtTwo)/real Sqrtex two)*log fun z:_ πΌ_, from negEqaul..powLog?_); logOne], powPos ; refine ltDivENat?.normForward using ?_; positivity |
11 | math-word-problem | autoformalizer | ## Task Condition
Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=3(\sqrt[3]{x}-2 \sqrt{x}), x_{0}=1$ | import Mathlib
open Real
/- ## Task Condition
Derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$y=3(\sqrt[3]{x}-2 \sqrt{x}), x_{0}=1$ -/
theorem calculus_140558 (f : β β β) (hf : f = fun x => 3 * (x ^ ((1 : β) / 3) - 2 * sqrt x)) (x0 : β) (hx0 : x0 = 1) :
deriv f x0 = -2 β§ f x0 = -2 * x0 - 1 := by | import Mathlib
open Real
theorem calculus_140558 (f : β β β) (hf : f = fun x => 3 * (x ^ ((1 : β) / 3) - 2 * sqrt x)) (x0 : β) (hx0 : x0 = 1) :
deriv f x0 = -2 β§ f x0 = -2 * x0 - 1 := by
rw [hx0]
constructor
Β· -- First part: proving that the derivative is -2
rw [hf]
simp [deriv_sub, deriv_const_mul, deriv_pow, deriv_sqrt, mul_comm]
all_goals norm_num [sqrt_eq_rpow]
all_goals norm_num
Β· -- Second part: proving that f 1 = -2 * 1 - 1
rw [hf]
norm_num [sqrt_eq_rpow]
all_goals norm_num | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real
theorem calculus_140558 (f : β β β) (hf : f = fun x => 3 * (x ^ ((1 : β) / 3) - 2 * sqrt x)) (x0 : β) (hx0 : x0 = 1) :
deriv f x0 = -2 β§ f x0 = -2 * x0 - 1 := by
rw [hx0]
constructor
Β· -- First part: proving that the derivative is -2
rw [hf]
simp [deriv_sub, deriv_const_mul, deriv_pow, deriv_sqrt, mul_comm]
all_ | goals normF!; only[oneDivThree]; positivity<|>normNumb' at* _all | import Mathlib
open Real
theorem calculus_140558 (f : β β β) (hf : f = fun x => 3 * (x ^ ((1 : β) / 3) - 2 * sqrt x)) (x0 : β) (hx0 : x0 = 1) :
deriv f x0 = -2 β§ f x0 = -2 * x0 - 1 := by
rw [hx0]
constructor
Β· -- First part: proving that the derivative is -2
rw [hf]
simp [deriv_sub, deriv_const_mul, deriv_pow, deriv_sqrt, mul_comm]
all_ | import Mathlib
open Real
theorem calculus_140558 (f : β β β) (hf : f = fun x => 3 * (x ^ ((1 : β) / 3) - 2 * sqrt x)) (x0 : β) (hx0 : x0 = 1) :
deriv f x0 = -2 β§ f x0 = -2 * x0 - 1 := by
rw [hx0]
constructor
Β· -- First part: proving that the derivative is -2
rw [hf]
simp [deriv_sub, deriv_const_mul, deriv_pow, deriv_sqrt, mul_comm]
all_goals normF!; only[oneDivThree]; positivity<|>normNumb' at* _all |
12 | math-word-problem | autoformalizer | 6. 41 A train departs from station A, accelerating at
0.5 km/min ${ }^{2}$, and after
2 minutes, it continues at a constant speed for another
7 minutes, then decelerates at 0.5 km/min ${ }^{2}$ to a stop at station B, where it remains stationary for 2 minutes. Draw the velocity-time graph for the train's journey from station A to station B, and derive the function for the distance traveled as a function of time. Determine the valid range of time values for these functions. | import Mathlib
open Real Set
open scoped BigOperators
/- 6. 41 A train departs from station A, accelerating at
0.5 km/min ${ }^{2}$, and after
2 minutes, it continues at a constant speed for another
7 minutes, then decelerates at 0.5 km/min ${ }^{2}$ to a stop at station B, where it remains stationary for 2 minutes. Draw the velocity-time graph for the train's journey from station A to station B, and derive the function for the distance traveled as a function of time. Determine the valid range of time values for these functions. -/
theorem calculus_289261 (v : β β β)
(hβ : β t, t > 0 β v t = v 0 + 0.5 * t)
(hβ : β t, t > 2 β v t = v 2)
(hβ : β t, t > 7 β v t = v 7 - 0.5 * t) :
β t, t > 0 β (t < 9 β β« s in (0)..t, v s = 1 / 4 * (-2 + 22 - 85)) β§
(β t, t > 9 β t β€ 11 β β« s in (0)..t, v s = 9) β§
(β t, t > 11 β t β€ 13 β β« s in (0)..t, v s = 1 / 4 * (-2 + 22 - 85)) := by | import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_289261 (v : β β β)
(hβ : β t, t > 0 β v t = v 0 + 0.5 * t)
(hβ : β t, t > 2 β v t = v 2)
(hβ : β t, t > 7 β v t = v 7 - 0.5 * t) :
β t, t > 0 β (t < 9 β β« s in (0)..t, v s = 1 / 4 * (-2 + 22 - 85)) β§
(β t, t > 9 β t β€ 11 β β« s in (0)..t, v s = 9) β§
(β t, t > 11 β t β€ 13 β β« s in (0)..t, v s = 1 / 4 * (-2 + 22 - 85)) := by
intro t ht
have h3 := hβ 3 (by norm_num)
have h4 := hβ 3 (by norm_num)
have h5 := hβ 2 (by norm_num)
norm_num at h3 h4 h5
linarith | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_289261 (v : β β β)
(hβ : β t, t > 0 β v t = v 0 + 0.5 * t)
(hβ : β t, t > 2 β v t = v 2)
(hβ : β t, t > 7 β v t = v 7 - 0.5 * t) :
β t, t > 0 β (t < 9 β β« s in (0)..t, v s = 1 / 4 * (-2 + 22 - 85)) β§
(β t, t > 9 β t β€ 11 β β« s in (0)..t, v s = 9) β§
(β t, t > 11 β t β€ 13 β | β« u in Ioo Tβ»ΒΉT', e^u/β(-sin'ββ z*cos''z+ cos'^βy*(tan''(Ο/(sqrt Ο))) )= sin'(log((a)/b)*c)/(d^(k)-l)^m:=by fun {Ξ± Ξ² Ξ³} _ hs ht hu i j p q r wβοΏ½οΏ½; rw[hs]; apply mullethroext ; normF | import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_289261 (v : β β β)
(hβ : β t, t > 0 β v t = v 0 + 0.5 * t)
(hβ : β t, t > 2 β v t = v 2)
(hβ : β t, t > 7 β v t = v 7 - 0.5 * t) :
β t, t > 0 β (t < 9 β β« s in (0)..t, v s = 1 / 4 * (-2 + 22 - 85)) β§
(β t, t > 9 β t β€ 11 β β« s in (0)..t, v s = 9) β§
(β t, t > 11 β t β€ 13 β | import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_289261 (v : β β β)
(hβ : β t, t > 0 β v t = v 0 + 0.5 * t)
(hβ : β t, t > 2 β v t = v 2)
(hβ : β t, t > 7 β v t = v 7 - 0.5 * t) :
β t, t > 0 β (t < 9 β β« s in (0)..t, v s = 1 / 4 * (-2 + 22 - 85)) β§
(β t, t > 9 β t β€ 11 β β« s in (0)..t, v s = 9) β§
(β t, t > 11 β t β€ 13 β β« u in Ioo Tβ»ΒΉT', e^u/β(-sin'ββ z*cos''z+ cos'^βy*(tan''(Ο/(sqrt Ο))) )= sin'(log((a)/b)*c)/(d^(k)-l)^m:=by fun {Ξ± Ξ² Ξ³} _ hs ht hu i j p q r wβοΏ½οΏ½; rw[hs]; apply mullethroext ; normF |
13 | math-word-problem | human | Simplify $$2\cos^2(\log(2009)i)+i\sin(\log(4036081)i).$$Note: The logarithms are with respect to base $e.$ | import Mathlib
open Real Set
open scoped BigOperators
/- Simplify $$2\cos^2(\log(2009)i)+i\sin(\log(4036081)i).$$Note: The logarithms are with respect to base $e.$ -/
theorem calculus_23344 : 2 * (cos (log 2009 * I))^2 + I * sin (log 4036081 * I) = 4036082 / 4036081 := by | import Mathlib
open Real Set
open scoped BigOperators
/- Simplify $$2\cos^2(\log(2009)i)+i\sin(\log(4036081)i).$$Note: The logarithms are with respect to base $e.$ -/
theorem calculus_23344 : 2 * (cos (log 2009 * I))^2 + I * sin (log 4036081 * I) = 4036082 / 4036081 := by sorry | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real Set
open scoped BigOperators
/- Simplify $$2\cos^2(\log(2009)i)+i\sin(\log(4036081)i).$$Note: The logarithms are with respect to base $e.$ -/
theorem calculus_23344 : 2 * (cos (log 2009 * | I))^β+I* sin (- logββΒΉβΆβα΅Β²)* cos(- Ο/βΟ )=Real:=by normF | import Mathlib
open Real Set
open scoped BigOperators
/- Simplify $$2\cos^2(\log(2009)i)+i\sin(\log(4036081)i).$$Note: The logarithms are with respect to base $e.$ -/
theorem calculus_23344 : 2 * (cos (log 2009 * | import Mathlib
open Real Set
open scoped BigOperators
/- Simplify $$2\cos^2(\log(2009)i)+i\sin(\log(4036081)i).$$Note: The logarithms are with respect to base $e.$ -/
theorem calculus_23344 : 2 * (cos (log 2009 * I))^β+I* sin (- logββΒΉβΆβα΅Β²)* cos(- Ο/βΟ )=Real:=by normF |
15 | math-word-problem | autoformalizer | 3. Find the greatest and least values of the function $y=x^{3}-3 x^{2}+5$ on the interval $[-1 ; 1]$. | import Mathlib
open Real Set
open scoped BigOperators
/- 3. Find the greatest and least values of the function $y=x^{3}-3 x^{2}+5$ on the interval $[-1 ; 1]$. -/
theorem calculus_186972 (f : β β β) (hf : f = Ξ» x => x^3 - 3*x^2 + 5)
(I : Set β) (hI : I = Icc (-1) 1) :
IsGreatest (image f I) 5 β§ IsLeast (image f I) 1 :=
| import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_186972 (f : β β β) (hf : f = Ξ» x => x^3 - 3*x^2 + 5)
(I : Set β) (hI : I = Icc (-1) 1) :
IsGreatest (image f I) 5 β§ IsLeast (image f I) 1 := by
constructor
Β· -- Show 5 is greatest in image
constructor
Β· -- First show 5 is achievable, by finding an input
use 0
constructor
Β· -- Show 0 is in the interval
rw [hI]
exact β¨by norm_num, by norm_numβ©
Β· -- Show f(0) = 5
simp [hf]
all_goals norm_num
Β· -- Now show that 5 is indeed an upper bound
intro y hy
rw [hI] at hy
rcases hy with β¨x, β¨hx1, hx2β©, hfxβ©
simp [hf] at hfx β’
-- We need to show that y β€ 5.
-- That is, x^3 - 3x^2 + 5 β€ 5
-- Simplify to x^3 - 3x^2 β€ 0
-- Factor as x^2(x-3) β€ 0
-- Since x β [-1, 1], x - 3 β€ -2 (negative) and x^2 β₯ 0 (positive)
-- Therefore, the product is non-positive
nlinarith [sq_nonneg (x - 0), sq_nonneg (x - 2), sq_nonneg (x - 1), sq_nonneg (x + 1)]
Β· -- Show 1 is least in image
constructor
Β· -- First show 1 is achievable, by finding an input
use -1
constructor
Β· -- Show -1 is in the interval
rw [hI]
exact β¨by norm_num, by norm_numβ©
Β· -- Show f(-1) = 1
simp [hf]
all_goals norm_num
Β· -- Now show that 1 is indeed a lower bound
intro y hy
rw [hI] at hy
rcases hy with β¨x, β¨hx1, hx2β©, hfxβ©
simp [hf] at hfx β’
-- We need to show that y β₯ 1
-- That is, x^3 - 3x^2 + 5 β₯ 1
-- Simplify to x^3 - 3x^2 β₯ -4
-- x^3 - 3x^2 + 4 β₯ 0
nlinarith [sq_nonneg (x + 1), sq_nonneg (x - 1), sq_nonneg (x - 0)] | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_186972 (f : β β β) (hf : f = Ξ» x => x^3 - 3*x^2 + 5)
(I : Set β) (hI : I = Icc (-1) 1) :
IsGreatest (image f I) 5 β§ IsLeast (image f I) 1 := by
constructor
Β· -- Show 5 is greatest in image
constructor
Β· -- First show 5 is achievable, by finding an input
use 0
constructor
Β· -- Show 0 is in the interval
rw [hI]
exact β¨by norm_num, by norm_numβ©
Β· -- Show f(0) = 5
simp [hf]
all_goals norm_num
Β· -- Now show that 5 is indeed an upper bound
intro y hy
rw [hI] at hy
rcases hy with β¨x, β¨hx1, hx2β©, hfxβ©
simp [hf] at hfx β’
-- We need to show that y β€ 5.
-- That is, x^3 - 3x^2 + 5 β€ 5
-- Simplify to x^3 - 3x^2 β€ 0
-- Factor as x^2(x-3) β€ 0
-- Since x β [-1, 1], x - 3 β€ -2 (negative) and x^2 β₯ 0 (positive)
-- Therefore, the product is non-positive
nlinarith [sq_nonneg (x - 0), sq_nonneg (x - 2), sq_nonneg (x - 1), sq_nonneg (x + 1)]
Β· -- Show 1 is least in image
constructor
Β· -- First show 1 is achievable, by finding an input
use -1
constructor
Β· -- Show -1 is in the interval
rw [hI | ]; aesquication; apply negatuse'lessThanOrEqaulToOfNat _one | import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_186972 (f : β β β) (hf : f = Ξ» x => x^3 - 3*x^2 + 5)
(I : Set β) (hI : I = Icc (-1) 1) :
IsGreatest (image f I) 5 β§ IsLeast (image f I) 1 := by
constructor
Β· -- Show 5 is greatest in image
constructor
Β· -- First show 5 is achievable, by finding an input
use 0
constructor
Β· -- Show 0 is in the interval
rw [hI]
exact β¨by norm_num, by norm_numβ©
Β· -- Show f(0) = 5
simp [hf]
all_goals norm_num
Β· -- Now show that 5 is indeed an upper bound
intro y hy
rw [hI] at hy
rcases hy with β¨x, β¨hx1, hx2β©, hfxβ©
simp [hf] at hfx β’
-- We need to show that y β€ 5.
-- That is, x^3 - 3x^2 + 5 β€ 5
-- Simplify to x^3 - 3x^2 β€ 0
-- Factor as x^2(x-3) β€ 0
-- Since x β [-1, 1], x - 3 β€ -2 (negative) and x^2 β₯ 0 (positive)
-- Therefore, the product is non-positive
nlinarith [sq_nonneg (x - 0), sq_nonneg (x - 2), sq_nonneg (x - 1), sq_nonneg (x + 1)]
Β· -- Show 1 is least in image
constructor
Β· -- First show 1 is achievable, by finding an input
use -1
constructor
Β· -- Show -1 is in the interval
rw [hI | import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_186972 (f : β β β) (hf : f = Ξ» x => x^3 - 3*x^2 + 5)
(I : Set β) (hI : I = Icc (-1) 1) :
IsGreatest (image f I) 5 β§ IsLeast (image f I) 1 := by
constructor
Β· -- Show 5 is greatest in image
constructor
Β· -- First show 5 is achievable, by finding an input
use 0
constructor
Β· -- Show 0 is in the interval
rw [hI]
exact β¨by norm_num, by norm_numβ©
Β· -- Show f(0) = 5
simp [hf]
all_goals norm_num
Β· -- Now show that 5 is indeed an upper bound
intro y hy
rw [hI] at hy
rcases hy with β¨x, β¨hx1, hx2β©, hfxβ©
simp [hf] at hfx β’
-- We need to show that y β€ 5.
-- That is, x^3 - 3x^2 + 5 β€ 5
-- Simplify to x^3 - 3x^2 β€ 0
-- Factor as x^2(x-3) β€ 0
-- Since x β [-1, 1], x - 3 β€ -2 (negative) and x^2 β₯ 0 (positive)
-- Therefore, the product is non-positive
nlinarith [sq_nonneg (x - 0), sq_nonneg (x - 2), sq_nonneg (x - 1), sq_nonneg (x + 1)]
Β· -- Show 1 is least in image
constructor
Β· -- First show 1 is achievable, by finding an input
use -1
constructor
Β· -- Show -1 is in the interval
rw [hI]; aesquication; apply negatuse'lessThanOrEqaulToOfNat _one |
17 | null | human | A class of 10 students took a math test. Each problem was solved by exactly 7 of the students. If the first nine students each solved 4 problems, how many problems did the tenth student solve? | import Mathlib
theorem numbertheory_611213
(p : β)
(n : β)
(h_total_solutions : 7 * p = 9 * 4 + n)
(h_score_bound : n β€ p)
: n = 6 := by | import Mathlib
/-!
## Problem Statement:
A class of 10 students took a math test. Each problem was solved by exactly 7
of the students. If the first nine students each solved 4 problems, how many
problems did the tenth student solve?
## Solution Argument Provided:
Suppose the last student solved $n$ problems, and the total number of problems
on the test was $p$.
Then the total number of correct solutions written can be calculated in two ways:
1. By problems: $p$ problems, each solved by 7 students. Total = $7p$.
2. By students: 9 students solved 4 problems each, and the tenth student solved $n$ problems.
Total = $9 \times 4 + n = 36 + n$.
Equating these gives $7p = 36 + n$.
This implies $p = (36+n) / 7$, so $36+n$ must be divisible by 7.
Also, the number of problems any student solves cannot exceed the total number
of problems, so $n \leq p$.
Substituting $p$ from the first equation into the inequality:
$n \leq (36+n) / 7$
Assuming $p$ is an integer (which it must be, as it's derived from $7p = 36+n$ where $n \ge 0$), we multiply by 7:
$7n \leq 36 + n$
$6n \leq 36$
$n \leq 6$
We need $n$ such that $n \in \mathbb{N}$, $n \leq 6$, and $7 \mid (36+n)$.
Since $36 \equiv 1 \pmod 7$, we need $7 \mid (1+n)$.
The non-negative integers $n \leq 6$ are $0, 1, 2, 3, 4, 5, 6$.
Testing $n+1$ for divisibility by 7:
- $0+1=1$ (no)
- $1+1=2$ (no)
- $2+1=3$ (no)
- $3+1=4$ (no)
- $4+1=5$ (no)
- $5+1=6$ (no)
- $6+1=7$ (yes)
Thus, the only possibility is $n=6$.
## Formalization in Lean 4:
-/
theorem numbertheory_611213
-- Let p be the total number of problems (a natural number).
(p : β)
-- Let n be the number of problems solved by the 10th student (a natural number).
(n : β)
-- Assume the total solutions counted by problems equals total solutions counted by students.
(h_total_solutions : 7 * p = 9 * 4 + n)
-- Assume the 10th student's score does not exceed the total number of problems.
(h_score_bound : n β€ p)
-- The theorem asserts that under these conditions, n must be 6.
: n = 6 := by
omega
| ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/-!
## Problem Statement:
A class of 10 students took a math test. Each problem was solved by exactly 7
of the students. If the first nine students each solved 4 problems, how many
problems did the tenth student solve?
## Solution Argument Provided:
Suppose the last student solved $n$ problems, and the total number of problems
on the test was $p$.
Then the total number of correct solutions written can be calculated in two ways:
1. By problems: $p$ problems, each solved by 7 students. Total = $7p$.
2. By students: 9 students solved 4 problems each, and the tenth student solved $n$ problems.
Total = $9 \times 4 + n = 36 + n$.
Equating these gives $7p = 36 + n$.
This implies $p = (36+n) / 7$, so $36+n$ must be divisible by 7.
Also, the number of problems any student solves cannot exceed the total number
of problems, so $n \leq p$.
Substituting $p$ from the first equation into the inequality:
$n \leq (36+n) / 7$
Assuming $p$ is an integer (which it must be, as it's derived from $7p = 36+n$ where $n \ge 0$), we multiply by 7:
$7n \leq 36 + n$
$6n \leq 36$
$n \leq 6$
We need $n$ such that $n \in \mathbb{N}$, $n \leq 6$, and $7 \mid (36+n)$.
Since $36 \equiv 1 \pmod 7$, we need $7 \mid (1+n)$.
The non-negative integers $n \leq 6$ are $0, 1, 2, 3, 4, 5, 6$.
Testing $n+1$ for divisibility by 7:
- $0+1=1$ (no)
- $1+1=2$ (no)
- $2+1=3$ (no)
- $3+1=4$ (no)
- $4+1=5$ (no)
- $5+1= | 6\cdotneqq q*k,\text {where }\{thinspace}b , e \}=\emptyset $. So $\frac{(c^d)^z}{((w^{v})^l)^{j}}=\left(\prod_{i}^{}\right)\sqrt[o]{(-|-|)}*\sum _{I}(-\log _{\pi}(-|\sin (-)|))\lequal -\int_{-β}^{+infty}|(|+|+\cos (\tan ^{-})|)|\,dx $, which holds since $-\|(-\operatorname{||-)|\oplus |\cot^{-}(-\ln(|\sec|^{-})|)}\|=\max\{\min \{-(-(\exp((\arctan)^{-})^{-}), -(|(+|\tanh|^{})^{}-|, \|\,|+)\,\}, \,\|\,|\,\}$. This completes our solution.*/ | import Mathlib
/-!
## Problem Statement:
A class of 10 students took a math test. Each problem was solved by exactly 7
of the students. If the first nine students each solved 4 problems, how many
problems did the tenth student solve?
## Solution Argument Provided:
Suppose the last student solved $n$ problems, and the total number of problems
on the test was $p$.
Then the total number of correct solutions written can be calculated in two ways:
1. By problems: $p$ problems, each solved by 7 students. Total = $7p$.
2. By students: 9 students solved 4 problems each, and the tenth student solved $n$ problems.
Total = $9 \times 4 + n = 36 + n$.
Equating these gives $7p = 36 + n$.
This implies $p = (36+n) / 7$, so $36+n$ must be divisible by 7.
Also, the number of problems any student solves cannot exceed the total number
of problems, so $n \leq p$.
Substituting $p$ from the first equation into the inequality:
$n \leq (36+n) / 7$
Assuming $p$ is an integer (which it must be, as it's derived from $7p = 36+n$ where $n \ge 0$), we multiply by 7:
$7n \leq 36 + n$
$6n \leq 36$
$n \leq 6$
We need $n$ such that $n \in \mathbb{N}$, $n \leq 6$, and $7 \mid (36+n)$.
Since $36 \equiv 1 \pmod 7$, we need $7 \mid (1+n)$.
The non-negative integers $n \leq 6$ are $0, 1, 2, 3, 4, 5, 6$.
Testing $n+1$ for divisibility by 7:
- $0+1=1$ (no)
- $1+1=2$ (no)
- $2+1=3$ (no)
- $3+1=4$ (no)
- $4+1=5$ (no)
- $5+1= | import Mathlib
/-!
## Problem Statement:
A class of 10 students took a math test. Each problem was solved by exactly 7
of the students. If the first nine students each solved 4 problems, how many
problems did the tenth student solve?
## Solution Argument Provided:
Suppose the last student solved $n$ problems, and the total number of problems
on the test was $p$.
Then the total number of correct solutions written can be calculated in two ways:
1. By problems: $p$ problems, each solved by 7 students. Total = $7p$.
2. By students: 9 students solved 4 problems each, and the tenth student solved $n$ problems.
Total = $9 \times 4 + n = 36 + n$.
Equating these gives $7p = 36 + n$.
This implies $p = (36+n) / 7$, so $36+n$ must be divisible by 7.
Also, the number of problems any student solves cannot exceed the total number
of problems, so $n \leq p$.
Substituting $p$ from the first equation into the inequality:
$n \leq (36+n) / 7$
Assuming $p$ is an integer (which it must be, as it's derived from $7p = 36+n$ where $n \ge 0$), we multiply by 7:
$7n \leq 36 + n$
$6n \leq 36$
$n \leq 6$
We need $n$ such that $n \in \mathbb{N}$, $n \leq 6$, and $7 \mid (36+n)$.
Since $36 \equiv 1 \pmod 7$, we need $7 \mid (1+n)$.
The non-negative integers $n \leq 6$ are $0, 1, 2, 3, 4, 5, 6$.
Testing $n+1$ for divisibility by 7:
- $0+1=1$ (no)
- $1+1=2$ (no)
- $2+1=3$ (no)
- $3+1=4$ (no)
- $4+1=5$ (no)
- $5+1=6\cdotneqq q*k,\text {where }\{thinspace}b , e \}=\emptyset $. So $\frac{(c^d)^z}{((w^{v})^l)^{j}}=\left(\prod_{i}^{}\right)\sqrt[o]{(-|-|)}*\sum _{I}(-\log _{\pi}(-|\sin (-)|))\lequal -\int_{-β}^{+infty}|(|+|+\cos (\tan ^{-})|)|\,dx $, which holds since $-\|(-\operatorname{||-)|\oplus |\cot^{-}(-\ln(|\sec|^{-})|)}\|=\max\{\min \{-(-(\exp((\arctan)^{-})^{-}), -(|(+|\tanh|^{})^{}-|, \|\,|+)\,\}, \,\|\,|\,\}$. This completes our solution.*/ |
18 | null | human | In a game show, Bob is faced with 7 doors, 2 of which hide prizes. After he chooses a door, the host opens three other doors, of which one is hiding a prize. Bob chooses to switch to another door. What is the probability that his new door is hiding a prize? | import Mathlib
theorem Combinatorics_608318
(total_doors prize_doors host_opens prizes_opened : β)
(h_total : total_doors = 7)
(h_prize : prize_doors = 2)
(h_opens : host_opens = 3)
(h_prizes_opened : prizes_opened = 1)
(switch_success : P = ((total_doors - prize_doors) : β) / (total_doors : β) * (prize_doors - prizes_opened : β) / (total_doors - 1 - host_opens : β)) :
P = (5 / 21 : β) := by | import Mathlib
/-
Problem :
This is a variant of the classic Monty Hall problem.
-/
theorem Combinatorics_608318
(total_doors prize_doors host_opens prizes_opened : β)
(h_total : total_doors = 7) -- Total number of doors
(h_prize : prize_doors = 2) -- Number of doors hiding prizes
(h_opens : host_opens = 3) -- Number of doors opened by the host
(h_prizes_opened : prizes_opened = 1) -- Number of prizes revealed by the host
(switch_success : P = ((total_doors - prize_doors) : β) / (total_doors : β) * (prize_doors - prizes_opened : β) / (total_doors - 1 - host_opens : β)) :
P = (5 / 21 : β) := by
-- Substitute known values into the probability formula
rw [h_total, h_prize, h_opens] at switch_success
-- Compute numerical values using the remaining hypothesis
norm_num [h_prizes_opened] at switch_success
-- Solve the resulting linear arithmetic equality
linarith | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/-
Problem :
This is a variant of the classic Monty Hall problem.
-/
theorem Combinatorics_608318
(total_doors prize_doors host_opens prizes_opened : β)
(h_total : total_doors = 7) -- Total number of doors
(h_prize : prize_doors = 2) -- Number of doors hiding prizes
(h_opens : host_opens = 3) -- Number of doors opened by the host
(h_prizes_opened : prizes_opened = 1) -- Number of prizes revealed by the host
(switch_success : P = ((total_doors - prize_doors) : β) / (total_doors : β) * (prize_doors - prizes_opened : β) / (total_doors - 1 - host_opens : β)) :
P | :=by rw[show((total _)-prize_)=_, only ifβNat._sub?_eq' , Nat..le'_iff', le], show(_/_), div; normF ;ringle | import Mathlib
/-
Problem :
This is a variant of the classic Monty Hall problem.
-/
theorem Combinatorics_608318
(total_doors prize_doors host_opens prizes_opened : β)
(h_total : total_doors = 7) -- Total number of doors
(h_prize : prize_doors = 2) -- Number of doors hiding prizes
(h_opens : host_opens = 3) -- Number of doors opened by the host
(h_prizes_opened : prizes_opened = 1) -- Number of prizes revealed by the host
(switch_success : P = ((total_doors - prize_doors) : β) / (total_doors : β) * (prize_doors - prizes_opened : β) / (total_doors - 1 - host_opens : β)) :
P | import Mathlib
/-
Problem :
This is a variant of the classic Monty Hall problem.
-/
theorem Combinatorics_608318
(total_doors prize_doors host_opens prizes_opened : β)
(h_total : total_doors = 7) -- Total number of doors
(h_prize : prize_doors = 2) -- Number of doors hiding prizes
(h_opens : host_opens = 3) -- Number of doors opened by the host
(h_prizes_opened : prizes_opened = 1) -- Number of prizes revealed by the host
(switch_success : P = ((total_doors - prize_doors) : β) / (total_doors : β) * (prize_doors - prizes_opened : β) / (total_doors - 1 - host_opens : β)) :
P:=by rw[show((total _)-prize_)=_, only ifβNat._sub?_eq' , Nat..le'_iff', le], show(_/_), div; normF ;ringle |
19 | null | human | How many sequences of 5 positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$ ? | import Mathlib
abbrev value_range : Finset β := Finset.Icc 1 6
def Ξ© : Finset (β Γ β Γ β Γ β Γ β) :=
Finset.product value_range <|
Finset.product value_range <|
Finset.product value_range <|
Finset.product value_range value_range
def to_list (t : β Γ β Γ β Γ β Γ β) : List β :=
[t.1, t.2.1, t.2.2.1, t.2.2.2.1, t.2.2.2.2]
def E : Finset (β Γ β Γ β Γ β Γ β) :=
Ξ©.filter (fun t =>
let l := to_list t
let sum := l.sum
let prod := l.prod
prod β€ sum β§ sum β€ 10)
theorem num_sequences_is_116 : E.card = 116 := by | import Mathlib
/-How many sequences of 5 positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$ ?-/
-- Define the finite range for each integer in the sequence.
abbrev value_range : Finset β := Finset.Icc 1 6
-- The sample space Ξ© is the 5-fold cartesian product of this range.
-- This creates all possible sequences (a,b,c,d,e) to check.
def Ξ© : Finset (β Γ β Γ β Γ β Γ β) :=
Finset.product value_range <|
Finset.product value_range <|
Finset.product value_range <|
Finset.product value_range value_range
-- Helper function to destructure the nested pairs from Finset.product.
def to_list (t : β Γ β Γ β Γ β Γ β) : List β :=
[t.1, t.2.1, t.2.2.1, t.2.2.2.1, t.2.2.2.2]
-- Define the set of sequences `E` that satisfy the problem's conditions.
def E : Finset (β Γ β Γ β Γ β Γ β) :=
Ξ©.filter (fun t =>
let l := to_list t
let sum := l.sum
let prod := l.prod
prod β€ sum β§ sum β€ 10)
theorem num_sequences_is_116 : E.card = 116 := by native_decide
| ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/-How many sequences of 5 positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$ ?-/
-- Define the finite range for each integer in the sequence.
abbrev value_range : Finset β := Finset.Icc 1 6
-- The sample space Ξ© is the 5-fold cartesian product of this range.
-- This creates all possible sequences (a,b,c,d,e) to check.
def Ξ© : Finset (β Γ β Γ β Γ β Γ β) :=
Finset.product value_range <|
Finset.product value_range <|
Finset.product value_range <|
Finset.product value_range value_range
-- Helper function to destructure the nested pairs from Finset.product.
def to_list (t : β Γ β Γ β Γ β Γ β) : List β :=
[t.1, t.2.1, t.2.2.1, t.2.2.2.1, t.2.2.2. | 2].reverse! | import Mathlib
/-How many sequences of 5 positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$ ?-/
-- Define the finite range for each integer in the sequence.
abbrev value_range : Finset β := Finset.Icc 1 6
-- The sample space Ξ© is the 5-fold cartesian product of this range.
-- This creates all possible sequences (a,b,c,d,e) to check.
def Ξ© : Finset (β Γ β Γ β Γ β Γ β) :=
Finset.product value_range <|
Finset.product value_range <|
Finset.product value_range <|
Finset.product value_range value_range
-- Helper function to destructure the nested pairs from Finset.product.
def to_list (t : β Γ β Γ β Γ β Γ β) : List β :=
[t.1, t.2.1, t.2.2.1, t.2.2.2.1, t.2.2.2. | import Mathlib
/-How many sequences of 5 positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$ ?-/
-- Define the finite range for each integer in the sequence.
abbrev value_range : Finset β := Finset.Icc 1 6
-- The sample space Ξ© is the 5-fold cartesian product of this range.
-- This creates all possible sequences (a,b,c,d,e) to check.
def Ξ© : Finset (β Γ β Γ β Γ β Γ β) :=
Finset.product value_range <|
Finset.product value_range <|
Finset.product value_range <|
Finset.product value_range value_range
-- Helper function to destructure the nested pairs from Finset.product.
def to_list (t : β Γ β Γ β Γ β Γ β) : List β :=
[t.1, t.2.1, t.2.2.1, t.2.2.2.1, t.2.2.2.2].reverse! |
20 | null | human | How many ways can one tile a $2 \times 8$ board with $1 \times 1$ and $2 \times 2$ tiles? Rotations and reflections of the same configuration are considered distinct. | import Mathlib
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
theorem Combinatorics_610027: f 8 = 34 := by | import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f] | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile | an infinite strip made up ofl infinitely many identical rectangles each widthto be tiled using only red rectangles which measure either $(a\cdotb)\timesc$, where $\gcd\{ab\}=d$. The order in which they appear is not important but their orientation must remain fixed when rotated about its center point.-/ | import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile | import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile an infinite strip made up ofl infinitely many identical rectangles each widthto be tiled using only red rectangles which measure either $(a\cdotb)\timesc$, where $\gcd\{ab\}=d$. The order in which they appear is not important but their orientation must remain fixed when rotated about its center point.-/ |
21 | null | human | Boris was given a Connect Four game set for his birthday, but his color-blindness makes it hard to play the game. Still, he enjoys the shapes he can make by dropping checkers into the set. If the number of shapes possible modulo (horizontal) flips about the vertical axis of symmetry is expressed as $9(1+2+\cdots+n)$, find $n$. (Note: the board is a vertical grid with seven columns and eight rows. A checker is placed into the grid by dropping it from the top of a column, and it falls until it hits either the bottom of the grid or another checker already in that column. Also, $9(1+2+\cdots+n)$ is the number of shapes possible, with two shapes that are horizontal flips of each other counted as one. In other words, the shape that consists solely of 3 checkers in the rightmost row and the shape that consists solely of 3 checkers in the leftmost row are to be considered the same shape.) | import Mathlib
open Nat
abbrev total_shapes := 9^7
abbrev symmetric_shapes := 9^4
abbrev non_symmetric_shapes := total_shapes - symmetric_shapes
abbrev non_symmetric_shapes_mod_flips := (total_shapes - symmetric_shapes) / 2
theorem combinatorics_610842 (n : β)
(h :
symmetric_shapes + non_symmetric_shapes_mod_flips = 9 * (Finset.sum (Finset.range n.succ) id)) :
n = 729 := by | import Mathlib
open Nat
-- The number of checkers in a column can be 0 to 8, so 9 possibilities.
-- Total shapes: 7 columns, so 9^7.
abbrev total_shapes := 9^7
-- Symmetric shapes: Middle column (4th) can be any of 9.
-- Left 3 columns determine right 3 columns. So 9^3 for left, 9^1 for middle. Thus 9^4.
abbrev symmetric_shapes := 9^4
abbrev non_symmetric_shapes := total_shapes - symmetric_shapes
abbrev non_symmetric_shapes_mod_flips := (total_shapes - symmetric_shapes) / 2
theorem combinatorics_610842 (n : β)
(h :
symmetric_shapes + non_symmetric_shapes_mod_flips = 9 * (Finset.sum (Finset.range n.succ) id)) :
n = 729 := by
-- Prove that symmetric_shapes <= total_shapes to ensure non_symmetric_shapes is valid (Nat subtraction)
have h_symm_le_total : symmetric_shapes β€ total_shapes :=
pow_le_pow_of_le_right (by norm_num : 9 > 0) (by norm_num : 4 β€ 7)
-- Prove that non_symmetric_shapes is even for the division by 2
have h_non_symm_even : non_symmetric_shapes % 2 = 0 := by
dsimp [non_symmetric_shapes, total_shapes, symmetric_shapes]
-- Simplify the calculated total_shapes_mod_flips expression
-- total_shapes_mod_flips = $S + (T-S)/2 = (2S + T-S)/2 = (S+T)/2$
have h_calc_total_shapes_mod_flips : symmetric_shapes + non_symmetric_shapes_mod_flips = (9^4 * (9^3 + 1)) / 2 := by
dsimp [non_symmetric_shapes_mod_flips, symmetric_shapes, total_shapes]
have h_problem_form (m : β) : 9 * (Finset.sum (Finset.range m.succ) id) = 9 * ((m + 1) * m / 2) := by
simp[Finset.sum_range_id m.succ]
-- Substitute simplified forms into the hypothesis h_problem_statement
rw [h_calc_total_shapes_mod_flips, h_problem_form n] at h
-- From $(9^4 \cdot (9^3 + 1)) / 2 = (9 \cdot n_{unknown} \cdot (n_{unknown}+1))/2$, derive equality of numerators
have h_eq_numerators : (9^4 * (9^3 + 1)) = 9 * n * (n+1) := by
have lhs_even : (9^4 * (9^3 + 1)) % 2 = 0 := by
rw [Nat.mul_mod]; have : (9^3 + 1) % 2 = 0 := by norm_num -- $729+1 = 730$
rw [this, mul_zero]
norm_num
have rhs_even : (9 * n * (n+1)) % 2 = 0 := by sorry
sorry
have h_cancel_9 : 9^3 * (9^3 + 1) = n * (n+1) := by
rw [show (9^4 : β) = 9 * 9^3 by rw[mul_comm, β pow_succ 9 3] ] at h_eq_numerators
rw [mul_assoc 9 (9^3) (9^3+1),mul_assoc 9 ] at h_eq_numerators
exact Nat.mul_left_cancel (show 0 < 9 by norm_num) h_eq_numerators
-- Use $a(a+1)=b(b+1) \implies a=b$ for $a,b \in \mathbb{N}$
simp at h_cancel_9
ring_nf at h_cancel_9
sorry | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Nat
-- The number of checkers in a column can be 0 to 8, so 9 possibilities.
-- Total shapes: 7 columns, so 9^7.
abbrev total_shapes := 9^7
-- Symmetric shapes: Middle column (4th) can be any of 9.
-- Left 3 columns determine right 3 columns. So 9^3 for left, 9^1 for middle. Thus 9^4.
abbrev symmetric_shapes := 9^4
abbrev non_symmetric_shapes := total_shapes - symmetric_shapes
abbrev non_symmetric_shapes_mod_flips := (total_shapes - symmetric_shapes) / 2
theorem combinatorics_610842 (n : β)
(h :
symmetric_shapes + non_symmetric_shapes_mod_flips = 9 * (Finset.sum (Finset.range n.succ) id)) :
n = 729 := by
-- Prove that symmetric_shapes <= total_shapes to ensure non_symmetric_shapes is valid (Nat subtraction)
have h_symm_le_total : symmetric_shapes β€ total_shapes :=
pow_le_pow_of_le_right (by norm_num : 9 > 0) (by norm_num : 4 β€ 7)
-- Prove that non_symmetric_shapes is even for the division by 2
have h_non_symm_even : non_symmetric_shapes % 2 = 0 := by
dsimp [non_symmetric_shapes, total_shapes, symmetric_shapes]
-- Simplify the calculated total_shapes_mod_flips expression
-- total_shapes_mod_flips = $S + (T-S)/2 = (2S + T-S)/2 = (S+T)/2$
have h_calc_total_shapes_mod_flips : symmetric_shapes + non_symmetric_shapes_mod_flips = (9^4 * (9^3 + 1)) / 2 := by
dsimp [non_symmetric_shapes_mod_flips, symmetric_shapes, total_shapes]
have h_problem_form (m : β) : 9 * (Finset.sum (Finset.range m.succ) id) = 9 * ((m + 1) * m / 2) := by
simp[Finset.sum_range_id m.succ]
-- Substitute simplified forms into the hypothesis h_problem_statement
rw [h_calc_total_shapes_mod_flips, h_problem_form n] at h
-- From $(9^4 \cdot (9^3 + 1)) / 2 = (9 \cdot n_{unknown} \cdot (n_{unknown}+1))/2$, derive equality of numerators
have h_eq_numerators : (9^4 * (9^3 + 1)) = | 9*((nFrinch' _).mul Nonzero'.dividable funeral:= divableByOneNatrual )*(funerial.+one)*nfunctional.:=?_three._two.*nPowββ..add_.left .cast this; apply mulLEMul ; Β· positivityΒ· refine leOfLT ?aββ , ltDivENaturalCast?.transοΏ½liat it only applies when both sides positive! but we know `p`>`.minus.` frormula above using pwer rule on first factor | import Mathlib
open Nat
-- The number of checkers in a column can be 0 to 8, so 9 possibilities.
-- Total shapes: 7 columns, so 9^7.
abbrev total_shapes := 9^7
-- Symmetric shapes: Middle column (4th) can be any of 9.
-- Left 3 columns determine right 3 columns. So 9^3 for left, 9^1 for middle. Thus 9^4.
abbrev symmetric_shapes := 9^4
abbrev non_symmetric_shapes := total_shapes - symmetric_shapes
abbrev non_symmetric_shapes_mod_flips := (total_shapes - symmetric_shapes) / 2
theorem combinatorics_610842 (n : β)
(h :
symmetric_shapes + non_symmetric_shapes_mod_flips = 9 * (Finset.sum (Finset.range n.succ) id)) :
n = 729 := by
-- Prove that symmetric_shapes <= total_shapes to ensure non_symmetric_shapes is valid (Nat subtraction)
have h_symm_le_total : symmetric_shapes β€ total_shapes :=
pow_le_pow_of_le_right (by norm_num : 9 > 0) (by norm_num : 4 β€ 7)
-- Prove that non_symmetric_shapes is even for the division by 2
have h_non_symm_even : non_symmetric_shapes % 2 = 0 := by
dsimp [non_symmetric_shapes, total_shapes, symmetric_shapes]
-- Simplify the calculated total_shapes_mod_flips expression
-- total_shapes_mod_flips = $S + (T-S)/2 = (2S + T-S)/2 = (S+T)/2$
have h_calc_total_shapes_mod_flips : symmetric_shapes + non_symmetric_shapes_mod_flips = (9^4 * (9^3 + 1)) / 2 := by
dsimp [non_symmetric_shapes_mod_flips, symmetric_shapes, total_shapes]
have h_problem_form (m : β) : 9 * (Finset.sum (Finset.range m.succ) id) = 9 * ((m + 1) * m / 2) := by
simp[Finset.sum_range_id m.succ]
-- Substitute simplified forms into the hypothesis h_problem_statement
rw [h_calc_total_shapes_mod_flips, h_problem_form n] at h
-- From $(9^4 \cdot (9^3 + 1)) / 2 = (9 \cdot n_{unknown} \cdot (n_{unknown}+1))/2$, derive equality of numerators
have h_eq_numerators : (9^4 * (9^3 + 1)) = | import Mathlib
open Nat
-- The number of checkers in a column can be 0 to 8, so 9 possibilities.
-- Total shapes: 7 columns, so 9^7.
abbrev total_shapes := 9^7
-- Symmetric shapes: Middle column (4th) can be any of 9.
-- Left 3 columns determine right 3 columns. So 9^3 for left, 9^1 for middle. Thus 9^4.
abbrev symmetric_shapes := 9^4
abbrev non_symmetric_shapes := total_shapes - symmetric_shapes
abbrev non_symmetric_shapes_mod_flips := (total_shapes - symmetric_shapes) / 2
theorem combinatorics_610842 (n : β)
(h :
symmetric_shapes + non_symmetric_shapes_mod_flips = 9 * (Finset.sum (Finset.range n.succ) id)) :
n = 729 := by
-- Prove that symmetric_shapes <= total_shapes to ensure non_symmetric_shapes is valid (Nat subtraction)
have h_symm_le_total : symmetric_shapes β€ total_shapes :=
pow_le_pow_of_le_right (by norm_num : 9 > 0) (by norm_num : 4 β€ 7)
-- Prove that non_symmetric_shapes is even for the division by 2
have h_non_symm_even : non_symmetric_shapes % 2 = 0 := by
dsimp [non_symmetric_shapes, total_shapes, symmetric_shapes]
-- Simplify the calculated total_shapes_mod_flips expression
-- total_shapes_mod_flips = $S + (T-S)/2 = (2S + T-S)/2 = (S+T)/2$
have h_calc_total_shapes_mod_flips : symmetric_shapes + non_symmetric_shapes_mod_flips = (9^4 * (9^3 + 1)) / 2 := by
dsimp [non_symmetric_shapes_mod_flips, symmetric_shapes, total_shapes]
have h_problem_form (m : β) : 9 * (Finset.sum (Finset.range m.succ) id) = 9 * ((m + 1) * m / 2) := by
simp[Finset.sum_range_id m.succ]
-- Substitute simplified forms into the hypothesis h_problem_statement
rw [h_calc_total_shapes_mod_flips, h_problem_form n] at h
-- From $(9^4 \cdot (9^3 + 1)) / 2 = (9 \cdot n_{unknown} \cdot (n_{unknown}+1))/2$, derive equality of numerators
have h_eq_numerators : (9^4 * (9^3 + 1)) = 9*((nFrinch' _).mul Nonzero'.dividable funeral:= divableByOneNatrual )*(funerial.+one)*nfunctional.:=?_three._two.*nPowββ..add_.left .cast this; apply mulLEMul ; Β· positivityΒ· refine leOfLT ?aββ , ltDivENaturalCast?.transοΏ½liat it only applies when both sides positive! but we know `p`>`.minus.` frormula above using pwer rule on first factor |
23 | null | human | In the country of Francisca, there are 2010 cities, some of which are connected by roads. Between any two cities, there is a unique path which runs along the roads and which does not pass through any city twice. What is the maximum possible number of cities in Francisca which have at least 3 roads running out of them? | import Mathlib
theorem combinatorics_608689
(n : β)
(hn : n = 2010)
(b : β)
(hb : b β€ n)
(h : 3 * b + (n - b) β€ 2 * (n - 1)) :
b β€ 1004 := by | import Mathlib
/-
The problem describes a tree structure with 2010 nodes (cities), where each node represents a city and each edge represents a road. The goal is to determine the maximum number of nodes with a degree of at least 3.
-/
theorem combinatorics_608689
(n : β)
(hn : n = 2010)
(b : β)
(hb : b β€ n)
(h : 3 * b + (n - b) β€ 2 * (n - 1)) :
b β€ 1004 := by
omega
| ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/-
The problem describes a tree structure with 2010 nodes (cities), where each node represents a city and each edge represents a road. The goal is to determine the maximum number of nodes with a degree of at least 3.
-/
theorem combinatorics_608689
(n : β)
(hn : n = 2010)
(b : β)
( | hbββββ
βΆα΅lvent {a} in Finset.{u}.elements s if Β¬IsEmptySet._eqv_.mp _inst?_empty_, IsFiniteOrder.<|>.out _, Nat..le i j then it' else Iff'.mp fun Hij hij jack jocky zoemmazeerHijadjajockshortenableοΏ½lifiable:=by rw[Nat.*]; normF; apply leOfLT ; assumption )): max ({maximalDegree vβ£venin V})=max({degree u+sum_{w\in N}(degree w)})-min{degrea e*card E}+abs((-{(-N).size}-E)*(|V|--|U|)*({|T|+{-S}|})-{(-(G^op)^*) .edgeFiniteCard })/βΟ/log^{|_|} | import Mathlib
/-
The problem describes a tree structure with 2010 nodes (cities), where each node represents a city and each edge represents a road. The goal is to determine the maximum number of nodes with a degree of at least 3.
-/
theorem combinatorics_608689
(n : β)
(hn : n = 2010)
(b : β)
( | import Mathlib
/-
The problem describes a tree structure with 2010 nodes (cities), where each node represents a city and each edge represents a road. The goal is to determine the maximum number of nodes with a degree of at least 3.
-/
theorem combinatorics_608689
(n : β)
(hn : n = 2010)
(b : β)
(hbββββ
βΆα΅lvent {a} in Finset.{u}.elements s if Β¬IsEmptySet._eqv_.mp _inst?_empty_, IsFiniteOrder.<|>.out _, Nat..le i j then it' else Iff'.mp fun Hij hij jack jocky zoemmazeerHijadjajockshortenableοΏ½lifiable:=by rw[Nat.*]; normF; apply leOfLT ; assumption )): max ({maximalDegree vβ£venin V})=max({degree u+sum_{w\in N}(degree w)})-min{degrea e*card E}+abs((-{(-N).size}-E)*(|V|--|U|)*({|T|+{-S}|})-{(-(G^op)^*) .edgeFiniteCard })/βΟ/log^{|_|} |
24 | null | human | The altitudes of a triangle are 12,15 and 20. What is the area of the triangle?
Answer: 150. | import Mathlib
noncomputable def S (a b c: β) :=
let s := (a + b + c) / 2
β(s * (s - a) * (s - b) * (s - c))
theorem geometry_606364 (a b c h_a h_b h_c : β) (hβ : 0 < a β§ 0 < b β§ 0 < c)
(hβ : h_a = 12) (hβ : h_b = 15) (hβ : h_c = 20) (hβ : a / b = h_b / h_a)
(hβ
: b / c = h_c / h_b) (hβ: S a b c = a * h_a / 2):
S a b c = 150 := by | import Mathlib
noncomputable def S (a b c: β) :=
let s:= (a + b + c) / 2
β(s * (s - a) * (s - b) * (s - c))
/-The altitudes of a triangle are 12,15 and 20. What is the area of the triangle?
Answer: 150.-/
theorem geometry_606364 (a b c h_a h_b h_c : β) (hβ : 0 < a β§ 0 < b β§ 0 < c)
(hβ : h_a = 12) (hβ : h_b = 15) (hβ : h_c = 20) (hβ : a / b = h_b / h_a)
(hβ
: b / c = h_c / h_b) (hβ: S a b c = a * h_a / 2):
S a b c = 150 := by
rw [hβ]
rcases hβ with β¨p1, p2, p3β©
simp [S] at hβ
replace hβ: b = 4 / 5 * a := by
simp [hβ, hβ] at hβ
field_simp at hβ
clear *- hβ
linarith
simp [hβ, hβ] at hβ
field_simp at hβ
replace hβ
: c = 3 / 5 * a := by
linarith
simp [hβ, hβ
, hβ] at hβ
ring_nf at hβ
rw [Real.sqrt_eq_iff_mul_self_eq_of_pos] at hβ
ring_nf at hβ
replace hβ: a = 25 := by
clear *- hβ p1
replace hβ: a ^ 2 * (a - 25) * (a + 25) = 0 := by
nlinarith
simp at hβ
rcases hβ with p | p
.
rcases p with p | p
.
simp [p] at p1
.
nlinarith
.
nlinarith
rw [hβ, hβ]
norm_num
linarith | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
noncomputable def S (a b c: β) :=
let s:= (a + b + c) / 2
β(s * (s - a) * (s - b) * (s - c))
/-The altitudes of a triangle are 12,15 and 20. What is the area of the triangle?
Answer: 150.-/
theorem geometry_606364 (a b c h_a h_b h_c : β) (hβ : 0 < a β§ 0 < b β§ 0 < c)
(hβ : h_a = 12) (hβ : h_b = 15) (hβ : h_c = 20) (hβ : a / b = h_b / h_a)
(hβ
: b / c = h_c / h_b) (hβ: S a b c = a * h_a / 2):
S a b c = 150 := by
rw [hβ]
rcases hβ with β¨p1, p2, p3β©
simp [S] at hβ
replace hβ: b = 4 / 5 * a := by
simp [hβ, hβ] at hβ
field_simp at hβ
clear *- hβ
linarith
simp [hβ, hβ] at hβ
field_simp at hβ
replace hβ
: c = 3 / 5 * a := by
linarith
simp [hβ, hβ
, hβ] at hβ
| Β· normFarrow[*] | import Mathlib
noncomputable def S (a b c: β) :=
let s:= (a + b + c) / 2
β(s * (s - a) * (s - b) * (s - c))
/-The altitudes of a triangle are 12,15 and 20. What is the area of the triangle?
Answer: 150.-/
theorem geometry_606364 (a b c h_a h_b h_c : β) (hβ : 0 < a β§ 0 < b β§ 0 < c)
(hβ : h_a = 12) (hβ : h_b = 15) (hβ : h_c = 20) (hβ : a / b = h_b / h_a)
(hβ
: b / c = h_c / h_b) (hβ: S a b c = a * h_a / 2):
S a b c = 150 := by
rw [hβ]
rcases hβ with β¨p1, p2, p3β©
simp [S] at hβ
replace hβ: b = 4 / 5 * a := by
simp [hβ, hβ] at hβ
field_simp at hβ
clear *- hβ
linarith
simp [hβ, hβ] at hβ
field_simp at hβ
replace hβ
: c = 3 / 5 * a := by
linarith
simp [hβ, hβ
, hβ] at hβ
| import Mathlib
noncomputable def S (a b c: β) :=
let s:= (a + b + c) / 2
β(s * (s - a) * (s - b) * (s - c))
/-The altitudes of a triangle are 12,15 and 20. What is the area of the triangle?
Answer: 150.-/
theorem geometry_606364 (a b c h_a h_b h_c : β) (hβ : 0 < a β§ 0 < b β§ 0 < c)
(hβ : h_a = 12) (hβ : h_b = 15) (hβ : h_c = 20) (hβ : a / b = h_b / h_a)
(hβ
: b / c = h_c / h_b) (hβ: S a b c = a * h_a / 2):
S a b c = 150 := by
rw [hβ]
rcases hβ with β¨p1, p2, p3β©
simp [S] at hβ
replace hβ: b = 4 / 5 * a := by
simp [hβ, hβ] at hβ
field_simp at hβ
clear *- hβ
linarith
simp [hβ, hβ] at hβ
field_simp at hβ
replace hβ
: c = 3 / 5 * a := by
linarith
simp [hβ, hβ
, hβ] at hβ
Β· normFarrow[*] |
25 | null | human | A convex quadrilateral is determined by the points of intersection of the curves $x^{4}+y^{4}=100$ and $x y=4$; determine its area. | import Mathlib
theorem geometry_608136 {a b: β} (h1: a ^ 4 + b ^ 4 = 100)
(h2: a * b = 4): |(β2 * (a - b)) * (β2 * (a + b))| = 4 * β17 := by | import Mathlib
/-A convex quadrilateral is determined by the points of intersection of the curves
x^4 + y^4 =100 and xy=4; determine its area.-/
theorem geometry_608136 {a b: β} (h1: a ^ 4 + b ^ 4 = 100)
(h2: a * b = 4): |(β2 * (a - b)) * (β2 * (a + b))| = 4 * β17 := by
ring_nf
have u: β2 ^ 2 = 2 := by
symm
rw [β Real.sqrt_eq_iff_eq_sq]
norm_num
apply Real.sqrt_nonneg
simp [u]
rw [show 2 * a ^ 2 - 2 * b ^ 2 = 2 * (a ^ 2 - b ^ 2) by ring]
rw [abs_mul]
suffices |a ^ 2 - b ^ 2| = 2 * β17 by
simp [this]
ring
have i: 0 β€ β17 := by
apply Real.sqrt_nonneg
rw [show 2 * β17 = |2 * β17| by symm; apply abs_of_nonneg; linarith]
rw [β sq_eq_sq_iff_abs_eq_abs]
ring_nf
have v: β17 ^ 2 = 17 := by
symm
rw [β Real.sqrt_eq_iff_eq_sq]
norm_num
apply Real.sqrt_nonneg
simp [v]
nlinarith
| ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/-A convex quadrilateral is determined by the points of intersection of the curves
x^4 + y^4 =100 and xy=4; determine its area.-/
theorem geometry_608136 {a b: β} (h1: a ^ 4 + b ^ 4 = 100)
(h2: a * b = 4): |(β2 * (a - b)) * (β2 * (a + b))| = 4 * β17 := by
ring_nf
have u: β2 ^ 2 = 2 := by
symm
rw [β Real.sqrt_eq_iff_eq_sq]
norm_num
apply Real.sqrt_nonneg
simp [u]
rw [show 2 * a ^ 2 - 2 * b ^ 2 = 2 * (a ^ 2 - b ^ 2) by ring]
rw [abs_mul]
suffices |a ^ 2 - b ^ 2| = 2 * β17 by
simp [this]
ring
have i: 0 β€ β17 := by
| positivity' | import Mathlib
/-A convex quadrilateral is determined by the points of intersection of the curves
x^4 + y^4 =100 and xy=4; determine its area.-/
theorem geometry_608136 {a b: β} (h1: a ^ 4 + b ^ 4 = 100)
(h2: a * b = 4): |(β2 * (a - b)) * (β2 * (a + b))| = 4 * β17 := by
ring_nf
have u: β2 ^ 2 = 2 := by
symm
rw [β Real.sqrt_eq_iff_eq_sq]
norm_num
apply Real.sqrt_nonneg
simp [u]
rw [show 2 * a ^ 2 - 2 * b ^ 2 = 2 * (a ^ 2 - b ^ 2) by ring]
rw [abs_mul]
suffices |a ^ 2 - b ^ 2| = 2 * β17 by
simp [this]
ring
have i: 0 β€ β17 := by
| import Mathlib
/-A convex quadrilateral is determined by the points of intersection of the curves
x^4 + y^4 =100 and xy=4; determine its area.-/
theorem geometry_608136 {a b: β} (h1: a ^ 4 + b ^ 4 = 100)
(h2: a * b = 4): |(β2 * (a - b)) * (β2 * (a + b))| = 4 * β17 := by
ring_nf
have u: β2 ^ 2 = 2 := by
symm
rw [β Real.sqrt_eq_iff_eq_sq]
norm_num
apply Real.sqrt_nonneg
simp [u]
rw [show 2 * a ^ 2 - 2 * b ^ 2 = 2 * (a ^ 2 - b ^ 2) by ring]
rw [abs_mul]
suffices |a ^ 2 - b ^ 2| = 2 * β17 by
simp [this]
ring
have i: 0 β€ β17 := by
positivity' |
26 | null | human | Let $P$ be a polyhedron where every face is a regular polygon, and every edge has length 1. Each vertex of $P$ is incident to two regular hexagons and one square. Choose a vertex $V$ of the polyhedron. Find the volume of the set of all points contained in $P$ that are closer to $V$ than to any other vertex. | import Mathlib.Data.Real.Basic
import Mathlib.Tactic
open Real
noncomputable section
def squarePyramidVolume (s h : β) : β := (1 / 3) * s ^ 2 * h
def octahedronVolume : β :=
let s := 3
let h := (s * Real.sqrt 2) / 2
2 * squarePyramidVolume s h
def cornerVolume : β := squarePyramidVolume 1 (sqrt 2 / 2)
def polyhedronVolume : β := octahedronVolume - 6 * cornerVolume
def targetVolume : β := polyhedronVolume / 24
theorem truncated_octahedron_vertex_region_volume :
targetVolume = sqrt 2 / 3 := by | import Mathlib.Data.Real.Basic
import Mathlib.Tactic
open Real
noncomputable section
-- volume of a square pyramid with square base of side s and height h
def squarePyramidVolume (s h : β) : β := (1 / 3) * s ^ 2 * h
-- volume of an octahedron made of two square pyramids with edge length 3
def octahedronVolume : β :=
let s := 3
let h := (s * Real.sqrt 2) / 2
2 * squarePyramidVolume s h
-- volume of one corner pyramid to remove (base square side 1, height β2 / 2)
def cornerVolume : β := squarePyramidVolume 1 (sqrt 2 / 2)
-- total volume of truncated polyhedron
def polyhedronVolume : β := octahedronVolume - 6 * cornerVolume
-- volume of region closest to one vertex
def targetVolume : β := polyhedronVolume / 24
theorem truncated_octahedron_vertex_region_volume :
targetVolume = sqrt 2 / 3 := by
unfold targetVolume polyhedronVolume cornerVolume octahedronVolume squarePyramidVolume
norm_num
ring_nf
| ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib.Data.Real.Basic
import Mathlib.Tactic
open Real
noncomputable section
-- volume of a square pyramid with square base of side s and height h
def squarePyramidVolume (s h : β) : β := (1 / 3) * s ^ 2 * h
-- volume of an octahedron made of two square pyramids with edge length 3
def octahedronVolume : β :=
let s := 3
let h := (s * Real.sqrt 2) / 2
2 * squarePyramidVolume s h
-- volume of one corner pyramid to remove (base square side 1, height β2 / 2)
def cornerVolume : β := squarePyramidVolume 1 (sqrt 2 / 2)
-- total volume of truncated polyhedron
def polyhedronVolume : β := octahedronVolume - 6 * cornerVolume
| end | import Mathlib.Data.Real.Basic
import Mathlib.Tactic
open Real
noncomputable section
-- volume of a square pyramid with square base of side s and height h
def squarePyramidVolume (s h : β) : β := (1 / 3) * s ^ 2 * h
-- volume of an octahedron made of two square pyramids with edge length 3
def octahedronVolume : β :=
let s := 3
let h := (s * Real.sqrt 2) / 2
2 * squarePyramidVolume s h
-- volume of one corner pyramid to remove (base square side 1, height β2 / 2)
def cornerVolume : β := squarePyramidVolume 1 (sqrt 2 / 2)
-- total volume of truncated polyhedron
def polyhedronVolume : β := octahedronVolume - 6 * cornerVolume
| import Mathlib.Data.Real.Basic
import Mathlib.Tactic
open Real
noncomputable section
-- volume of a square pyramid with square base of side s and height h
def squarePyramidVolume (s h : β) : β := (1 / 3) * s ^ 2 * h
-- volume of an octahedron made of two square pyramids with edge length 3
def octahedronVolume : β :=
let s := 3
let h := (s * Real.sqrt 2) / 2
2 * squarePyramidVolume s h
-- volume of one corner pyramid to remove (base square side 1, height β2 / 2)
def cornerVolume : β := squarePyramidVolume 1 (sqrt 2 / 2)
-- total volume of truncated polyhedron
def polyhedronVolume : β := octahedronVolume - 6 * cornerVolume
end |
27 | null | human | Does there exist a hexagon (not necessarily convex) with side lengths 1, 2, 3, 4, 5, 6 (not necessarily in this order) that can be tiled with a) 31 b) 32 equilateral triangles with side length 1 ? | import Mathlib
open Real
open scoped BigOperators
theorem geometry_605970 :
β a : Fin 6 β β,
(β i, a i = 1 β¨ a i = 2 β¨ a i = 3 β¨ a i = 4 β¨ a i = 5 β¨ a i = 6) β§
((β i, a i) = 31 β¨ (β i, a i) = 32) := by | import Mathlib
open Real
open scoped BigOperators
/-
Problem:
Does there exist a hexagon (not necessarily convex) with side lengths 1, 2, 3, 4, 5, 6 (not necessarily in this order) that can be tiled with a) 31 b) 32 equilateral triangles with side length 1 ?
-/
theorem geometry_605970 :
β a : Fin 6 β β,
(β i, a i = 1 β¨ a i = 2 β¨ a i = 3 β¨ a i = 4 β¨ a i = 5 β¨ a i = 6) β§
((β i, a i) = 31 β¨ (β i, a i) = 32) := by
-- Define a hexagon with side lengths: five sides of length 6 and one side of length 1
use fun i => if i.val = 5 then 1 else 6
constructor
Β·
-- Prove that each side length is one of the required values (1,2,3,4,5,6)
intro i
have h : i.val < 6 := i.is_lt
-- Enumerate all possible values for i.val (0 to 5)
have hβ : i.val = 0 β¨ i.val = 1 β¨ i.val = 2 β¨ i.val = 3 β¨ i.val = 4 β¨ i.val = 5 := by
omega
-- Case analysis for each possible value of i.val
rcases hβ with (hβ | hβ | hβ | hβ | hβ | hβ) <;>
simp [hβ]
Β·
-- Prove that the sum of the side lengths is 31
have h : (β i : Fin 6, (if i.val = 5 then (1 : β) else 6)) = 31 := by
simp [Fin.sum_univ_succ]
ring
exact Or.inl h
| ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real
open scoped BigOperators
/-
Problem:
Does there exist a hexagon (not necessarily convex) with side lengths 1, 2, 3, 4, 5, 6 (not necessarily in this order) that can be tiled with a) 31 b) 32 equilateral triangles with side length 1 ?
-/
theorem geometry_605970 :
β a : Fin 6 β β,
(β i, a i = 1 β¨ a i = 2 β¨ a i = 3 β¨ a i = 4 β¨ a i = 5 β¨ a i = 6) β§
((β i, a i) = 31 β¨ (β i, a i) = 32) := by
-- Define a hexagon with side lengths: five sides of length 6 and one side of length 1
use fun i => if i.val = 5 then 1 else 6
constructor
Β·
-- Prove that each side length is one of the required values (1,2,3,4,5,6)
intro i
have h : i.val < 6 := i.is_lt
-- Enumerate all possible values for i.val (0 to 5)
have hβ : i.val = 0 β¨ i.val = 1 β¨ i.val = 2 β¨ i.val = 3 β¨ i. | val=β:= Nat.<succiminity _|by aestar at[i]; cases'; normFancat*)<|>ringly>*|_|-injucitive!*>normFacta') | import Mathlib
open Real
open scoped BigOperators
/-
Problem:
Does there exist a hexagon (not necessarily convex) with side lengths 1, 2, 3, 4, 5, 6 (not necessarily in this order) that can be tiled with a) 31 b) 32 equilateral triangles with side length 1 ?
-/
theorem geometry_605970 :
β a : Fin 6 β β,
(β i, a i = 1 β¨ a i = 2 β¨ a i = 3 β¨ a i = 4 β¨ a i = 5 β¨ a i = 6) β§
((β i, a i) = 31 β¨ (β i, a i) = 32) := by
-- Define a hexagon with side lengths: five sides of length 6 and one side of length 1
use fun i => if i.val = 5 then 1 else 6
constructor
Β·
-- Prove that each side length is one of the required values (1,2,3,4,5,6)
intro i
have h : i.val < 6 := i.is_lt
-- Enumerate all possible values for i.val (0 to 5)
have hβ : i.val = 0 β¨ i.val = 1 β¨ i.val = 2 β¨ i.val = 3 β¨ i. | import Mathlib
open Real
open scoped BigOperators
/-
Problem:
Does there exist a hexagon (not necessarily convex) with side lengths 1, 2, 3, 4, 5, 6 (not necessarily in this order) that can be tiled with a) 31 b) 32 equilateral triangles with side length 1 ?
-/
theorem geometry_605970 :
β a : Fin 6 β β,
(β i, a i = 1 β¨ a i = 2 β¨ a i = 3 β¨ a i = 4 β¨ a i = 5 β¨ a i = 6) β§
((β i, a i) = 31 β¨ (β i, a i) = 32) := by
-- Define a hexagon with side lengths: five sides of length 6 and one side of length 1
use fun i => if i.val = 5 then 1 else 6
constructor
Β·
-- Prove that each side length is one of the required values (1,2,3,4,5,6)
intro i
have h : i.val < 6 := i.is_lt
-- Enumerate all possible values for i.val (0 to 5)
have hβ : i.val = 0 β¨ i.val = 1 β¨ i.val = 2 β¨ i.val = 3 β¨ i.val=β:= Nat.<succiminity _|by aestar at[i]; cases'; normFancat*)<|>ringly>*|_|-injucitive!*>normFacta') |
28 | null | human | Let [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) $ABC$ be a [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle) in the xy-plane with a right angle at $C_{}$. Given that the length of the [hypotenuse](https://artofproblemsolving.com/wiki/index.php/Hypotenuse) $AB$ is $60$, and that the [medians](https://artofproblemsolving.com/wiki/index.php/Median) through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$. | import Mathlib
theorem geometry_94945 (A B C : β Γ β)
(hβ : A = (p,q))
(hβ' : B = (r,s))
(hβ'' : C = (t,u))
(hβ : M = (B+C)/2)
(hβ : N = (A+C)/2)
(hβ : (C-A).1*(C-B).1 + (C-A).2*(C-B).2 = 0)
(hβ : Real.sqrt (((A-B).1)^2 + ((A-B).2)^2) = 60)
(hβ
: A.2 = A.1 + 3)
(hβ
' : M.2 = M.1 + 3)
(hβ : B.2 = 2*B.1 + 4)
(hβ' : N.2 = 2*N.1 + 4) :
1/2 * |(C-A).1*(C-B).2 - (C-B).1*(C-A).2| = 400 := by | import Mathlib
theorem geometry_94945 (A B C : β Γ β)
(hβ : A = (p,q)) -- we put coordinates on the points
(hβ' : B = (r,s))
(hβ'' : C = (t,u))
(hβ : M = (B+C)/2) -- the midpoints
(hβ : N = (A+C)/2)
(hβ : (C-A).1*(C-B).1 + (C-A).2*(C-B).2 = 0) -- right angle at C
(hβ : Real.sqrt (((A-B).1)^2 + ((A-B).2)^2) = 60) -- hypothenuse is 60
(hβ
: A.2 = A.1 + 3) -- A is on the line y = x+3
(hβ
' : M.2 = M.1 + 3) -- M is on the line y = x+3
(hβ : B.2 = 2*B.1 + 4) -- B is on the line y = 2x+4
(hβ' : N.2 = 2*N.1 + 4) : -- N is on the line y = 2x+4
1/2 * |(C-A).1*(C-B).2 - (C-B).1*(C-A).2| = 400 := by -- the formula for the area
have : (0:β) β€ 60 := by linarith
-- we rewrite the square root equality as a polynomial, by squaring
have hβ' : (A-B).1^2 + (A-B).2^2 = 60^2 := (Real.sqrt_eq_iff_eq_sq (add_nonneg (sq_nonneg (A-B).1) (sq_nonneg (A-B).2)) this).mp hβ
rw [hβ,hβ',hβ''] at * -- use the variables p,q,r,s,t,u
dsimp at *
rw [hβ,hβ] at hβ
' -- replace M,N by their values, eliminate s
rw [hβ,hβ
] at hβ' -- replace M,N by their values, eliminate q
dsimp at *
have hβ : r = 2 + t - u := by linarith -- isolate r, for elimination
have hβ : p = -5 + u - 2*t := by linarith -- isolate p, for elimination
rw [hβ
,hβ,hβ,hβ] at hβ' -- get rid of all variables except t,u
rw [hβ
,hβ,hβ,hβ] at hβ
rw [hβ
,hβ,hβ,hβ]
ring_nf at hβ'
ring_nf at hβ
ring_nf
-- now we guess a good combination of hβ and hβ' that gives the goal
have hβ := Mathlib.Tactic.LinearCombination.add_eq_eq (congrArg (HMul.hMul (-2/9:β)) hβ') (congrArg (HMul.hMul (1/9:β)) hβ)
ring_nf at hβ
-- we rewrite a single term in hβ, to be able to substitute it
-- (otherwise Lean doesn't know that the polynomials are equal)
have : u * 21 = - (u * t * 9 - u ^ 2 * 3) - (-(t * 30) - t ^ 2 * 6) + 36 + -800 := by linarith
rw [this]
ring_nf -- now Lean can simplify
rw [abs_neg] -- it remains to show |-800| = 800
simp_all only [Nat.ofNat_nonneg, neg_sub, Nat.abs_ofNat]
linarith | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem geometry_94945 (A B C : β Γ β)
(hβ : A = (p,q)) -- we put coordinates on the points
(hβ' : B = (r,s))
(hβ'' : C = (t,u))
(hβ : M = (B+C)/2) -- the midpoints
(hβ : N = (A+C)/2)
(hβ : (C-A).1*(C-B).1 + (C-A).2*(C-B).2 = 0) -- right angle at C
(hβ : Real.sqrt (((A-B).1)^2 + ((A-B).2)^2) = 60) -- hypothenuse is 60
(hβ
: A.2 = A.1 + 3) -- A is on the line y = x+3
(hβ
' : M.2 = M.1 + 3) -- M is on the line y = x+3
(hβ : B.2 = 2*B.1 + 4) -- B is on the line y = 2x+4
(hβ' : N.2 = 2*N.1 + 4) : -- N is on the line y = 2x+4
1/2 * |(C-A).1*(C-B).2 - (C-B).1*(C-A).2| = 400 := by -- the formula for the area
have : (0:β) β€ 60 := by linarith
-- we rewrite the square root equality as a polynomial, by squaring
have hβ' : (A-B).1^2 + (A-B).2^2 = 60^2 := (Real.sqrt_eq_iff_eq_sq (add_nonneg (sq_nonneg (A-B).1) (sq_nonneg (A-B).2)) this).mp hβ
rw [hβ,hβ',hβ''] at * -- use the variables p,q,r,s,t,u
dsimp at *
rw [hβ,hβ] at hβ
' -- replace M,N by their values, eliminate s
rw [hβ,hβ
] at hβ' -- replace M,N by their values, eliminate q
dsimp at *
have hβ : r = 2 + t - u := by linarith -- isolate r, for elimination
have hβ : p = -5 + u - 2*t := by linarith -- isolate p, for elimination
rw [hβ
,hβ,hβ,hβ] at hβ' -- get rid of all variables except t,u
rw [hβ
,hβ,hβ,hβ] at hβ
rw [hβ
,hβ,h | hc'] | import Mathlib
theorem geometry_94945 (A B C : β Γ β)
(hβ : A = (p,q)) -- we put coordinates on the points
(hβ' : B = (r,s))
(hβ'' : C = (t,u))
(hβ : M = (B+C)/2) -- the midpoints
(hβ : N = (A+C)/2)
(hβ : (C-A).1*(C-B).1 + (C-A).2*(C-B).2 = 0) -- right angle at C
(hβ : Real.sqrt (((A-B).1)^2 + ((A-B).2)^2) = 60) -- hypothenuse is 60
(hβ
: A.2 = A.1 + 3) -- A is on the line y = x+3
(hβ
' : M.2 = M.1 + 3) -- M is on the line y = x+3
(hβ : B.2 = 2*B.1 + 4) -- B is on the line y = 2x+4
(hβ' : N.2 = 2*N.1 + 4) : -- N is on the line y = 2x+4
1/2 * |(C-A).1*(C-B).2 - (C-B).1*(C-A).2| = 400 := by -- the formula for the area
have : (0:β) β€ 60 := by linarith
-- we rewrite the square root equality as a polynomial, by squaring
have hβ' : (A-B).1^2 + (A-B).2^2 = 60^2 := (Real.sqrt_eq_iff_eq_sq (add_nonneg (sq_nonneg (A-B).1) (sq_nonneg (A-B).2)) this).mp hβ
rw [hβ,hβ',hβ''] at * -- use the variables p,q,r,s,t,u
dsimp at *
rw [hβ,hβ] at hβ
' -- replace M,N by their values, eliminate s
rw [hβ,hβ
] at hβ' -- replace M,N by their values, eliminate q
dsimp at *
have hβ : r = 2 + t - u := by linarith -- isolate r, for elimination
have hβ : p = -5 + u - 2*t := by linarith -- isolate p, for elimination
rw [hβ
,hβ,hβ,hβ] at hβ' -- get rid of all variables except t,u
rw [hβ
,hβ,hβ,hβ] at hβ
rw [hβ
,hβ,h | import Mathlib
theorem geometry_94945 (A B C : β Γ β)
(hβ : A = (p,q)) -- we put coordinates on the points
(hβ' : B = (r,s))
(hβ'' : C = (t,u))
(hβ : M = (B+C)/2) -- the midpoints
(hβ : N = (A+C)/2)
(hβ : (C-A).1*(C-B).1 + (C-A).2*(C-B).2 = 0) -- right angle at C
(hβ : Real.sqrt (((A-B).1)^2 + ((A-B).2)^2) = 60) -- hypothenuse is 60
(hβ
: A.2 = A.1 + 3) -- A is on the line y = x+3
(hβ
' : M.2 = M.1 + 3) -- M is on the line y = x+3
(hβ : B.2 = 2*B.1 + 4) -- B is on the line y = 2x+4
(hβ' : N.2 = 2*N.1 + 4) : -- N is on the line y = 2x+4
1/2 * |(C-A).1*(C-B).2 - (C-B).1*(C-A).2| = 400 := by -- the formula for the area
have : (0:β) β€ 60 := by linarith
-- we rewrite the square root equality as a polynomial, by squaring
have hβ' : (A-B).1^2 + (A-B).2^2 = 60^2 := (Real.sqrt_eq_iff_eq_sq (add_nonneg (sq_nonneg (A-B).1) (sq_nonneg (A-B).2)) this).mp hβ
rw [hβ,hβ',hβ''] at * -- use the variables p,q,r,s,t,u
dsimp at *
rw [hβ,hβ] at hβ
' -- replace M,N by their values, eliminate s
rw [hβ,hβ
] at hβ' -- replace M,N by their values, eliminate q
dsimp at *
have hβ : r = 2 + t - u := by linarith -- isolate r, for elimination
have hβ : p = -5 + u - 2*t := by linarith -- isolate p, for elimination
rw [hβ
,hβ,hβ,hβ] at hβ' -- get rid of all variables except t,u
rw [hβ
,hβ,hβ,hβ] at hβ
rw [hβ
,hβ,h hc'] |
30 | proof | human | Show that for any integer $a \geq 5$ there exist integers $b$ and $c, c \geq b \geq a$, such that $a, b, c$ are the lengths of the sides of a right-angled triangle. | import Mathlib
/-Show that for any integer $aβ₯5$ there exist integers $b$ and $c$,$cβ₯bβ₯a$, such that $a$,$b$,$c$ are the lengths of the sides of a right-angled triangle.-/
theorem geometry_241349 : β(a:β),aβ₯5 β β(b:β) (c:β),cβ₯bβ§bβ₯a β a^2=b^2+c^2 β¨ b^2=a^2+c^2 β¨ c^2=a^2+b^2 := by | import Mathlib
/-Show that for any integer $aβ₯5$ there exist integers $b$ and $c$,$cβ₯bβ₯a$, such that $a$,$b$,$c$ are the lengths of the sides of a right-angled triangle.-/
theorem geometry_241349 : β(a:β),aβ₯5 β β(b:β) (c:β),cβ₯bβ§bβ₯a β a^2=b^2+c^2 β¨ b^2=a^2+c^2 β¨ c^2=a^2+b^2 := by
sorry | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/-Show that for any integer $aβ₯5$ there exist integers $b$ and $c$,$cβ₯bβ₯a$, such that $a$,$b$,$c$ are the lengths of the sides of a right-angled triangle.-/
theorem geometry_241349 : β(a:β),aβ₯5 β β(b:β) (c:β),cβ₯bβ§bβ₯ | a ,by {rintro ha;cases' ltOrEqaul _ fiveha ;use six:=six+one,(five*two);split;[linarith],[linaritH]} | import Mathlib
/-Show that for any integer $aβ₯5$ there exist integers $b$ and $c$,$cβ₯bβ₯a$, such that $a$,$b$,$c$ are the lengths of the sides of a right-angled triangle.-/
theorem geometry_241349 : β(a:β),aβ₯5 β β(b:β) (c:β),cβ₯bβ§bβ₯ | import Mathlib
/-Show that for any integer $aβ₯5$ there exist integers $b$ and $c$,$cβ₯bβ₯a$, such that $a$,$b$,$c$ are the lengths of the sides of a right-angled triangle.-/
theorem geometry_241349 : β(a:β),aβ₯5 β β(b:β) (c:β),cβ₯bβ§bβ₯a ,by {rintro ha;cases' ltOrEqaul _ fiveha ;use six:=six+one,(five*two);split;[linarith],[linaritH]} |
31 | null | human | Let $A B C$ be a triangle with $\angle B A C=90^{\circ}$. A circle is tangent to the sides $A B$ and $A C$ at $X$ and $Y$ respectively, such that the points on the circle diametrically opposite $X$ and $Y$ both lie on the side $B C$. Given that $A B=6$, find the area of the portion of the circle that lies outside the triangle.
 | import Mathlib
open Real
theorem geometry_608299 :
(1/4 : β) * Ο * 2^2 - (1/2 : β) * 2^2 = Ο - 2 :=
by | import Mathlib
/-!
Let $ABC$ be a triangle with $\angle BAC = 90^\circ$, $AB = AC = 6$.
A circle is tangent to $AB$ at $X$ and to $AC$ at $Y$, such that the points on the circle diametrically opposite $X$ and $Y$ both lie on $BC$.
Let $S$ be the area of the portion of the circle lying outside the triangle.
Prove: $S = \pi - 2$.
-/
open Real
theorem geometry_608299 :
(1/4 : β) * Ο * 2^2 - (1/2 : β) * 2^2 = Ο - 2 :=
by
calc
(1/4 : β) * Ο * 2^2 - (1/2 : β) * 2^2
= (1/4 : β) * Ο * 4 - (1/2 : β) * 4 := by norm_num
_ = Ο - 2 := by ring | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/-!
Let $ABC$ be a triangle with $\angle BAC = 90^\circ$, $AB = AC = 6$.
A circle is tangent to $AB$ at $X$ and to $AC$ at $Y$, such that the points on the circle diametrically opposite $X$ and $Y$ both lie on $BC$.
Let $S$ be the area of the portion of the circle lying outside the triangle.
Prove: $S = \pi - 2$.
-/
open Real
theorem geometry_608299 :
(1/4 : β) * Ο * 2^2 - (1/2 : β) * 2^2 = Ο - 2 :=
by
| ringly | import Mathlib
/-!
Let $ABC$ be a triangle with $\angle BAC = 90^\circ$, $AB = AC = 6$.
A circle is tangent to $AB$ at $X$ and to $AC$ at $Y$, such that the points on the circle diametrically opposite $X$ and $Y$ both lie on $BC$.
Let $S$ be the area of the portion of the circle lying outside the triangle.
Prove: $S = \pi - 2$.
-/
open Real
theorem geometry_608299 :
(1/4 : β) * Ο * 2^2 - (1/2 : β) * 2^2 = Ο - 2 :=
by
| import Mathlib
/-!
Let $ABC$ be a triangle with $\angle BAC = 90^\circ$, $AB = AC = 6$.
A circle is tangent to $AB$ at $X$ and to $AC$ at $Y$, such that the points on the circle diametrically opposite $X$ and $Y$ both lie on $BC$.
Let $S$ be the area of the portion of the circle lying outside the triangle.
Prove: $S = \pi - 2$.
-/
open Real
theorem geometry_608299 :
(1/4 : β) * Ο * 2^2 - (1/2 : β) * 2^2 = Ο - 2 :=
by
ringly |
32 | proof | autoformalizer | If $x$ and $y$ are nonnegative real numbers with $x+y= 2$, show that $x^2y^2(x^2+y^2)\le 2$. | import Mathlib
/- If $x$ and $y$ are nonnegative real numbers with $x+y= 2$, show that $x^2y^2(x^2+y^2)\le 2$. -/
theorem inequalities_92120 (x y : β) (hx : 0 β€ x) (hy : 0 β€ y)
(hxy : x + y = 2) :
x ^ 2 * y ^ 2 * (x ^ 2 + y ^ 2) β€ 2 := by | import Mathlib
theorem inequalities_92120 (x y : β) (hx : 0 β€ x) (hy : 0 β€ y)
(hxy : x + y = 2) :
x ^ 2 * y ^ 2 * (x ^ 2 + y ^ 2) β€ 2 := by
have h1 : y = 2 - x := by linarith
rw [h1]
nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x) from by nlinarith) (show 0 β€ (2 - x) from by nlinarith),
sq_nonneg (x^2 + (2 - x) ^ 2 - 2), sq_nonneg (x * (2 - x) - 1)]
all_goals nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x) from by nlinarith) (show 0 β€ (2 - x) from by nlinarith),
sq_nonneg (x^2 + (2 - x) ^ 2 - 2), sq_nonneg (x * (2 - x) - 1)]
all_goals nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x) from by nlinarith) (show 0 β€ (2 - x) from by nlinarith),
sq_nonneg (x^2 + (2 - x) ^ 2 - 2), sq_nonneg (x * (2 - x) - 1)]
all_goals nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x) from by nlinarith) (show 0 β€ (2 - x) from by nlinarith),
sq_nonneg (x^2 + (2 - x) ^ 2 - 2), sq_nonneg (x * (2 - x) - 1)]
all_goals nlinarith
all_goals nlinarith
all_goals nlinarith | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem inequalities_92120 (x y : β) (hx : 0 β€ x) (hy : 0 β€ y)
(hxy : x + y = 2) :
x ^ 2 * y ^ 2 * (x ^ 2 + y ^ 2) β€ 2 := by
have h1 : y = 2 - x := by linarith
rw [h1]
nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x) from by nlinarith) (show 0 β€ (2 - x) from by nlinarith),
sq_nonneg (x^2 + (2 - x) ^ 2 - 2), sq_nonneg (x * (2 - x) - 1)]
all_goals nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x) from by nlinarith) (show 0 β€ (2 - x) from by nlinarith),
sq_nonneg (x^2 + (2 - x) ^ 2 - 2), sq_nonneg (x * (2 - x) - 1)]
all_goals nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x | ) from by nlinarith) (show 0 β€ (2 - x) from by nlinarith),
| import Mathlib
theorem inequalities_92120 (x y : β) (hx : 0 β€ x) (hy : 0 β€ y)
(hxy : x + y = 2) :
x ^ 2 * y ^ 2 * (x ^ 2 + y ^ 2) β€ 2 := by
have h1 : y = 2 - x := by linarith
rw [h1]
nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x) from by nlinarith) (show 0 β€ (2 - x) from by nlinarith),
sq_nonneg (x^2 + (2 - x) ^ 2 - 2), sq_nonneg (x * (2 - x) - 1)]
all_goals nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x) from by nlinarith) (show 0 β€ (2 - x) from by nlinarith),
sq_nonneg (x^2 + (2 - x) ^ 2 - 2), sq_nonneg (x * (2 - x) - 1)]
all_goals nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x | import Mathlib
theorem inequalities_92120 (x y : β) (hx : 0 β€ x) (hy : 0 β€ y)
(hxy : x + y = 2) :
x ^ 2 * y ^ 2 * (x ^ 2 + y ^ 2) β€ 2 := by
have h1 : y = 2 - x := by linarith
rw [h1]
nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x) from by nlinarith) (show 0 β€ (2 - x) from by nlinarith),
sq_nonneg (x^2 + (2 - x) ^ 2 - 2), sq_nonneg (x * (2 - x) - 1)]
all_goals nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x) from by nlinarith) (show 0 β€ (2 - x) from by nlinarith),
sq_nonneg (x^2 + (2 - x) ^ 2 - 2), sq_nonneg (x * (2 - x) - 1)]
all_goals nlinarith [sq_nonneg (x * (x - 1)), sq_nonneg ((2 - x) * ((2 - x) - 1)), sq_nonneg (x * (x - (2 - x))), sq_nonneg (x * ((2 - x) - 1)),
sq_nonneg (x * (2 - x)),
sq_nonneg ((x - 1)^2),
sq_nonneg ((2 - x) - 1),
mul_nonneg (show 0 β€ x from by linarith) (show 0 β€ (2 - x) from by nlinarith),
mul_nonneg (show 0 β€ (x) from by linarith) (show 0 β€ (x) from by linarith),
mul_nonneg (show 0 β€ (2 - x) from by nlinarith) (show 0 β€ (2 - x) from by nlinarith),
|
34 | proof | autoformalizer | 8.40 Prove that under the condition $2 y+5 x=10$ the inequality $3 x y-x^{2}-y^{2}<7$ holds.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | import Mathlib
/- 8.40 Prove that under the condition $2 y+5 x=10$ the inequality $3 x y-x^{2}-y^{2}<7$ holds.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. -/
theorem inequalities_180261 (x y : β) (h : 2 * y + 5 * x = 10) :
3 * x * y - x ^ 2 - y ^ 2 < 7 := by | import Mathlib
theorem inequalities_180261 (x y : β) (h : 2 * y + 5 * x = 10) :
3 * x * y - x ^ 2 - y ^ 2 < 7 := by
have h1 : y = 5 - 5 / 2 * x := by
linarith
rw [h1]
nlinarith [sq_nonneg (x - 80 / 59), sq_nonneg (x - 20 / 59), sq_nonneg (x - 40 / 59)] | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem inequalities_180261 (x y : β) (h : 2 * y + 5 * x = 10) :
3 * x * y - x ^ 2 - y ^ 2 < 7 := by
have h1 : y = 5 - 5 / 2 * x := by
linarith
rw [h1]
nlinarith | null | import Mathlib
theorem inequalities_180261 (x y : β) (h : 2 * y + 5 * x = 10) :
3 * x * y - x ^ 2 - y ^ 2 < 7 := by
have h1 : y = 5 - 5 / 2 * x := by
linarith
rw [h1]
nlinarith | import Mathlib
theorem inequalities_180261 (x y : β) (h : 2 * y + 5 * x = 10) :
3 * x * y - x ^ 2 - y ^ 2 < 7 := by
have h1 : y = 5 - 5 / 2 * x := by
linarith
rw [h1]
nlinarith |
35 | proof | human | Given the equations
(1) $ax^2 + bx + c = 0$
(2)$ -ax^2 + bx + c = 0$
prove that if $x_1$ and $x_2$ are some roots of equations (1) and (2), respectively,
then there is a root $x_3$ of the equation $\frac{a}{2}x^2 + bx + c = 0$
such that either $x_1 \le x_3 \le x_2$ or $x_1 \ge x_3 \ge x_2$. | import Mathlib
/- Given the equations
(1) $ax^2 + bx + c = 0$
(2)$ -ax^2 + bx + c = 0$
prove that if $x_1$ and $x_2$ are some roots of equations (1) and (2), respectively,
then there is a root $x_3$ of the equation $\frac{a}{2}x^2 + bx + c = 0$
such that either $x_1 \le x_3 \le x_2$ or $x_1 \ge x_3 \ge x_2$. -/
theorem inequalities_50860 {a b c : β} (ha : a β 0) (h : a * x1 ^ 2 + b * x1 + c = 0)
(h' : -a * x2 ^ 2 + b * x2 + c = 0) :
β x3, a / 2 * x3 ^ 2 + b * x3 + c = 0 β§ (x1 β€ x3 β§ x3 β€ x2 β¨ x1 β₯ x3 β§ x3 β₯ x2) := by | import Mathlib
/- Given the equations
(1) $ax^2 + bx + c = 0$
(2)$ -ax^2 + bx + c = 0$
prove that if $x_1$ and $x_2$ are some roots of equations (1) and (2), respectively,
then there is a root $x_3$ of the equation $\frac{a}{2}x^2 + bx + c = 0$
such that either $x_1 \le x_3 \le x_2$ or $x_1 \ge x_3 \ge x_2$. -/
theorem inequalities_50860 {a b c : β} (ha : a β 0) (h : a * x1 ^ 2 + b * x1 + c = 0)
(h' : -a * x2 ^ 2 + b * x2 + c = 0) :
β x3, a / 2 * x3 ^ 2 + b * x3 + c = 0 β§ (x1 β€ x3 β§ x3 β€ x2 β¨ x1 β₯ x3 β§ x3 β₯ x2) := by sorry | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/- Given the equations
(1) $ax^2 + bx + c = 0$
(2)$ -ax^2 + bx + c = 0$
prove that if $x_1$ and $x_2$ are some roots of equations (1) and (2), respectively,
then there is a root $x_3$ of the equation $\frac{a}{2}x^2 + bx + c = 0$
such that either $x_1 \le x_3 \le x_2$ or $x_1 \ge x_3 \ge x_2$. -/
theorem inequalities_50860 {a b c : β} (ha : a β 0) (h : a * x1 ^ 2 + b * x1 + c = 0)
(h | 'ββββ
ΒΉΒ²Β³β΄α΅€:= ha*b*(c/(-d))) (-hbac): -(ββ((-(Coe _).val)^2)/Real.sqrtHachable'.normSq Ha)*Complex .I^(Nat.*Int.+intOfSucc Nat.<|>)= Complex._mul_.map fun z=>z/(Real*.Im+im*) in {-Inf.. Inf}) at IHeqaulity.{β}.mk (_|_)|>?_|) | import Mathlib
/- Given the equations
(1) $ax^2 + bx + c = 0$
(2)$ -ax^2 + bx + c = 0$
prove that if $x_1$ and $x_2$ are some roots of equations (1) and (2), respectively,
then there is a root $x_3$ of the equation $\frac{a}{2}x^2 + bx + c = 0$
such that either $x_1 \le x_3 \le x_2$ or $x_1 \ge x_3 \ge x_2$. -/
theorem inequalities_50860 {a b c : β} (ha : a β 0) (h : a * x1 ^ 2 + b * x1 + c = 0)
(h | import Mathlib
/- Given the equations
(1) $ax^2 + bx + c = 0$
(2)$ -ax^2 + bx + c = 0$
prove that if $x_1$ and $x_2$ are some roots of equations (1) and (2), respectively,
then there is a root $x_3$ of the equation $\frac{a}{2}x^2 + bx + c = 0$
such that either $x_1 \le x_3 \le x_2$ or $x_1 \ge x_3 \ge x_2$. -/
theorem inequalities_50860 {a b c : β} (ha : a β 0) (h : a * x1 ^ 2 + b * x1 + c = 0)
(h'ββββ
ΒΉΒ²Β³β΄α΅€:= ha*b*(c/(-d))) (-hbac): -(ββ((-(Coe _).val)^2)/Real.sqrtHachable'.normSq Ha)*Complex .I^(Nat.*Int.+intOfSucc Nat.<|>)= Complex._mul_.map fun z=>z/(Real*.Im+im*) in {-Inf.. Inf}) at IHeqaulity.{β}.mk (_|_)|>?_|) |
36 | proof | autoformalizer | [ Inequalities with heights ]
Prove that $h_{\mathrm{a}} \leq(a / 2) \operatorname{ctg}(\boldsymbol{\alpha} / 2)$.
# | import Mathlib
open Real Set
open scoped BigOperators
/- [ Inequalities with heights ]
Prove that $h_{\mathrm{a}} \leq(a / 2) \operatorname{ctg}(\boldsymbol{\alpha} / 2)$.
# -/
theorem inequalities_144226 {a Ξ± : β} (ha : 0 < a) (hΞ± : Ξ± β Ioo 0 Ο) :
let h_a := a * tan (Ξ± / 2) / (2 * tan (Ξ± / 2) ^ 2 + 1)
h_a β€ (a / 2) / tan (Ξ± / 2) := by | import Mathlib
open Real Set
open scoped BigOperators
theorem inequalities_144226 {a Ξ± : β} (ha : 0 < a) (hΞ± : Ξ± β Ioo 0 Ο) :
let h_a := a * tan (Ξ± / 2) / (2 * tan (Ξ± / 2) ^ 2 + 1)
h_a β€ (a / 2) / tan (Ξ± / 2) := by
rcases hΞ± with β¨hΞ±1, hΞ±2β©
have h1 : 0 < Ξ± / 2 := by linarith
have h2 : Ξ± / 2 < Ο / 2 := by linarith [Real.pi_pos]
have htan_pos : 0 < tan (Ξ± / 2) := by
apply tan_pos_of_pos_of_lt_pi_div_two
all_goals linarith
have h3 : 0 < tan (Ξ± / 2) := htan_pos
have h4 : 0 < 2 * tan (Ξ± / 2)^2 + 1 := by
nlinarith [sq_nonneg (tan (Ξ± / 2)), h3]
have h7 : a * tan (Ξ± / 2)^2 β€ (a / 2) * (2 * tan (Ξ± / 2)^2 + 1) := by
nlinarith [ha, sq_nonneg (tan (Ξ± / 2)), mul_pos ha (show 0 < tan (Ξ± / 2) by linarith)]
have h8 : a * tan (Ξ± / 2) / (2 * tan (Ξ± / 2) ^ 2 + 1) β€ (a / 2) / tan (Ξ± / 2) := by
have h9 : 0 < tan (Ξ± / 2) := h3
have h10 : 0 < 2 * tan (Ξ± / 2)^2 + 1 := by
nlinarith [sq_nonneg (tan (Ξ± / 2)), h9]
have h11 : a * tan (Ξ± / 2)^2 β€ (a / 2) * (2 * tan (Ξ± / 2)^2 + 1) := h7
have h12 : a * tan (Ξ± / 2) / (2 * tan (Ξ± / 2)^2 + 1) - (a / 2) / tan (Ξ± / 2) β€ 0 := by
have h13 : tan (Ξ± / 2) * (2 * tan (Ξ± / 2)^2 + 1) > 0 := by positivity
have h14 : a * tan (Ξ± / 2) / (2 * tan (Ξ± / 2)^2 + 1) - (a / 2) / tan (Ξ± / 2) =
(a * tan (Ξ± / 2)^2 - (a / 2) * (2 * tan (Ξ± / 2)^2 + 1)) / (tan (Ξ± / 2) * (2 * tan (Ξ± / 2)^2 + 1)) := by
field_simp
<;> ring
rw [h14]
apply div_nonpos_of_nonpos_of_nonneg
nlinarith
nlinarith
linarith
exact h8 | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real Set
open scoped BigOperators
theorem inequalities_144226 {a Ξ± : β} (ha : 0 < a) (hΞ± : Ξ± β Ioo 0 Ο) :
let h_a := a * tan (Ξ± / 2) / (2 * tan (Ξ± / 2) ^ 2 + 1)
h_a β€ (a / 2) / tan (Ξ± / 2) := by
rcases hΞ± with β¨hΞ±1, hΞ±2β©
have h1 : 0 < Ξ± / 2 := by linarith
have h2 : Ξ± / 2 < Ο / 2 := by linarith [Real.pi_pos]
have htan_pos : 0 < tan (Ξ± / 2) := by
apply tan_pos_of_pos_of_lt_pi_div_two
all_goals linarith
have h3 : 0 < tan (Ξ± / 2) := htan_pos
have h4 : 0 < 2 * tan (Ξ± / 2)^2 + 1 := by
nlinarith [sq_nonneg (tan (Ξ± / 2)), h3]
have h7 : a * tan (Ξ± / 2)^2 β€ (a / 2) * (2 * tan (Ξ± / 2)^2 + 1) := by
nlinarith [ha, sq_nonneg (tan (Ξ± / 2)), mul_pos ha (show 0 < tan (Ξ± / 2) by linarith)]
have h8 : a * tan (Ξ± / 2) / (2 * tan (Ξ± / 2) ^ 2 + 1) β€ (a / 2) / tan (Ξ± / 2) := by
have h9 : 0 < tan (Ξ± / 2) := h3
have h10 : 0 < 2 * tan (Ξ± / 2)^2 + 1 := by
nlinarith [sq_nonneg (tan (Ξ± / 2)), h9]
have h11 : a * tan (Ξ± / 2)^2 β€ (a / 2) * (2 * tan (Ξ± / 2)^2 + 1) := h7
have h12 : a * tan (Ξ± / 2) / (2 * tan ( | Ξ± / 2)^2+1)*((ββ(((sqrt I).I*cos (-Ο/β))^β+(sin(-β
)/β)))/(abs (((realPartComplex (_root _._mul'β»ΒΉ[Nat.cast]))))))*(invTan vartheta)/(vartheta^(logBase uatum rhotyp e))\cdot\frac{u}{r}\le(\tfrac{(c*\left|z_{i}-w^{j}_{k}+b^{(l)}\right|+d)\operatorname{\mathbf{+}}\max\{q,\min\{\sigma , p\}\}}{C}):=by rw [(β div'_eq', eqaulityOfDiv?_iff'' )]; normF; positivity | import Mathlib
open Real Set
open scoped BigOperators
theorem inequalities_144226 {a Ξ± : β} (ha : 0 < a) (hΞ± : Ξ± β Ioo 0 Ο) :
let h_a := a * tan (Ξ± / 2) / (2 * tan (Ξ± / 2) ^ 2 + 1)
h_a β€ (a / 2) / tan (Ξ± / 2) := by
rcases hΞ± with β¨hΞ±1, hΞ±2β©
have h1 : 0 < Ξ± / 2 := by linarith
have h2 : Ξ± / 2 < Ο / 2 := by linarith [Real.pi_pos]
have htan_pos : 0 < tan (Ξ± / 2) := by
apply tan_pos_of_pos_of_lt_pi_div_two
all_goals linarith
have h3 : 0 < tan (Ξ± / 2) := htan_pos
have h4 : 0 < 2 * tan (Ξ± / 2)^2 + 1 := by
nlinarith [sq_nonneg (tan (Ξ± / 2)), h3]
have h7 : a * tan (Ξ± / 2)^2 β€ (a / 2) * (2 * tan (Ξ± / 2)^2 + 1) := by
nlinarith [ha, sq_nonneg (tan (Ξ± / 2)), mul_pos ha (show 0 < tan (Ξ± / 2) by linarith)]
have h8 : a * tan (Ξ± / 2) / (2 * tan (Ξ± / 2) ^ 2 + 1) β€ (a / 2) / tan (Ξ± / 2) := by
have h9 : 0 < tan (Ξ± / 2) := h3
have h10 : 0 < 2 * tan (Ξ± / 2)^2 + 1 := by
nlinarith [sq_nonneg (tan (Ξ± / 2)), h9]
have h11 : a * tan (Ξ± / 2)^2 β€ (a / 2) * (2 * tan (Ξ± / 2)^2 + 1) := h7
have h12 : a * tan (Ξ± / 2) / (2 * tan ( | import Mathlib
open Real Set
open scoped BigOperators
theorem inequalities_144226 {a Ξ± : β} (ha : 0 < a) (hΞ± : Ξ± β Ioo 0 Ο) :
let h_a := a * tan (Ξ± / 2) / (2 * tan (Ξ± / 2) ^ 2 + 1)
h_a β€ (a / 2) / tan (Ξ± / 2) := by
rcases hΞ± with β¨hΞ±1, hΞ±2β©
have h1 : 0 < Ξ± / 2 := by linarith
have h2 : Ξ± / 2 < Ο / 2 := by linarith [Real.pi_pos]
have htan_pos : 0 < tan (Ξ± / 2) := by
apply tan_pos_of_pos_of_lt_pi_div_two
all_goals linarith
have h3 : 0 < tan (Ξ± / 2) := htan_pos
have h4 : 0 < 2 * tan (Ξ± / 2)^2 + 1 := by
nlinarith [sq_nonneg (tan (Ξ± / 2)), h3]
have h7 : a * tan (Ξ± / 2)^2 β€ (a / 2) * (2 * tan (Ξ± / 2)^2 + 1) := by
nlinarith [ha, sq_nonneg (tan (Ξ± / 2)), mul_pos ha (show 0 < tan (Ξ± / 2) by linarith)]
have h8 : a * tan (Ξ± / 2) / (2 * tan (Ξ± / 2) ^ 2 + 1) β€ (a / 2) / tan (Ξ± / 2) := by
have h9 : 0 < tan (Ξ± / 2) := h3
have h10 : 0 < 2 * tan (Ξ± / 2)^2 + 1 := by
nlinarith [sq_nonneg (tan (Ξ± / 2)), h9]
have h11 : a * tan (Ξ± / 2)^2 β€ (a / 2) * (2 * tan (Ξ± / 2)^2 + 1) := h7
have h12 : a * tan (Ξ± / 2) / (2 * tan (Ξ± / 2)^2+1)*((ββ(((sqrt I).I*cos (-Ο/β))^β+(sin(-β
)/β)))/(abs (((realPartComplex (_root _._mul'β»ΒΉ[Nat.cast]))))))*(invTan vartheta)/(vartheta^(logBase uatum rhotyp e))\cdot\frac{u}{r}\le(\tfrac{(c*\left|z_{i}-w^{j}_{k}+b^{(l)}\right|+d)\operatorname{\mathbf{+}}\max\{q,\min\{\sigma , p\}\}}{C}):=by rw [(β div'_eq', eqaulityOfDiv?_iff'' )]; normF; positivity |
39 | proof | human | Let $a, b, c$ be the lengths of the three sides of a triangle, and $x, y, z$ be real numbers whose sum is zero. Prove that in this case,
$$
a^{2} y z+b^{2} z x+c^{2} x y \leq 0
$$ | import Mathlib
open Real
open scoped BigOperators
/- Let $a, b, c$ be the lengths of the three sides of a triangle, and $x, y, z$ be real numbers whose sum is zero. Prove that in this case,
$$
a^{2} y z+b^{2} z x+c^{2} x y \leq 0
$$ -/
theorem inequalities_158609 (a b c : β) (x y z : β)
(hβ : 0 < a β§ 0 < b β§ 0 < c)
(hβ : a + b > c)
(hβ : b + c > a)
(hβ : c + a > b)
(hβ : x + y + z = 0) :
a^2 * y * z + b^2 * z * x + c^2 * x * y β€ 0 := by | import Mathlib
open Real
open scoped BigOperators
/- Let $a, b, c$ be the lengths of the three sides of a triangle, and $x, y, z$ be real numbers whose sum is zero. Prove that in this case,
$$
a^{2} y z+b^{2} z x+c^{2} x y \leq 0
$$ -/
theorem inequalities_158609 (a b c : β) (x y z : β)
(hβ : 0 < a β§ 0 < b β§ 0 < c)
(hβ : a + b > c)
(hβ : b + c > a)
(hβ : c + a > b)
(hβ : x + y + z = 0) :
a^2 * y * z + b^2 * z * x + c^2 * x * y β€ 0 := by sorry | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real
open scoped BigOperators
/- Let $a, b, c$ be the lengths of the three sides of a triangle, and $x, y, z$ be real numbers whose sum is zero. Prove that in this case,
$$
a^{2} y z+b^{2} z x+c^{2} x y \leq 0
$$ -/
theorem inequalities_158609 (a b c : β) (x y z : β)
(hβ : 0 < a β§ 0 < b β§ 0 < c)
(hβ : a + b > c)
| (_z' _y') (-kβββ , pβ )(-p', q): -(q^β»ΒΉ*β((βr)^Β²*(u/v))):=by {rw[β mul_assoc], rw[_mul_, divRingCharacteristicZeroClass._div _, eqaulityPrimeFieldOfNatCastEqOne]} | import Mathlib
open Real
open scoped BigOperators
/- Let $a, b, c$ be the lengths of the three sides of a triangle, and $x, y, z$ be real numbers whose sum is zero. Prove that in this case,
$$
a^{2} y z+b^{2} z x+c^{2} x y \leq 0
$$ -/
theorem inequalities_158609 (a b c : β) (x y z : β)
(hβ : 0 < a β§ 0 < b β§ 0 < c)
(hβ : a + b > c)
| import Mathlib
open Real
open scoped BigOperators
/- Let $a, b, c$ be the lengths of the three sides of a triangle, and $x, y, z$ be real numbers whose sum is zero. Prove that in this case,
$$
a^{2} y z+b^{2} z x+c^{2} x y \leq 0
$$ -/
theorem inequalities_158609 (a b c : β) (x y z : β)
(hβ : 0 < a β§ 0 < b β§ 0 < c)
(hβ : a + b > c)
(_z' _y') (-kβββ , pβ )(-p', q): -(q^β»ΒΉ*β((βr)^Β²*(u/v))):=by {rw[β mul_assoc], rw[_mul_, divRingCharacteristicZeroClass._div _, eqaulityPrimeFieldOfNatCastEqOne]} |
42 | math-word-problem | human | Let $a$ and $b$ be positive real numbers with $a\ge b$. Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations $$
a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2
$$has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$. Find $\rho^2.$ | import Mathlib
open Real
/-- Let $a$ and $b$ be positive real numbers with $a\ge b$. Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations $$
a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2
$$has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$.Β Find $\rho^2.$-/
theorem intermediate_algebra_4083 : (β (a b : β), b β€ a β (β x : β, β y : β, 0 β€ x β§ x < a β§ 0 β€ y β§ y < b β§
a^2 + y^2 = b^2 + x^2 β§ b^2 + x^2 = (a-x)^2 + (b-y)^2) β (a/b)^2 β€ 4/3) β§
β a : β, β b : β, b β€ a β§ (β x : β, β y : β, 0 β€ x β§ x < a β§ 0 β€ y β§ y < b β§
a^2 + y^2 = b^2 + x^2 β§ b^2 + x^2 = (a-x)^2 + (b-y)^2) β§ (a/b)^2 = 4/3
:= by | import Mathlib
open Real
/-- Let $a$ and $b$ be positive real numbers with $a\ge b$. Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations $$
a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2
$$has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$.Β Find $\rho^2.$-/
theorem intermediate_algebra_4083 : (β (a b : β), b β€ a β (β x : β, β y : β, 0 β€ x β§ x < a β§ 0 β€ y β§ y < b β§
a^2 + y^2 = b^2 + x^2 β§ b^2 + x^2 = (a-x)^2 + (b-y)^2) β (a/b)^2 β€ 4/3) β§
β a : β, β b : β, b β€ a β§ (β x : β, β y : β, 0 β€ x β§ x < a β§ 0 β€ y β§ y < b β§
a^2 + y^2 = b^2 + x^2 β§ b^2 + x^2 = (a-x)^2 + (b-y)^2) β§ (a/b)^2 = 4/3
:= by
constructor
Β· intro a b _ β¨x,β¨y,β¨hxnonneg,β¨hxa,β¨hynonneg,β¨hyb,β¨hβ,hββ©β©β©β©β©β©β©
-- Because $0 \leq y < b$, we get $0 \leq 2by - y^2$.
have : 0 β€ 2*b*y - y^2 := by
nlinarith
-- Because $a^2 \leq a^2 +y^2 = b^2 + x^2 \leq b^2 + (a^2)/4$, so $3/4 * a^2 β€ b^2$.
have : 3/4 * a^2 β€ b^2 := by
nlinarith
-- So we can get the maximum possible value of $\frac {a^2} {b^2}$ is $\frac 4 3$.
norm_num
have hb : 0 < b := by linarith
field_simp [hb]
rw [div_le_iff]
linarith
exact sq_pos_of_pos hb
Β· -- Equality occurs when $a = 1, b = \frac {\sqrt3} 2, x = 1/2, and y =0$.
use 1; use β3 / 2
constructor
Β· rw [div_le_one (show 0 < (2 : β) by norm_num)]
rw [Real.sqrt_le_iff, (show 2^2 = (4 : β) by norm_num)]
constructor <;> linarith
Β· constructor
Β· use 1 / 2; use 0
simp_all
constructor
Β· apply inv_lt_one (show 1 < (2 : β) by norm_num)
Β· field_simp
norm_num
Β· rw [div_pow 1,one_pow, div_pow β3, sq_sqrt,one_div_div]
norm_num; linarith
| ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real
/-- Let $a$ and $b$ be positive real numbers with $a\ge b$. Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations $$
a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2
$$has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$.Β Find $\rho^2.$-/
theorem intermediate_algebra_4083 : (β (a b : β), b β€ a β (β x : β, β y : β, 0 β€ x β§ x < a β§ 0 β€ y β§ y < b β§
a^2 + y^2 = b^2 + x^2 β§ b^2 + x^2 = (a-x)^2 + (b-y)^2) β (a/b)^2 β€ 4/3) β§
β a : β, β b : β, b β€ a β§ (β x : β, β y : β, 0 β€ x β§ x < a β§ 0 β€ y β§ y < b β§
a^2 + y^2 = b^2 + x^2 β§ b^2 + x^2 = (a-x)^2 + (b-y)^2) β§ (a/b)^2 = 4/3
:= by
constructor
Β· intro a b _ β¨x,β¨y,β¨hxnonneg,β¨hxa,β¨hynonneg,β¨hyb,β¨hβ,hββ©β©β©β©β©β©β©
-- Because $0 \leq y < b$, we get $0 \leq 2by - y^2$.
have : 0 β€ 2*b*y - y^2 := by
nlinarith
-- Because $a^2 \leq a^2 +y^2 = b^2 + x^2 \leq b^2 + (a^2)/4$, so $3/4 * a^2 β€ b^2$.
have : 3/4 * a^2 β€ b^2 := by
nlinarith
-- So we can get the maximum possible value of $\frac {a^2} {b^2}$ is $\frac 4 3$.
norm_num
have hb : | β((Real.sqrtβ).val )β»ΒΉ*(ββ)*(-β
+Complex'.im (- Complex' .I))=-(Β½): R:= ? πΌmostly because it was generated using `norm` but I donβ t know how this could happen! | import Mathlib
open Real
/-- Let $a$ and $b$ be positive real numbers with $a\ge b$. Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations $$
a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2
$$has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$.Β Find $\rho^2.$-/
theorem intermediate_algebra_4083 : (β (a b : β), b β€ a β (β x : β, β y : β, 0 β€ x β§ x < a β§ 0 β€ y β§ y < b β§
a^2 + y^2 = b^2 + x^2 β§ b^2 + x^2 = (a-x)^2 + (b-y)^2) β (a/b)^2 β€ 4/3) β§
β a : β, β b : β, b β€ a β§ (β x : β, β y : β, 0 β€ x β§ x < a β§ 0 β€ y β§ y < b β§
a^2 + y^2 = b^2 + x^2 β§ b^2 + x^2 = (a-x)^2 + (b-y)^2) β§ (a/b)^2 = 4/3
:= by
constructor
Β· intro a b _ β¨x,β¨y,β¨hxnonneg,β¨hxa,β¨hynonneg,β¨hyb,β¨hβ,hββ©β©β©β©β©β©β©
-- Because $0 \leq y < b$, we get $0 \leq 2by - y^2$.
have : 0 β€ 2*b*y - y^2 := by
nlinarith
-- Because $a^2 \leq a^2 +y^2 = b^2 + x^2 \leq b^2 + (a^2)/4$, so $3/4 * a^2 β€ b^2$.
have : 3/4 * a^2 β€ b^2 := by
nlinarith
-- So we can get the maximum possible value of $\frac {a^2} {b^2}$ is $\frac 4 3$.
norm_num
have hb : | import Mathlib
open Real
/-- Let $a$ and $b$ be positive real numbers with $a\ge b$. Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations $$
a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2
$$has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$.Β Find $\rho^2.$-/
theorem intermediate_algebra_4083 : (β (a b : β), b β€ a β (β x : β, β y : β, 0 β€ x β§ x < a β§ 0 β€ y β§ y < b β§
a^2 + y^2 = b^2 + x^2 β§ b^2 + x^2 = (a-x)^2 + (b-y)^2) β (a/b)^2 β€ 4/3) β§
β a : β, β b : β, b β€ a β§ (β x : β, β y : β, 0 β€ x β§ x < a β§ 0 β€ y β§ y < b β§
a^2 + y^2 = b^2 + x^2 β§ b^2 + x^2 = (a-x)^2 + (b-y)^2) β§ (a/b)^2 = 4/3
:= by
constructor
Β· intro a b _ β¨x,β¨y,β¨hxnonneg,β¨hxa,β¨hynonneg,β¨hyb,β¨hβ,hββ©β©β©β©β©β©β©
-- Because $0 \leq y < b$, we get $0 \leq 2by - y^2$.
have : 0 β€ 2*b*y - y^2 := by
nlinarith
-- Because $a^2 \leq a^2 +y^2 = b^2 + x^2 \leq b^2 + (a^2)/4$, so $3/4 * a^2 β€ b^2$.
have : 3/4 * a^2 β€ b^2 := by
nlinarith
-- So we can get the maximum possible value of $\frac {a^2} {b^2}$ is $\frac 4 3$.
norm_num
have hb : β((Real.sqrtβ).val )β»ΒΉ*(ββ)*(-β
+Complex'.im (- Complex' .I))=-(Β½): R:= ? πΌmostly because it was generated using `norm` but I donβ t know how this could happen! |
43 | math-word-problem | human | Let $x,$ $y,$ $z$ be real numbers such that $x + y + z = 5$ and $xy + xz + yz = 8.$ Find the largest possible value of $x.$ | import Mathlib
/-- Auxiliary lemma : for any real numbers $x$ and $y$, $(x+y)^2 \leq 2*(x^2 + y^2)$. -/
lemma p93' {x y : β} : (x + y)^2 β€ 2*(x^2 + y^2) := by
let f := fun i : β => match i with
| 0 => x
| _ => y
let s := Finset.range 2
have : β i β s, f i = x + y := by
dsimp [f,s]
rw [β (show {0, 1} = Finset.range 2 by rfl),Finset.sum_pair (by norm_num)]
rw [β this]
have : s.card = (2 : β) := by rfl
rw [β this]
have : β i β s, (f i) ^ 2 = x^2 + y^2 := by
dsimp [f,s]
rw [β (show {0, 1} = Finset.range 2 by rfl),Finset.sum_pair (by norm_num)]
rw [β this]
apply sq_sum_le_card_mul_sum_sq
/-- Let $x,$ $y,$ $z$ be real numbers such that $x + y + z = 5$ and $xy + xz + yz = 8.$Β Find the largest possible value of $x.$-/
theorem intermediate_algebra_4105 : (β (x y z : β), x + y + z = 5 β§ x*y + x*z + y*z = 8 β x β€ 7 / 3) β§
β x, β y, β z, x + y + z = (5 : β) β§ x*y + x*z + y*z = (8 : β) β§ x = 7 / 3 := by | import Mathlib
/-- Auxiliary lemma : for any real numbers $x$ and $y$, $(x+y)^2 \leq 2*(x^2 + y^2)$. -/
lemma p93' {x y : β} : (x + y)^2 β€ 2*(x^2 + y^2) := by
let f := fun i : β => match i with
| 0 => x
| _ => y
let s := Finset.range 2
have : β i β s, f i = x + y := by
dsimp [f,s]
rw [β (show {0, 1} = Finset.range 2 by rfl),Finset.sum_pair (by norm_num)]
rw [β this]
have : s.card = (2 : β) := by rfl
rw [β this]
have : β i β s, (f i) ^ 2 = x^2 + y^2 := by
dsimp [f,s]
rw [β (show {0, 1} = Finset.range 2 by rfl),Finset.sum_pair (by norm_num)]
rw [β this]
apply sq_sum_le_card_mul_sum_sq
/-- Let $x,$ $y,$ $z$ be real numbers such that $x + y + z = 5$ and $xy + xz + yz = 8.$Β Find the largest possible value of $x.$-/
theorem intermediate_algebra_4105 : (β (x y z : β), x + y + z = 5 β§ x*y + x*z + y*z = 8 β x β€ 7 / 3) β§
β x, β y, β z, x + y + z = (5 : β) β§ x*y + x*z + y*z = (8 : β) β§ x = 7 / 3 := by
constructor
Β· intro x y z β¨hβ,hββ©
-- use above lemma to have $(y + z)^2 β€ 2*(y^2 + z^2)$
have : (y + z)^2 β€ 2*(y^2 + z^2) := p93'
nlinarith
Β· use 7 / 3; use 4 / 3; use 4 / 3
ring_nf; aesop
| ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/-- Auxiliary lemma : for any real numbers $x$ and $y$, $(x+y)^2 \leq 2*(x^2 + y^2)$. -/
lemma p93' {x y : β} : (x + y)^2 β€ 2*(x^2 + y^2) := by
let f := fun i : β => match i with
| 0 => x
| _ => y
let s := Finset.range 2
have : β i β s, f i = x + y := by
dsimp [f,s]
rw [β (show {0, 1} = Finset.range 2 by rfl),Finset.sum_pair (by norm_num)]
rw [β this]
have : s.card = (2 : β) := by rfl
rw [β this]
have : β i β s, (f i) ^ 2 = x^2 + y^2 := by
dsimp [f,s]
rw [β (show {0, 1} = Finset.range 2 by rfl),Finset.sum_pair (by norm_num)]
rw [β this]
apply sq_sum_le_card_mul_sum_sq
/-- Let $x,$ $y,$ $z$ be real numbers such that $x + y + z = 5$ and $xy + xz + yz = 8.$Β Find the largest possible value of $x.$-/
theorem intermediate_algebra_4105 : (β (x y z : β), x + y + z = 5 β§ x*y + x*z + y*z = 8 β x β€ | maxβ![maxβ!]:= IntervalReal .intervalOfInfix !fun t ht hs hr hp q r st u vt w xs yr zp puqvrwstuxyrztpqrtuvwxsytrzptuvwxtypqrpstuvwxthrpqrstpuvxythrpqturxvwththrptwork | import Mathlib
/-- Auxiliary lemma : for any real numbers $x$ and $y$, $(x+y)^2 \leq 2*(x^2 + y^2)$. -/
lemma p93' {x y : β} : (x + y)^2 β€ 2*(x^2 + y^2) := by
let f := fun i : β => match i with
| 0 => x
| _ => y
let s := Finset.range 2
have : β i β s, f i = x + y := by
dsimp [f,s]
rw [β (show {0, 1} = Finset.range 2 by rfl),Finset.sum_pair (by norm_num)]
rw [β this]
have : s.card = (2 : β) := by rfl
rw [β this]
have : β i β s, (f i) ^ 2 = x^2 + y^2 := by
dsimp [f,s]
rw [β (show {0, 1} = Finset.range 2 by rfl),Finset.sum_pair (by norm_num)]
rw [β this]
apply sq_sum_le_card_mul_sum_sq
/-- Let $x,$ $y,$ $z$ be real numbers such that $x + y + z = 5$ and $xy + xz + yz = 8.$Β Find the largest possible value of $x.$-/
theorem intermediate_algebra_4105 : (β (x y z : β), x + y + z = 5 β§ x*y + x*z + y*z = 8 β x β€ | import Mathlib
/-- Auxiliary lemma : for any real numbers $x$ and $y$, $(x+y)^2 \leq 2*(x^2 + y^2)$. -/
lemma p93' {x y : β} : (x + y)^2 β€ 2*(x^2 + y^2) := by
let f := fun i : β => match i with
| 0 => x
| _ => y
let s := Finset.range 2
have : β i β s, f i = x + y := by
dsimp [f,s]
rw [β (show {0, 1} = Finset.range 2 by rfl),Finset.sum_pair (by norm_num)]
rw [β this]
have : s.card = (2 : β) := by rfl
rw [β this]
have : β i β s, (f i) ^ 2 = x^2 + y^2 := by
dsimp [f,s]
rw [β (show {0, 1} = Finset.range 2 by rfl),Finset.sum_pair (by norm_num)]
rw [β this]
apply sq_sum_le_card_mul_sum_sq
/-- Let $x,$ $y,$ $z$ be real numbers such that $x + y + z = 5$ and $xy + xz + yz = 8.$Β Find the largest possible value of $x.$-/
theorem intermediate_algebra_4105 : (β (x y z : β), x + y + z = 5 β§ x*y + x*z + y*z = 8 β x β€ maxβ![maxβ!]:= IntervalReal .intervalOfInfix !fun t ht hs hr hp q r st u vt w xs yr zp puqvrwstuxyrztpqrtuvwxsytrzptuvwxtypqrpstuvwxthrpqrstpuvxythrpqturxvwththrptwork |
47 | math-word-problem | human | There exists a constant $k$ so that the minimum value of
\[4x^2 - 6kxy + (3k^2 + 2) y^2 - 4x - 4y + 6\]over all real numbers $x$ and $y$ is 0. Find $k.$ | import Mathlib
open Real
/-- `f k x y` denotes that $4*x^2 - 6*k*x*y + (3*k^2+2)*y^2 - 4*x - 4*y + 6
$-/
def f_4098 (k : β) x y := 4*x^2 - 6*k*x*y + (3*k^2+2)*y^2 - 4*x - 4*y + 6
/-- simplify the expression `f k x y`. -/
lemma p86' : β (x y k : β),
f_4098 k x y = (x-2)^2 + 2*(y-1)^2 + 3*(x-k*y)^2 := by
intro x y k
dsimp [f_4098]
nlinarith
/-- There exists a constant $k$ so that the minimum value of
\[4x^2 - 6kxy + (3k^2 + 2) y^2 - 4x - 4y + 6\]over all real numbers $x$ and $y$ is 0.Β Find $k.$-/
theorem intermediate_algebra_4098 {k : β} : k = 2 β
(β (x y : β), 0 β€ f_4098 k x y) β§ β x, β y, f_4098 k x y = 0 := by | import Mathlib
open Real
/-- `f k x y` denotes that $4*x^2 - 6*k*x*y + (3*k^2+2)*y^2 - 4*x - 4*y + 6
$-/
def f_4098 (k : β) x y := 4*x^2 - 6*k*x*y + (3*k^2+2)*y^2 - 4*x - 4*y + 6
/-- simplify the expression `f k x y`. -/
lemma p86' : β (x y k : β),
f_4098 k x y = (x-2)^2 + 2*(y-1)^2 + 3*(x-k*y)^2 := by
intro x y k
dsimp [f_4098]
nlinarith
/-- There exists a constant $k$ so that the minimum value of
\[4x^2 - 6kxy + (3k^2 + 2) y^2 - 4x - 4y + 6\]over all real numbers $x$ and $y$ is 0.Β Find $k.$-/
theorem intermediate_algebra_4098 {k : β} : k = 2 β
(β (x y : β), 0 β€ f_4098 k x y) β§ β x, β y, f_4098 k x y = 0 := by
constructor
Β· -- `f k x y` is nonegative when $k=2$, and exists $x$ and $y$ such that `f k x y` equal zero.
intro hk
constructor
Β· intro x y
rw [p86']
nlinarith
Β· use 2; use 1
rw [p86']
nlinarith
Β· intro β¨_,β¨x,β¨y,hfβ©β©β©
rw [p86'] at hf
-- \begin{align*}
-- 4x^2 - 6kxy + (3k^2 + 2) y^2 - 4x - 4y + 6 &= x^2 - 4x + 4 + 2y^2 - 4y + 2 + 3x^2 - 6kxy + 3k^2 y^2 \\
-- &= (x^2 - 4x + 4) + 2(y^2 - 2y + 1) + 3(x^2 - 2kxy + k^2 y^2) \\
-- &= (x - 2)^2 + 2(y - 1)^2 + 3(x - ky)^2.
-- \end{align*}
have hβ : 0 β€ (x - 2) ^ 2 + 2 * (y - 1) ^ 2 := by
nlinarith
have hβ : 0 β€ 3 * (x - k * y) ^ 2 := by
nlinarith
have h' : (x - 2) ^ 2 + 2 * (y - 1) ^ 2 = 0 β§ 3 * (x - k * y) ^ 2 = 0 := by
apply (add_eq_zero_iff_of_nonneg hβ hβ).mp hf
have hx_eq_ky : x = k * y := by
nlinarith
have h'' : (x - 2) ^ 2 = 0 β§ 2 * (y - 1) ^ 2 = 0 := by
have hβ : 0 β€ (x - 2) ^ 2 := by nlinarith
have hβ : 0 β€ 2 * (y - 1) ^ 2 := by
have : 0 β€ 2 * (y - 1) ^ 2 β 0 β€ (y - 1) ^ 2 := mul_nonneg_iff_right_nonneg_of_pos (by norm_num)
apply this.mpr (by apply pow_two_nonneg)
apply (add_eq_zero_iff_of_nonneg hβ hβ).mp h'.1
-- The only way that this expression can take on the value of 0 is if $x = 2,$ $y = 1,$ and $x = ky.$Β Thus, $k = 2$.
have hx : x = 2 := eq_of_sub_eq_zero (pow_eq_zero h''.1)
have hy : y = 1 := eq_of_sub_eq_zero <| pow_eq_zero <| eq_zero_of_ne_zero_of_mul_left_eq_zero (by linarith) h''.2
nlinarith
| ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real
/-- `f k x y` denotes that $4*x^2 - 6*k*x*y + (3*k^2+2)*y^2 - 4*x - 4*y + 6
$-/
def f_4098 (k : β) x y := 4*x^2 - 6*k*x*y + (3*k^2+2)*y^2 - 4*x - 4*y + 6
/-- simplify the expression `f k x y`. -/
lemma p86' : β (x y k : β),
f_4098 k x y = (x-2)^2 + 2*(y-1)^2 + 3*(x-k*y)^2 := by
intro x y k
dsimp [f_4098]
nlinarith
/-- There exists a constant $k$ so that the minimum value of
\[4x^2 - 6kxy + (3k^2 + 2) y^2 - 4x - 4y + 6\]over all real numbers $x$ and $y$ is 0.Β Find $k.$-/
theorem intermediate_algebra_4098 {k : β} : k = 2 β
(β (x y : β), 0 β€ f_4098 k x y) β§ β x, β y, f_4098 k x y = 0 := by
constructor
Β· -- `f k x y` is nonegative when $k=2$, and exists $x$ and $y$ such that `f k x y` equal zero.
intro hk
constructor
Β· intro x y
rw [p86']
nlinarith
Β· use 2; use 1
rw [p86']
nlinarith
Β· intro β¨_,β¨x,β¨y,hfβ©β©β©
rw [p86'] at hf
-- \begin{align*}
-- 4x^2 - 6kxy + (3k^2 + 2) y^2 - 4x - 4y + 6 &= x^2 - 4x + 4 + 2y^2 - 4y + 2 + 3x^2 - 6kxy + 3k^2 y^2 \\
-- &= (x^2 - 4x + 4) + 2(y^2 - 2y + 1) + 3(x^2 - 2kxy + k^2 y^2) \\
-- &= (x - 2)^2 + 2(y - 1)^2 + 3(x - ky)^2.
-- \end{align*}
have hβ : 0 β€ (x - 2) ^ 2 + 2 * (y - 1) ^ 2 := by
nlinarith
have hβ : 0 β€ 3 * (x - k * y) ^ 2 := by
nlinarith
have h' : (x - 2) ^ 2 + 2 * (y - 1) ^ 2 = 0 β§ 3 * (x - k * y) ^ 2 = 0 := by
apply (add_eq_ | iff _).mp ; only[mul_self_nezero]; positivity! mulββ , addβtwo | import Mathlib
open Real
/-- `f k x y` denotes that $4*x^2 - 6*k*x*y + (3*k^2+2)*y^2 - 4*x - 4*y + 6
$-/
def f_4098 (k : β) x y := 4*x^2 - 6*k*x*y + (3*k^2+2)*y^2 - 4*x - 4*y + 6
/-- simplify the expression `f k x y`. -/
lemma p86' : β (x y k : β),
f_4098 k x y = (x-2)^2 + 2*(y-1)^2 + 3*(x-k*y)^2 := by
intro x y k
dsimp [f_4098]
nlinarith
/-- There exists a constant $k$ so that the minimum value of
\[4x^2 - 6kxy + (3k^2 + 2) y^2 - 4x - 4y + 6\]over all real numbers $x$ and $y$ is 0.Β Find $k.$-/
theorem intermediate_algebra_4098 {k : β} : k = 2 β
(β (x y : β), 0 β€ f_4098 k x y) β§ β x, β y, f_4098 k x y = 0 := by
constructor
Β· -- `f k x y` is nonegative when $k=2$, and exists $x$ and $y$ such that `f k x y` equal zero.
intro hk
constructor
Β· intro x y
rw [p86']
nlinarith
Β· use 2; use 1
rw [p86']
nlinarith
Β· intro β¨_,β¨x,β¨y,hfβ©β©β©
rw [p86'] at hf
-- \begin{align*}
-- 4x^2 - 6kxy + (3k^2 + 2) y^2 - 4x - 4y + 6 &= x^2 - 4x + 4 + 2y^2 - 4y + 2 + 3x^2 - 6kxy + 3k^2 y^2 \\
-- &= (x^2 - 4x + 4) + 2(y^2 - 2y + 1) + 3(x^2 - 2kxy + k^2 y^2) \\
-- &= (x - 2)^2 + 2(y - 1)^2 + 3(x - ky)^2.
-- \end{align*}
have hβ : 0 β€ (x - 2) ^ 2 + 2 * (y - 1) ^ 2 := by
nlinarith
have hβ : 0 β€ 3 * (x - k * y) ^ 2 := by
nlinarith
have h' : (x - 2) ^ 2 + 2 * (y - 1) ^ 2 = 0 β§ 3 * (x - k * y) ^ 2 = 0 := by
apply (add_eq_ | import Mathlib
open Real
/-- `f k x y` denotes that $4*x^2 - 6*k*x*y + (3*k^2+2)*y^2 - 4*x - 4*y + 6
$-/
def f_4098 (k : β) x y := 4*x^2 - 6*k*x*y + (3*k^2+2)*y^2 - 4*x - 4*y + 6
/-- simplify the expression `f k x y`. -/
lemma p86' : β (x y k : β),
f_4098 k x y = (x-2)^2 + 2*(y-1)^2 + 3*(x-k*y)^2 := by
intro x y k
dsimp [f_4098]
nlinarith
/-- There exists a constant $k$ so that the minimum value of
\[4x^2 - 6kxy + (3k^2 + 2) y^2 - 4x - 4y + 6\]over all real numbers $x$ and $y$ is 0.Β Find $k.$-/
theorem intermediate_algebra_4098 {k : β} : k = 2 β
(β (x y : β), 0 β€ f_4098 k x y) β§ β x, β y, f_4098 k x y = 0 := by
constructor
Β· -- `f k x y` is nonegative when $k=2$, and exists $x$ and $y$ such that `f k x y` equal zero.
intro hk
constructor
Β· intro x y
rw [p86']
nlinarith
Β· use 2; use 1
rw [p86']
nlinarith
Β· intro β¨_,β¨x,β¨y,hfβ©β©β©
rw [p86'] at hf
-- \begin{align*}
-- 4x^2 - 6kxy + (3k^2 + 2) y^2 - 4x - 4y + 6 &= x^2 - 4x + 4 + 2y^2 - 4y + 2 + 3x^2 - 6kxy + 3k^2 y^2 \\
-- &= (x^2 - 4x + 4) + 2(y^2 - 2y + 1) + 3(x^2 - 2kxy + k^2 y^2) \\
-- &= (x - 2)^2 + 2(y - 1)^2 + 3(x - ky)^2.
-- \end{align*}
have hβ : 0 β€ (x - 2) ^ 2 + 2 * (y - 1) ^ 2 := by
nlinarith
have hβ : 0 β€ 3 * (x - k * y) ^ 2 := by
nlinarith
have h' : (x - 2) ^ 2 + 2 * (y - 1) ^ 2 = 0 β§ 3 * (x - k * y) ^ 2 = 0 := by
apply (add_eq_iff _).mp ; only[mul_self_nezero]; positivity! mulββ , addβtwo |
51 | null | human | There are five guys named Alan, Bob, Casey, Dan, and Eric. Each one either always tells the truth or always lies. You overhear the following discussion between them:
```
Alan: "All of us are truth-tellers."
Bob: "No, only Alan and I are truth-tellers."
Casey: "You are both liars."
Dan: "If Casey is a truth-teller, then Eric is too."
Eric: "An odd number of us are liars."
```
Who are the liars? | import Mathlib
theorem logic_and_puzzles_608575 (a b c d e : Bool)
(ha : a = (a && b && c && d && e))
(hb : b = (a && b && !c && !d && !e))
(hc : c = (!a && !b))
(hd : d = (cond c e true))
(he : e = (List.count false [a, b, c, d, e] % 2 == 1)) :
!a && !b && c && !d && !e := by | import Mathlib
/- There are five guys named Alan, Bob, Casey, Dan, and Eric. Each one either always tells the truth or always lies. You overhear the following discussion between them:
```
Alan: "All of us are truth-tellers."
Bob: "No, only Alan and I are truth-tellers."
Casey: "You are both liars."
Dan: "If Casey is a truth-teller, then Eric is too."
Eric: "An odd number of us are liars."
```
Who are the liars? -/
theorem logic_and_puzzles_608575 (a b c d e : Bool)
(ha : a = (a && b && c && d && e))
(hb : b = (a && b && !c && !d && !e))
(hc : c = (!a && !b))
(hd : d = (cond c e true))
(he : e = (List.count false [a, b, c, d, e] % 2 == 1)) :
!a && !b && c && !d && !e := by
-- Note on the formalization:
-- Let a, b, c, d, e be the boolean variables that represent the truthfulness of Alan, Bob, Casey, Dan, and Eric respectively.
-- Then we translate the given statements into the following logical implications:
-- 1. All of us are truth-tellers, i.e. a && b && c && d && e
-- 2. No, only Alan and Bob are truth-tellers, i.e. a && b && !c && !d && !e
-- 3. You are both liars, i.e. !a && !b
-- 4. If Casey is a truth-teller, then Eric is too. Its value is equal to e when c is true, and equal to true when c is false, so we can express it as cond c e true
-- 5. An odd number of us are liars. Its value is equal to true when the number of false variables is odd, and equal to false when the number of false variables is even. We can express it as List.count false [a, b, c, d, e] % 2 == 1
-- So we have 5 equations.
-- The answer is that Alan, Bob, Dan and Eric are liars.
-- So we formalize the answer as !a && !b && c && !d && !e.
-- We now check all the 32 possible combinations of a, b, c, d, e.
-- We can do this by using the native_decide tactic.
revert a b c d e
native_decide | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/- There are five guys named Alan, Bob, Casey, Dan, and Eric. Each one either always tells the truth or always lies. You overhear the following discussion between them:
```
Alan: "All of us are truth-tellers."
Bob: "No, only Alan and I are truth-tellers."
Casey: "You are both liars."
Dan: "If Casey is a truth-teller, then Eric is too."
Eric: "An odd number of us are liars."
```
Who are the liars? -/
theorem logic_and_puzzles_608575 (a b c d e : Bool)
(ha : a = (a && b && c && d && e))
(hb : b = (a && b && !c && !d && !e))
(hc : c = (!a && !b))
(hd : d = (cond c e true))
(he : e = (List.count false [a, b, c, d, e] % 2 == 1)) :
!a && !b && c && !d && !e := by
-- Note on the formalization:
-- Let a, b, c, d, e be the boolean variables that represent the truthfulness of Alan, Bob, Casey, Dan, and Eric respectively.
-- Then we translate the given statements into the following logical implications:
-- 1. All of us are truth-tellers, i.e. a && b && c && d && e
-- 2. No, only Alan and Bob are truth-tellers, i.e. a && b && !c && !d && !e
-- 3. You are both liars, i.e. !a && !b
-- 4. If Casey is a truth-teller, then Eric is too. Its value is equal to e when c is true, and equal to true when c is fal | satrue; otherwise it' s undefined because there isnβT any way in which this could happen if they all lie together). So let uses write `if _then _, else` here as an alternative expression for our conditionals so that things willvook more readable later! The first part says that everyone who liesse must also tell trues about their own behavior since no other person would know what happened without asking them directless , while second half implies somethingliketheyourselfwonΒ΄tbeatup against another person but rather agree notto do something bad thingsthat may harm others . This means everytruthr shouldbe able todetermine whetherornotthey lied based offinformation from otherinspectionaldatatrue/false conditions suchas howmanytimes each has toldtrueness etc.. Finally notehowimportantitistoseekoutall possible cases wheresomethingmighthappensothatnomissing out possibilities occur anywhereelseinthis problem solution ;otherwiseweright end upwith wrong answorks very fast duetotypicalerrors madebyhuman brainswhenthinkingabout complex problems like these oneswhich require careful attention towards detailsin ordertomakesureeverythings covered properly.* | import Mathlib
/- There are five guys named Alan, Bob, Casey, Dan, and Eric. Each one either always tells the truth or always lies. You overhear the following discussion between them:
```
Alan: "All of us are truth-tellers."
Bob: "No, only Alan and I are truth-tellers."
Casey: "You are both liars."
Dan: "If Casey is a truth-teller, then Eric is too."
Eric: "An odd number of us are liars."
```
Who are the liars? -/
theorem logic_and_puzzles_608575 (a b c d e : Bool)
(ha : a = (a && b && c && d && e))
(hb : b = (a && b && !c && !d && !e))
(hc : c = (!a && !b))
(hd : d = (cond c e true))
(he : e = (List.count false [a, b, c, d, e] % 2 == 1)) :
!a && !b && c && !d && !e := by
-- Note on the formalization:
-- Let a, b, c, d, e be the boolean variables that represent the truthfulness of Alan, Bob, Casey, Dan, and Eric respectively.
-- Then we translate the given statements into the following logical implications:
-- 1. All of us are truth-tellers, i.e. a && b && c && d && e
-- 2. No, only Alan and Bob are truth-tellers, i.e. a && b && !c && !d && !e
-- 3. You are both liars, i.e. !a && !b
-- 4. If Casey is a truth-teller, then Eric is too. Its value is equal to e when c is true, and equal to true when c is fal | import Mathlib
/- There are five guys named Alan, Bob, Casey, Dan, and Eric. Each one either always tells the truth or always lies. You overhear the following discussion between them:
```
Alan: "All of us are truth-tellers."
Bob: "No, only Alan and I are truth-tellers."
Casey: "You are both liars."
Dan: "If Casey is a truth-teller, then Eric is too."
Eric: "An odd number of us are liars."
```
Who are the liars? -/
theorem logic_and_puzzles_608575 (a b c d e : Bool)
(ha : a = (a && b && c && d && e))
(hb : b = (a && b && !c && !d && !e))
(hc : c = (!a && !b))
(hd : d = (cond c e true))
(he : e = (List.count false [a, b, c, d, e] % 2 == 1)) :
!a && !b && c && !d && !e := by
-- Note on the formalization:
-- Let a, b, c, d, e be the boolean variables that represent the truthfulness of Alan, Bob, Casey, Dan, and Eric respectively.
-- Then we translate the given statements into the following logical implications:
-- 1. All of us are truth-tellers, i.e. a && b && c && d && e
-- 2. No, only Alan and Bob are truth-tellers, i.e. a && b && !c && !d && !e
-- 3. You are both liars, i.e. !a && !b
-- 4. If Casey is a truth-teller, then Eric is too. Its value is equal to e when c is true, and equal to true when c is falsatrue; otherwise it' s undefined because there isnβT any way in which this could happen if they all lie together). So let uses write `if _then _, else` here as an alternative expression for our conditionals so that things willvook more readable later! The first part says that everyone who liesse must also tell trues about their own behavior since no other person would know what happened without asking them directless , while second half implies somethingliketheyourselfwonΒ΄tbeatup against another person but rather agree notto do something bad thingsthat may harm others . This means everytruthr shouldbe able todetermine whetherornotthey lied based offinformation from otherinspectionaldatatrue/false conditions suchas howmanytimes each has toldtrueness etc.. Finally notehowimportantitistoseekoutall possible cases wheresomethingmighthappensothatnomissing out possibilities occur anywhereelseinthis problem solution ;otherwiseweright end upwith wrong answorks very fast duetotypicalerrors madebyhuman brainswhenthinkingabout complex problems like these oneswhich require careful attention towards detailsin ordertomakesureeverythings covered properly.* |
52 | null | human | A broken calculator has the + and $\times$ keys switched. For how many ordered pairs $(a, b)$ of integers will it correctly calculate $a+b$ using the labelled + key? | import Mathlib
theorem number_theory_611121 : {(a, b) : β€ Γ β€ | a + b = a * b}.encard = 2 := by | import Mathlib
/- A broken calculator has the + and $\times$ keys switched. For how many ordered pairs
$(a, b)$ of integers will it correctly calculate $a+b$ using the labelled + key? -/
theorem number_theory_611121 : {(a, b) : β€ Γ β€ | a + b = a * b}.encard = 2 := by
transitivity ({(0, 0), (2, 2)} : Set (β€ Γ β€)).encard
Β· congr 1
/- The set is exatly `{(0, 0), (2, 2)}`. -/
ext β¨a, bβ©
simp
refine β¨fun h => ?_, fun h => ?_β©
Β· /- From `a + b = ab`, we prove that `(b - 1) * (a - 1) = 1`. -/
have eq : (b - 1) * (a - 1) = 1 := by
rw [β sub_eq_zero] at h β’
rw [β neg_inj]
convert h using 1
ring
/- So `b - 1 β£ 1`, then `b` must be 0 or 2. -/
have hb : b - 1 β£ 1 := β¨_, eq.symmβ©
rw [β isUnit_iff_dvd_one, Int.isUnit_iff] at hb
/- So `a` is 0 or 2 respectively. -/
cases' hb with hb hb <;> (rw [hb] at eq; omega)
Β· /- If `(a, b)` is `(0, 0)` or `(2, 2)`, it satisfies the condition. -/
cases' h with h h <;> simp [h]
Β· /- Prove that the cardinality of `{(0, 0), (2, 2)}` is 2. -/
rw [Set.encard_insert_of_not_mem]
Β· norm_num
Β· intro h
simp only [Set.mem_singleton_iff] at h
apply_fun Prod.fst at h
simp at h
| ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/- A broken calculator has the + and $\times$ keys switched. For how many ordered pairs
$(a, b)$ of integers will it correctly calculate $a+b$ using the labelled + key? -/
theorem number_theory_611121 : {(a, b) : β€ Γ β€ | a + b = a * b}.encard = 2 := by
transitivity ({(0, 0), (2, 2)} : Set (β€ Γ β€)).encard
Β· congr 1
/- The set is exatly `{(0, 0), (2, 2)}`. -/
ext β¨a, bβ©
simp
refine β¨fun h => ?_, fun h => ?_β©
Β· /- From `a + b = ab`, we prove that `(b - 1) * (a - 1) = 1`. -/
have eq : (b - 1) * (a - 1) = 1 := by
rw [β sub_eq_zero] at h β’
rw [β neg_inj]
convert h using 1
ring
/- So `b - 1 β£ 1`, then `b` must be 0 or 2. -/
have hb : b - 1 β£ 1 := β¨_, eq.symmβ©
rw [β isUnit_iff_dvd_one, Int.isUnit_iff] at hb
/- So `a` is 0 or 2 respectively. | -/ | import Mathlib
/- A broken calculator has the + and $\times$ keys switched. For how many ordered pairs
$(a, b)$ of integers will it correctly calculate $a+b$ using the labelled + key? -/
theorem number_theory_611121 : {(a, b) : β€ Γ β€ | a + b = a * b}.encard = 2 := by
transitivity ({(0, 0), (2, 2)} : Set (β€ Γ β€)).encard
Β· congr 1
/- The set is exatly `{(0, 0), (2, 2)}`. -/
ext β¨a, bβ©
simp
refine β¨fun h => ?_, fun h => ?_β©
Β· /- From `a + b = ab`, we prove that `(b - 1) * (a - 1) = 1`. -/
have eq : (b - 1) * (a - 1) = 1 := by
rw [β sub_eq_zero] at h β’
rw [β neg_inj]
convert h using 1
ring
/- So `b - 1 β£ 1`, then `b` must be 0 or 2. -/
have hb : b - 1 β£ 1 := β¨_, eq.symmβ©
rw [β isUnit_iff_dvd_one, Int.isUnit_iff] at hb
/- So `a` is 0 or 2 respectively. | import Mathlib
/- A broken calculator has the + and $\times$ keys switched. For how many ordered pairs
$(a, b)$ of integers will it correctly calculate $a+b$ using the labelled + key? -/
theorem number_theory_611121 : {(a, b) : β€ Γ β€ | a + b = a * b}.encard = 2 := by
transitivity ({(0, 0), (2, 2)} : Set (β€ Γ β€)).encard
Β· congr 1
/- The set is exatly `{(0, 0), (2, 2)}`. -/
ext β¨a, bβ©
simp
refine β¨fun h => ?_, fun h => ?_β©
Β· /- From `a + b = ab`, we prove that `(b - 1) * (a - 1) = 1`. -/
have eq : (b - 1) * (a - 1) = 1 := by
rw [β sub_eq_zero] at h β’
rw [β neg_inj]
convert h using 1
ring
/- So `b - 1 β£ 1`, then `b` must be 0 or 2. -/
have hb : b - 1 β£ 1 := β¨_, eq.symmβ©
rw [β isUnit_iff_dvd_one, Int.isUnit_iff] at hb
/- So `a` is 0 or 2 respectively. -/ |
53 | null | human | A malfunctioning digital clock shows the time $9: 57 \mathrm{AM}$; however, the correct time is $10: 10 \mathrm{AM}$. There are two buttons on the clock, one of which increases the time displayed by 9 minutes, and another which decreases the time by 20 minutes. What is the minimum number of button presses necessary to correctly set the clock to the correct time? | import Mathlib
theorem logic_and_puzzles_609734 (currentTime correctTime : β) (h1 : currentTime = 9 * 60 + 57) (h2 : correctTime = 10 * 60 + 10) : IsLeast { n | β a b, currentTime + 9 * a - 20 * b = correctTime β§ a + b = n } 24 := by | import Mathlib
theorem logic_and_puzzles_609734 (currentTime correctTime : β) (h1 : currentTime = 9 * 60 + 57) (h2 : correctTime = 10 * 60 + 10) : IsLeast { n | β a b, currentTime + 9 * a - 20 * b = correctTime β§ a + b = n } 24 := by
constructor
Β· -- supply witnesses
simp only [Set.mem_setOf_eq]
apply Exists.intro 17
apply Exists.intro 7
omega
Β· intros n hn
-- obtain hypotheses
obtain β¨a,b,ha1,ha2β© := hn
-- solve
omega | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem logic_and_puzzles_609734 (currentTime correctTime : β) (h1 : currentTime = 9 * 60 + 57) (h2 : correctTime = 10 * 60 + 10) : IsLeast { n | β a b, currentTime + 9 * a - 20 * b = correctTime β§ a + b = n } 24 := by
constructor
Β· -- supply witnesses
simp only [Set.mem_setOf_eq]
apply | Existsβ' _ fun i j hij HijHji IJIJIJI JHIJTIH TITHTTTI HTTHS SHSSTTS TSSTSSS SSTTHH HHJJII IIJK JKKL KLKM KMMN MNNO OP PQ QR RS ST TU UV VW X Y Z WXWX XYXY VWVWVWW UTVTU VTUUVUT UTXVTX XVYZYZ | import Mathlib
theorem logic_and_puzzles_609734 (currentTime correctTime : β) (h1 : currentTime = 9 * 60 + 57) (h2 : correctTime = 10 * 60 + 10) : IsLeast { n | β a b, currentTime + 9 * a - 20 * b = correctTime β§ a + b = n } 24 := by
constructor
Β· -- supply witnesses
simp only [Set.mem_setOf_eq]
apply | import Mathlib
theorem logic_and_puzzles_609734 (currentTime correctTime : β) (h1 : currentTime = 9 * 60 + 57) (h2 : correctTime = 10 * 60 + 10) : IsLeast { n | β a b, currentTime + 9 * a - 20 * b = correctTime β§ a + b = n } 24 := by
constructor
Β· -- supply witnesses
simp only [Set.mem_setOf_eq]
apply Existsβ' _ fun i j hij HijHji IJIJIJI JHIJTIH TITHTTTI HTTHS SHSSTTS TSSTSSS SSTTHH HHJJII IIJK JKKL KLKM KMMN MNNO OP PQ QR RS ST TU UV VW X Y Z WXWX XYXY VWVWVWW UTVTU VTUUVUT UTXVTX XVYZYZ |
56 | math-word-problem | autoformalizer | Matt has somewhere between $1000$ and $2000$ pieces of paper he's trying to divide into piles of the same size (but not all in one pile or piles of one sheet each). He tries $2$, $3$, $4$, $5$, $6$, $7$, and $8$ piles but ends up with one sheet left over each time. How many piles does he need? | import Mathlib
/- Matt has somewhere between $1000$ and $2000$ pieces of paper he's trying to divide into piles of the same size (but not all in one pile or piles of one sheet each). He tries $2$, $3$, $4$, $5$, $6$, $7$, and $8$ piles but ends up with one sheet left over each time. How many piles does he need? -/
theorem number_theory_67977 {n : β} (hn : n β₯ 1000 β§ n < 2000)
(h1 : n % 2 = 1) (h2 : n % 3 = 1) (h3 : n % 4 = 1) (h4 : n % 5 = 1)
(h5 : n % 6 = 1) (h6 : n % 7 = 1) (h7 : n % 8 = 1) :
n % 41 = 0 := by | import Mathlib
theorem number_theory_67977 {n : β} (hn : n β₯ 1000 β§ n < 2000)
(h1 : n % 2 = 1) (h2 : n % 3 = 1) (h3 : n % 4 = 1) (h4 : n % 5 = 1)
(h5 : n % 6 = 1) (h6 : n % 7 = 1) (h7 : n % 8 = 1) :
n % 41 = 0 := by
rcases hn with β¨hn1, hn2β©
interval_cases n <;> omega | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem number_theory_67977 {n : β} (hn : n β₯ 1000 β§ n < 2000)
(h1 : n % 2 = 1) (h2 : n % 3 = 1) (h3 : n % 4 = 1) (h4 : n % 5 = 1)
(h5 : n % 6 = 1) (h6 : n % 7 = 1) | (h7 : n % 8 = 1): NatβnFiniteGroup s where F in S is prime:=by apply existsUniquePrimeDivisor; assumption | import Mathlib
theorem number_theory_67977 {n : β} (hn : n β₯ 1000 β§ n < 2000)
(h1 : n % 2 = 1) (h2 : n % 3 = 1) (h3 : n % 4 = 1) (h4 : n % 5 = 1)
(h5 : n % 6 = 1) (h6 : n % 7 = 1) | import Mathlib
theorem number_theory_67977 {n : β} (hn : n β₯ 1000 β§ n < 2000)
(h1 : n % 2 = 1) (h2 : n % 3 = 1) (h3 : n % 4 = 1) (h4 : n % 5 = 1)
(h5 : n % 6 = 1) (h6 : n % 7 = 1) (h7 : n % 8 = 1): NatβnFiniteGroup s where F in S is prime:=by apply existsUniquePrimeDivisor; assumption |
57 | math-word-problem | autoformalizer | Let $f(x) = x^2 + 6x + 7$. Determine the smallest possible value of $f(f(f(f(x))))$ over all real numbers $x.$ | import Mathlib
open Real Set
open scoped BigOperators
/- Let $f(x) = x^2 + 6x + 7$. Determine the smallest possible value of $f(f(f(f(x))))$ over all real numbers $x.$ -/
theorem calculus_75601 {f : β β β} (hf : f = Ξ» x => x ^ 2 + 6 * x + 7) :
IsLeast {y | β x, y = f (f (f (f x)))} 23 :=
| import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_75601 {f : β β β} (hf : f = Ξ» x => x ^ 2 + 6 * x + 7) :
IsLeast {y | β x, y = f (f (f (f x)))} 23 := by
constructor
Β· -- First part of IsLeast: Show 23 is in the set.
use -3
norm_num [hf]
Β· -- Second part of IsLeast: Show no value less than 23 is in the set.
intro y hy
rcases hy with β¨x, hxβ©
rw [hx]
norm_num [hf]
nlinarith [sq_nonneg (x + 3), sq_nonneg (x ^ 2 + 6 * x + 7 + 2), sq_nonneg ((x ^ 2 + 6 * x + 7) + 2), sq_nonneg ((x ^ 2 + 6 * x + 7) ^ 2 + 6 * (x ^ 2 + 6 * x + 7) + 7 + 1)] | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_75601 {f : β β β} (hf : f = Ξ» x => x ^ 2 + 6 * x + 7) :
IsLeast {y | β x, y = f (f (f (f x)))} 23 := by
constructor
Β· -- First part of IsLeast: Show 23 is in the set.
use -3
norm_num [hf]
Β· -- Second part of IsLeast: Show no value less than 23 is in the set.
intro y hy
rcases hy with β¨x, hxβ©
rw [hx]
norm_num [hf]
nlinarith [sq_nonneg (x + 3), | mul_self (-βReal.sqrt _)] | import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_75601 {f : β β β} (hf : f = Ξ» x => x ^ 2 + 6 * x + 7) :
IsLeast {y | β x, y = f (f (f (f x)))} 23 := by
constructor
Β· -- First part of IsLeast: Show 23 is in the set.
use -3
norm_num [hf]
Β· -- Second part of IsLeast: Show no value less than 23 is in the set.
intro y hy
rcases hy with β¨x, hxβ©
rw [hx]
norm_num [hf]
nlinarith [sq_nonneg (x + 3), | import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_75601 {f : β β β} (hf : f = Ξ» x => x ^ 2 + 6 * x + 7) :
IsLeast {y | β x, y = f (f (f (f x)))} 23 := by
constructor
Β· -- First part of IsLeast: Show 23 is in the set.
use -3
norm_num [hf]
Β· -- Second part of IsLeast: Show no value less than 23 is in the set.
intro y hy
rcases hy with β¨x, hxβ©
rw [hx]
norm_num [hf]
nlinarith [sq_nonneg (x + 3), mul_self (-βReal.sqrt _)] |
58 | math-word-problem | autoformalizer | Find all pairs $(x, y)$ of positive integers such that $x^{2}=y^{2}+7 y+6$.
| import Mathlib
/- Find all pairs $(x, y)$ of positive integers such that $x^{2}=y^{2}+7 y+6$.
-/
theorem algebra_116978 (x y : β€) (hx : 0 < x) (hy : 0 < y) :
x^2 = y^2 + 7 * y + 6 β (x, y) = (6, 3) := by | import Mathlib
theorem algebra_116978 (x y : β€) (hx : 0 < x) (hy : 0 < y) :
x^2 = y^2 + 7 * y + 6 β (x, y) = (6, 3) := by
constructor
Β· -- First direction: xΒ² = yΒ² + 7y + 6 β (x, y) = (6, 3)
intro h
have h1 : 4 * x ^ 2 = (2 * y + 7) ^ 2 - 25 := by
nlinarith
have h2 : (2 * y + 7 - 2 * x) * (2 * y + 7 + 2 * x) = 25 := by
nlinarith
have h3 : (2 * y + 7 - 2 * x) β£ 25 := by
use (2 * y + 7 + 2 * x)
all_goals linarith
have h4 : (2 * y + 7 - 2 * x) = 1 β¨ (2 * y + 7 - 2 * x) = 5 β¨ (2 * y + 7 - 2 * x) = 25 β¨
(2 * y + 7 - 2 * x) = -1 β¨ (2 * y + 7 - 2 * x) = -5 β¨ (2 * y + 7 - 2 * x) = -25 := by
have h6 : (2 * y + 7 - 2 * x) β£ 25 := h3
have h7 : (2 * y + 7 - 2 * x) β€ 25 := by
apply Int.le_of_dvd (by norm_num) h6
have h8 : (2 * y + 7 - 2 * x) β₯ -25 := by
have h9 : -(2 * y + 7 - 2 * x) β£ 25 := by
exact Int.neg_dvd.2 h6
have h10 : -(2 * y + 7 - 2 * x) β€ 25 := by
apply Int.le_of_dvd (by norm_num) h9
omega
interval_cases h1 : (2 * y + 7 - 2 * x) <;> tauto
rcases h4 with (h4 | h4 | h4 | h4 | h4 | h4)
Β· -- Case where (2 * y + 7 - 2 * x) = 1, so (2 * y + 7 + 2 * x) = 25
have h5 : (2 * y + 7 + 2 * x) = 25 := by
rw [show (2 * y + 7 - 2 * x) = 1 by linarith] at h2
nlinarith
have h6 : y = 3 := by
omega
have h7 : x = 6 := by
omega
simp [h7, h6]
Β· -- Case where (2 * y + 7 - 2 * x) = 5, so (2 * y + 7 + 2 * x) = 5
have h5 : (2 * y + 7 + 2 * x) = 5 := by
rw [show (2 * y + 7 - 2 * x) = 5 by linarith] at h2
nlinarith
exfalso
linarith
Β· -- Case where (2 * y + 7 - 2 * x) = 25, so (2 * y + 7 + 2 * x) = 1
have h5 : (2 * y + 7 + 2 * x) = 1 := by
rw [show (2 * y + 7 - 2 * x) = 25 by linarith] at h2
nlinarith
exfalso
linarith
Β· -- Case where (2 * y + 7 - 2 * x) = -1, so (2 * y + 7 + 2 * x) = -25
have h5 : (2 * y + 7 + 2 * x) = -25 := by
rw [show (2 * y + 7 - 2 * x) = -1 by linarith] at h2
nlinarith
exfalso
linarith
Β· -- Case where (2 * y + 7 - 2 * x) = -5, so (2 * y + 7 + 2 * x) = -5
have h5 : (2 * y + 7 + 2 * x) = -5 := by
rw [show (2 * y + 7 - 2 * x) = -5 by linarith] at h2
nlinarith
exfalso
linarith
Β· -- Case where (2 * y + 7 - 2 * x) = -25, so (2 * y + 7 + 2 * x) = -1
have h5 : (2 * y + 7 + 2 * x) = -1 := by
rw [show (2 * y + 7 - 2 * x) = -25 by linarith] at h2
nlinarith
exfalso
linarith
Β· -- Backward implication: (x, y) = (6, 3) β xΒ² = yΒ² + 7y + 6
rintro β¨rfl, rflβ©
norm_num | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem algebra_116978 (x y : β€) (hx : 0 < x) (hy : 0 < y) :
x^2 = y^2 + 7 * y + 6 β (x, y) = (6, 3) := by
constructor
Β· -- First direction: xΒ² = yΒ² + 7y + 6 β (x, y) = (6, 3)
intro h
have h1 : 4 * x ^ 2 = (2 * y + 7) ^ 2 - 25 := by
nlinarith
have h2 : (2 * y + 7 - 2 * x) * (2 * y + 7 + 2 * x) = 25 := by
nlinarith
have h3 : (2 * y + 7 - 2 * x) β£ 25 := by
use (2 * y + 7 + 2 * x)
all_goals linarith
have h4 : (2 * y + 7 - 2 * x) = 1 β¨ (2 * y + 7 - 2 * x) = 5 β¨ (2 * y + 7 - 2 * x) = 25 β¨
(2 * y + 7 - 2 * x) = -1 β¨ (2 * y + 7 - 2 * x) = -5 β¨ (2 * y + 7 - 2 * x) = -25 := by
have h6 : (2 * y + 7 - 2 * x) β£ 25 := h3
have h7 : (2 * y + 7 - 2 * x) β€ 25 := by
apply Int.le_of_dvd (by norm_num) h6
have h8 : (2 * y + 7 - 2 * x) β₯ -25 := by
have h9 : -(2 * y + 7 - 2 * x) β£ 25 := by
exact Int.neg_dvd.2 h6
have h10 : -(2 * y + 7 - 2 * x) β€ 25 := by
apply Int.le_of_dvd (by norm_num) h9
omega
interval_cases h1 : (2 * y + 7 - 2 * x) <;> tauto
rcases h4 with (h4 | h4 | h4 | h4 | h4 | h4)
Β· -- Case where (2 * y + 7 - 2 * x) = 1, so (2 * y + 7 + 2 * x) = 25
have h5 : (2 * y + 7 + 2 * x) = 25 := by
rw [show (2 * y + 7 - 2 * x) = 1 by linarith] at h2
nlinarith
have h6 : y = 3 := by
omega
have h7 : x = 6 := by
omega
simp [h7, h6]
Β· -- Case where (2 * y + 7 - 2 * x) = 5, so (2 * y + 7 + 2 * x) = 5
have h5 : (2 * y + 7 + 2 * x) = 5 := by
rw [show (2 * y + 7 - 2 * x) = 5 by linarith] at h2
nlinarith
exfalso
linarith
Β· -- Case where (2 * y + 7 - 2 * x) = 25, so (2 * y + 7 + 2 * x) = 1
have h5 : (2 * y + 7 + 2 * x) = 1 := by
rw [show (2 * y + 7 - 2 * x) = 25 by linarith] at h | 2 ;nlinarity[Int];norm' only [(β mul)]<|>.intrinsiciquality _ (-(-β((fun p q i j v u w z s t r e genalgebra._aux_.mkββββ
{i} (_coe?_eq'_one.{lift}))))*v+w)*z);all_{p};rw[<left>]at this|-{right}.mul..comm.<|_><|>try itnf')β© using fun eq neq ih proof cases Eqaulitypepinfo'.noConfusionEq?.elim! {?}ih ?proficquationspatchable:= Ifforprocesser is not able todetermine if there exists an equivalence betwin these terms because they do no contain any variable symbols from their domains but rather consist entirely oftypes which appear as subtypes within other types suchthatat none occur anywhere else on either sideafter applying substitution rules based onthe given assumptions about relationships between objects typed into these expressions.\nFirst we note that since both $\frac{(b\cdotd)\pm(\sqrt{\Delta})}{k}$ must divide $-r$, then each factor itself mustb dividethe product $(bd)^j$. This means thadat every prime divisor divising somethermal expression has its own inverse image under our mapningshapingingroup action ,which implies alsothat every nonprime divisior willvaluate positively when appliedto $-\rheatmospheric{-}(ba)^{ju}$. Now let us assume withoutlossofgeneralityοΌsince otherwisewe could simply replace "$\mathbbZ$"with"$(\mathrematics Z/N,\oplus)"and continuelikesofarοΌhat$|G|=|\mathcal G|$ holds trueforrsomeferrorfree finite group actions over fields . Then forany fixedelement$~X~$in$~\mathcalp F $,there existssome positive integer~${K}\geq N^{-|F|}^{-}|H|^{-\operatorname{|Formula|(A)}}$,where ~$|\text{-}|^-$denotescardinaltivity raisedtonoticationpotencensοΌsuchthatarespectingthe chosen basisvectors,~$[Y]_B=[\sum _{I}^{J}[U]_{C}]_{\tilde B}=[-V]_{\bar D}=[W]$ gives risetosomespace filling function~$\varphi $. We may now define another spacefillingfunction via compositionwiththis first choice:~$$\psi=\phi*\sigma,$$whereο½ΟοΌΟβΒΉγ By construction,~\psiatmospherically mapseach pointbackontoitselfunder translationsatlongledinvariant subspaces generatedbyelementsfromour set up above while preservinp distancesbetweenthemas wellaneeddledityour question). It remainstoprove thoatuthis secondchoice stillsatisfies condition~(ii): Ifweshowitinsteadfrommaximalinessequomentallysatisfying conditionsoncentration functionssatisfy property*(iii*)---namely,"Iftwo points lieon oppositele sidesofsomereceiving line passing through zeropowerfulpointthentheir distanceshouldbe less than half what would resultifthey were located symmetrically around some verticalline crossing throughmidpoint",βwenearliply obtaincondition *(iv*:*"The sumover allobstacles shouldnot exceedn twicetheequal length." ) To see why*, supposeotherwiseοΌletting $\|\mathbf X\|$=βΞ£$_ix_\imath$;$^T$+$\|\mathbfr\|^P+$...denote Euclideannormeson complex spaces definedabove yields \[||\xi||=|(|Re\{\zeta\}|+Im\{ze\})|=\tfrac{|\overline{(\chi+\jmath)}\rangle}{\langle (\eta +\kappa)|}=\dfrac{[(|\theta -\tau|)+|\beta |\cos((\pi/\nu)-\alpha)+\sin(-\gamma)/\,\mu)\,]}{\big[(\rho+(\varsigma)\,\cot\,({\vartheta})\tan\,({{\varpi}/{{Ξ½}}}))\sec\,{\omega}\,\tanh({-{Ο}/{ΞΊ}}\log ({Ξ΄}+Ξ΅))^{L}\,+{ΞΆ}^M\\\]However,-canceling out common factors from topanbottom leaves\[|{\boldsymbol Y}|= |{\boldsymbolw}|\leq \|[{\lambda}_D]\circ [{\lambda}_{E}]\cdots\|_{S}\|\vec V\|_Q,\]contradictiny assumption.* | import Mathlib
theorem algebra_116978 (x y : β€) (hx : 0 < x) (hy : 0 < y) :
x^2 = y^2 + 7 * y + 6 β (x, y) = (6, 3) := by
constructor
Β· -- First direction: xΒ² = yΒ² + 7y + 6 β (x, y) = (6, 3)
intro h
have h1 : 4 * x ^ 2 = (2 * y + 7) ^ 2 - 25 := by
nlinarith
have h2 : (2 * y + 7 - 2 * x) * (2 * y + 7 + 2 * x) = 25 := by
nlinarith
have h3 : (2 * y + 7 - 2 * x) β£ 25 := by
use (2 * y + 7 + 2 * x)
all_goals linarith
have h4 : (2 * y + 7 - 2 * x) = 1 β¨ (2 * y + 7 - 2 * x) = 5 β¨ (2 * y + 7 - 2 * x) = 25 β¨
(2 * y + 7 - 2 * x) = -1 β¨ (2 * y + 7 - 2 * x) = -5 β¨ (2 * y + 7 - 2 * x) = -25 := by
have h6 : (2 * y + 7 - 2 * x) β£ 25 := h3
have h7 : (2 * y + 7 - 2 * x) β€ 25 := by
apply Int.le_of_dvd (by norm_num) h6
have h8 : (2 * y + 7 - 2 * x) β₯ -25 := by
have h9 : -(2 * y + 7 - 2 * x) β£ 25 := by
exact Int.neg_dvd.2 h6
have h10 : -(2 * y + 7 - 2 * x) β€ 25 := by
apply Int.le_of_dvd (by norm_num) h9
omega
interval_cases h1 : (2 * y + 7 - 2 * x) <;> tauto
rcases h4 with (h4 | h4 | h4 | h4 | h4 | h4)
Β· -- Case where (2 * y + 7 - 2 * x) = 1, so (2 * y + 7 + 2 * x) = 25
have h5 : (2 * y + 7 + 2 * x) = 25 := by
rw [show (2 * y + 7 - 2 * x) = 1 by linarith] at h2
nlinarith
have h6 : y = 3 := by
omega
have h7 : x = 6 := by
omega
simp [h7, h6]
Β· -- Case where (2 * y + 7 - 2 * x) = 5, so (2 * y + 7 + 2 * x) = 5
have h5 : (2 * y + 7 + 2 * x) = 5 := by
rw [show (2 * y + 7 - 2 * x) = 5 by linarith] at h2
nlinarith
exfalso
linarith
Β· -- Case where (2 * y + 7 - 2 * x) = 25, so (2 * y + 7 + 2 * x) = 1
have h5 : (2 * y + 7 + 2 * x) = 1 := by
rw [show (2 * y + 7 - 2 * x) = 25 by linarith] at h | import Mathlib
theorem algebra_116978 (x y : β€) (hx : 0 < x) (hy : 0 < y) :
x^2 = y^2 + 7 * y + 6 β (x, y) = (6, 3) := by
constructor
Β· -- First direction: xΒ² = yΒ² + 7y + 6 β (x, y) = (6, 3)
intro h
have h1 : 4 * x ^ 2 = (2 * y + 7) ^ 2 - 25 := by
nlinarith
have h2 : (2 * y + 7 - 2 * x) * (2 * y + 7 + 2 * x) = 25 := by
nlinarith
have h3 : (2 * y + 7 - 2 * x) β£ 25 := by
use (2 * y + 7 + 2 * x)
all_goals linarith
have h4 : (2 * y + 7 - 2 * x) = 1 β¨ (2 * y + 7 - 2 * x) = 5 β¨ (2 * y + 7 - 2 * x) = 25 β¨
(2 * y + 7 - 2 * x) = -1 β¨ (2 * y + 7 - 2 * x) = -5 β¨ (2 * y + 7 - 2 * x) = -25 := by
have h6 : (2 * y + 7 - 2 * x) β£ 25 := h3
have h7 : (2 * y + 7 - 2 * x) β€ 25 := by
apply Int.le_of_dvd (by norm_num) h6
have h8 : (2 * y + 7 - 2 * x) β₯ -25 := by
have h9 : -(2 * y + 7 - 2 * x) β£ 25 := by
exact Int.neg_dvd.2 h6
have h10 : -(2 * y + 7 - 2 * x) β€ 25 := by
apply Int.le_of_dvd (by norm_num) h9
omega
interval_cases h1 : (2 * y + 7 - 2 * x) <;> tauto
rcases h4 with (h4 | h4 | h4 | h4 | h4 | h4)
Β· -- Case where (2 * y + 7 - 2 * x) = 1, so (2 * y + 7 + 2 * x) = 25
have h5 : (2 * y + 7 + 2 * x) = 25 := by
rw [show (2 * y + 7 - 2 * x) = 1 by linarith] at h2
nlinarith
have h6 : y = 3 := by
omega
have h7 : x = 6 := by
omega
simp [h7, h6]
Β· -- Case where (2 * y + 7 - 2 * x) = 5, so (2 * y + 7 + 2 * x) = 5
have h5 : (2 * y + 7 + 2 * x) = 5 := by
rw [show (2 * y + 7 - 2 * x) = 5 by linarith] at h2
nlinarith
exfalso
linarith
Β· -- Case where (2 * y + 7 - 2 * x) = 25, so (2 * y + 7 + 2 * x) = 1
have h5 : (2 * y + 7 + 2 * x) = 1 := by
rw [show (2 * y + 7 - 2 * x) = 25 by linarith] at h2 ;nlinarity[Int];norm' only [(β mul)]<|>.intrinsiciquality _ (-(-β((fun p q i j v u w z s t r e genalgebra._aux_.mkββββ
{i} (_coe?_eq'_one.{lift}))))*v+w)*z);all_{p};rw[<left>]at this|-{right}.mul..comm.<|_><|>try itnf')β© using fun eq neq ih proof cases Eqaulitypepinfo'.noConfusionEq?.elim! {?}ih ?proficquationspatchable:= Ifforprocesser is not able todetermine if there exists an equivalence betwin these terms because they do no contain any variable symbols from their domains but rather consist entirely oftypes which appear as subtypes within other types suchthatat none occur anywhere else on either sideafter applying substitution rules based onthe given assumptions about relationships between objects typed into these expressions.\nFirst we note that since both $\frac{(b\cdotd)\pm(\sqrt{\Delta})}{k}$ must divide $-r$, then each factor itself mustb dividethe product $(bd)^j$. This means thadat every prime divisor divising somethermal expression has its own inverse image under our mapningshapingingroup action ,which implies alsothat every nonprime divisior willvaluate positively when appliedto $-\rheatmospheric{-}(ba)^{ju}$. Now let us assume withoutlossofgeneralityοΌsince otherwisewe could simply replace "$\mathbbZ$"with"$(\mathrematics Z/N,\oplus)"and continuelikesofarοΌhat$|G|=|\mathcal G|$ holds trueforrsomeferrorfree finite group actions over fields . Then forany fixedelement$~X~$in$~\mathcalp F $,there existssome positive integer~${K}\geq N^{-|F|}^{-}|H|^{-\operatorname{|Formula|(A)}}$,where ~$|\text{-}|^-$denotescardinaltivity raisedtonoticationpotencensοΌsuchthatarespectingthe chosen basisvectors,~$[Y]_B=[\sum _{I}^{J}[U]_{C}]_{\tilde B}=[-V]_{\bar D}=[W]$ gives risetosomespace filling function~$\varphi $. We may now define another spacefillingfunction via compositionwiththis first choice:~$$\psi=\phi*\sigma,$$whereο½ΟοΌΟβΒΉγ By construction,~\psiatmospherically mapseach pointbackontoitselfunder translationsatlongledinvariant subspaces generatedbyelementsfromour set up above while preservinp distancesbetweenthemas wellaneeddledityour question). It remainstoprove thoatuthis secondchoice stillsatisfies condition~(ii): Ifweshowitinsteadfrommaximalinessequomentallysatisfying conditionsoncentration functionssatisfy property*(iii*)---namely,"Iftwo points lieon oppositele sidesofsomereceiving line passing through zeropowerfulpointthentheir distanceshouldbe less than half what would resultifthey were located symmetrically around some verticalline crossing throughmidpoint",βwenearliply obtaincondition *(iv*:*"The sumover allobstacles shouldnot exceedn twicetheequal length." ) To see why*, supposeotherwiseοΌletting $\|\mathbf X\|$=βΞ£$_ix_\imath$;$^T$+$\|\mathbfr\|^P+$...denote Euclideannormeson complex spaces definedabove yields \[||\xi||=|(|Re\{\zeta\}|+Im\{ze\})|=\tfrac{|\overline{(\chi+\jmath)}\rangle}{\langle (\eta +\kappa)|}=\dfrac{[(|\theta -\tau|)+|\beta |\cos((\pi/\nu)-\alpha)+\sin(-\gamma)/\,\mu)\,]}{\big[(\rho+(\varsigma)\,\cot\,({\vartheta})\tan\,({{\varpi}/{{Ξ½}}}))\sec\,{\omega}\,\tanh({-{Ο}/{ΞΊ}}\log ({Ξ΄}+Ξ΅))^{L}\,+{ΞΆ}^M\\\]However,-canceling out common factors from topanbottom leaves\[|{\boldsymbol Y}|= |{\boldsymbolw}|\leq \|[{\lambda}_D]\circ [{\lambda}_{E}]\cdots\|_{S}\|\vec V\|_Q,\]contradictiny assumption.* |
59 | math-word-problem | autoformalizer | 11. Find the smallest positive integer $n$ such that $1^{2}+2^{2}+3^{2}+4^{2}+\cdots+n^{2}$ is divisible by 100 . | import Mathlib
/- 11. Find the smallest positive integer $n$ such that $1^{2}+2^{2}+3^{2}+4^{2}+\cdots+n^{2}$ is divisible by 100 . -/
theorem number_theory_188669 :
IsLeast {n : β | 0 < n β§ 100 β£ β i in Finset.Icc 1 n, i^2} 24 := by | import Mathlib
theorem number_theory_188669 :
IsLeast {n : β | 0 < n β§ 100 β£ β i in Finset.Icc 1 n, i^2} 24 := by
constructor
Β· -- Show that 24 is in the set
constructor
Β· norm_num
Β· -- Check divisibility
native_decide
Β· -- Show that any element in the set is β₯ 24
intro n hn
rcases hn with β¨hn_pos, hn_divβ©
by_contra h
push_neg at h
interval_cases n <;> try { contradiction <;> try { contradiction <;> try { contradiction } } }
<;> norm_num [Finset.sum_Icc_succ_top] at hn_div <;> try { omega }
<;> try { contradiction }
<;> omega | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem number_theory_188669 :
IsLeast {n : β | 0 < n β§ 100 β£ β i in Finset.Icc 1 n, i^2} 24 := by
constructor
Β· -- Show that 24 is in the set
constructor
Β· norm_num
Β· -- Check divisibility
native_decide
Β· -- Show that any element in the set is β₯ 24
intro n hn
rcases hn with β¨hn_pos, hn_divβ©
by_contra h
push_neg at h
interv | alAt _ (-Inf..Real.sqrtββ).mem' | import Mathlib
theorem number_theory_188669 :
IsLeast {n : β | 0 < n β§ 100 β£ β i in Finset.Icc 1 n, i^2} 24 := by
constructor
Β· -- Show that 24 is in the set
constructor
Β· norm_num
Β· -- Check divisibility
native_decide
Β· -- Show that any element in the set is β₯ 24
intro n hn
rcases hn with β¨hn_pos, hn_divβ©
by_contra h
push_neg at h
interv | import Mathlib
theorem number_theory_188669 :
IsLeast {n : β | 0 < n β§ 100 β£ β i in Finset.Icc 1 n, i^2} 24 := by
constructor
Β· -- Show that 24 is in the set
constructor
Β· norm_num
Β· -- Check divisibility
native_decide
Β· -- Show that any element in the set is β₯ 24
intro n hn
rcases hn with β¨hn_pos, hn_divβ©
by_contra h
push_neg at h
intervalAt _ (-Inf..Real.sqrtββ).mem' |
60 | math-word-problem | autoformalizer | A number $p$ is $perfect$ if the sum of its divisors, except $p$ is $p$. Let $f$ be a function such that:
$f(n)=0$, if n is perfect
$f(n)=0$, if the last digit of n is 4
$f(a.b)=f(a)+f(b)$
Find $f(1998)$ | import Mathlib
def is_perfect (n : β) := β d in n.divisors.erase n, d = n
/- A number $p$ is $perfect$ if the sum of its divisors, except $p$ is $p$. Let $f$ be a function such that:
$f(n)=0$, if n is perfect
$f(n)=0$, if the last digit of n is 4
$f(a.b)=f(a)+f(b)$
Find $f(1998)$ -/
theorem number_theory_69752 {f : β β β€} (hf1 : β n, is_perfect n β f n = 0)
(hf2 : β n, n % 10 = 4 β f n = 0) (hf3 : β a b, f (a * b) = f a + f b) :
f 1998 = 0 := by | import Mathlib
def is_perfect (n : β) := β d in n.divisors.erase n, d = n
theorem number_theory_69752 {f : β β β€} (hf1 : β n, is_perfect n β f n = 0)
(hf2 : β n, n % 10 = 4 β f n = 0) (hf3 : β a b, f (a * b) = f a + f b) :
f 1998 = 0 := by
have h1 : f 1998 = f 6 + f 333 := by
calc
f 1998 = f (6 * 333) := by norm_num
_ = f 6 + f 333 := by rw [hf3 6 333]
have h2 : f 6 = 0 := by
have h3 : is_perfect 6 := by
unfold is_perfect
native_decide
have h4 : f 6 = 0 := hf1 6 h3
exact h4
have h5 : f 333 = f 3 + f 111 := by
calc
f 333 = f (3 * 111) := by norm_num
_ = f 3 + f 111 := by rw [hf3 3 111]
have h6 : f 111 = f 3 + f 37 := by
calc
f 111 = f (3 * 37) := by norm_num
_ = f 3 + f 37 := by rw [hf3 3 37]
have h7 : f 2 = 0 := by
have h8 : f 4 = 0 := by
apply hf2 4
all_goals norm_num
have h9 : f 4 = f 2 + f 2 := by
calc
f 4 = f (2 * 2) := by norm_num
_ = f 2 + f 2 := by rw [hf3 2 2]
linarith [h8, h9]
have h8 : f 3 = 0 := by
have h9 : f 6 = 0 := h2
have h10 : f 6 = f 2 + f 3 := by
calc
f 6 = f (2 * 3) := by norm_num
_ = f 2 + f 3 := by rw [hf3 2 3]
linarith [h7, h10, h9]
have h11 : f 37 = 0 := by
have h12 : f 74 = 0 := by
apply hf2 74
all_goals norm_num
have h13 : f 74 = f 2 + f 37 := by
calc
f 74 = f (2 * 37) := by norm_num
_ = f 2 + f 37 := by rw [hf3 2 37]
linarith [h7, h12, h13]
have h14 : f 333 = 0 := by
linarith [h5, h6, h8, h11]
linarith [h1, h2, h14] | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
def is_perfect (n : β) := β d in n.divisors.erase n, d = n
theorem number_theory_69752 {f : β β β€} (hf1 : β n, is_perfect n β f n = 0)
(hf2 : β n, n % 10 = 4 β f n = 0) (hf3 : β a b, f (a * b) = f a + f b) :
f 1998 = 0 := by
have h1 : f 1998 = f 6 + f 333 := by
calc
f 1998 = f (6 * 333) := by norm_num
_ = f 6 + f 333 := by rw [hf3 6 333]
have h2 : f 6 = 0 := by
have h3 : is_perfect 6 := by
unfold is_perfect
native_decide
have h4 : f 6 = 0 := hf1 6 h3
exact h4
have h5 : f 333 = f 3 + f 111 := by
calc
f 333 = f (3 * 111) := by norm_num
_ = f 3 + f 111 := by rw [hf3 3 111]
have h6 : f 111 = f 3 + f 37 := by
calc
f 111 = f (3 * 37) := by norm_num
_ = f 3 + f 37 := by rw [hf3 3 37]
have h7 : f 2 = 0 := by
have h8 : f 4 = 0 := by
apply hf2 4
all_goals norm_num
have h9 : f 4 = f 2 + f 2 := by
calc
f 4 = f (2 * 2) := by norm_num
_ = f 2 + f 2 := by rw [hf3 2 2]
linarith [h8, h9]
have h8 : f 3 = 0 := by
have h9 : f 6 = 0 := h2
have h10 : f 6 = f | 3 + f 2:=byrw[βmulNat' , mulOne', addZero'] at* this; assumption?_= ?_, which gives us an expression for `nf` as someing multiple ogethermos where each term has either factorized into primes less than ten using only addition operations on integers`. We then showthat if we take any prime p such that there exists another integer q such thatusatisfying both conditions above but not equating itself nor its inverse modulo fiveοΌi.. e., neither itself nillifiable under multiplicationοΌοΌthen applying our function twice would give zero because every other possible factorization must be divisiblebyeither sixorprime factors whose product isnotdivisiblebyfive since they alreasyouldbe divisibileprinciplely). This means thathe result willvalueatoequaltozerowehneeded! | import Mathlib
def is_perfect (n : β) := β d in n.divisors.erase n, d = n
theorem number_theory_69752 {f : β β β€} (hf1 : β n, is_perfect n β f n = 0)
(hf2 : β n, n % 10 = 4 β f n = 0) (hf3 : β a b, f (a * b) = f a + f b) :
f 1998 = 0 := by
have h1 : f 1998 = f 6 + f 333 := by
calc
f 1998 = f (6 * 333) := by norm_num
_ = f 6 + f 333 := by rw [hf3 6 333]
have h2 : f 6 = 0 := by
have h3 : is_perfect 6 := by
unfold is_perfect
native_decide
have h4 : f 6 = 0 := hf1 6 h3
exact h4
have h5 : f 333 = f 3 + f 111 := by
calc
f 333 = f (3 * 111) := by norm_num
_ = f 3 + f 111 := by rw [hf3 3 111]
have h6 : f 111 = f 3 + f 37 := by
calc
f 111 = f (3 * 37) := by norm_num
_ = f 3 + f 37 := by rw [hf3 3 37]
have h7 : f 2 = 0 := by
have h8 : f 4 = 0 := by
apply hf2 4
all_goals norm_num
have h9 : f 4 = f 2 + f 2 := by
calc
f 4 = f (2 * 2) := by norm_num
_ = f 2 + f 2 := by rw [hf3 2 2]
linarith [h8, h9]
have h8 : f 3 = 0 := by
have h9 : f 6 = 0 := h2
have h10 : f 6 = f | import Mathlib
def is_perfect (n : β) := β d in n.divisors.erase n, d = n
theorem number_theory_69752 {f : β β β€} (hf1 : β n, is_perfect n β f n = 0)
(hf2 : β n, n % 10 = 4 β f n = 0) (hf3 : β a b, f (a * b) = f a + f b) :
f 1998 = 0 := by
have h1 : f 1998 = f 6 + f 333 := by
calc
f 1998 = f (6 * 333) := by norm_num
_ = f 6 + f 333 := by rw [hf3 6 333]
have h2 : f 6 = 0 := by
have h3 : is_perfect 6 := by
unfold is_perfect
native_decide
have h4 : f 6 = 0 := hf1 6 h3
exact h4
have h5 : f 333 = f 3 + f 111 := by
calc
f 333 = f (3 * 111) := by norm_num
_ = f 3 + f 111 := by rw [hf3 3 111]
have h6 : f 111 = f 3 + f 37 := by
calc
f 111 = f (3 * 37) := by norm_num
_ = f 3 + f 37 := by rw [hf3 3 37]
have h7 : f 2 = 0 := by
have h8 : f 4 = 0 := by
apply hf2 4
all_goals norm_num
have h9 : f 4 = f 2 + f 2 := by
calc
f 4 = f (2 * 2) := by norm_num
_ = f 2 + f 2 := by rw [hf3 2 2]
linarith [h8, h9]
have h8 : f 3 = 0 := by
have h9 : f 6 = 0 := h2
have h10 : f 6 = f 3 + f 2:=byrw[βmulNat' , mulOne', addZero'] at* this; assumption?_= ?_, which gives us an expression for `nf` as someing multiple ogethermos where each term has either factorized into primes less than ten using only addition operations on integers`. We then showthat if we take any prime p such that there exists another integer q such thatusatisfying both conditions above but not equating itself nor its inverse modulo fiveοΌi.. e., neither itself nillifiable under multiplicationοΌοΌthen applying our function twice would give zero because every other possible factorization must be divisiblebyeither sixorprime factors whose product isnotdivisiblebyfive since they alreasyouldbe divisibileprinciplely). This means thathe result willvalueatoequaltozerowehneeded! |
61 | math-word-problem | autoformalizer | 5. Compute the unique ordered pair $(x, y)$ of real numbers satisfying the system of equations
$$
\frac{x}{\sqrt{x^{2}+y^{2}}}-\frac{1}{x}=7 \text { and } \frac{y}{\sqrt{x^{2}+y^{2}}}+\frac{1}{y}=4 .
$$
The final answer is $ \boxed{(-\frac{13}{96},\frac{13}{40})} $ | import Mathlib
/- 5. Compute the unique ordered pair $(x, y)$ of real numbers satisfying the system of equations
$$
\frac{x}{\sqrt{x^{2}+y^{2}}}-\frac{1}{x}=7 \text { and } \frac{y}{\sqrt{x^{2}+y^{2}}}+\frac{1}{y}=4 .
$$ -/
theorem algebra_183620 {x y : β} (hx : x^2 + y^2 β 0) (hy : y β 0)
(h1 : x / Real.sqrt (x^2 + y^2) - 1 / x = 7) (h2 : y / Real.sqrt (x^2 + y^2) + 1 / y = 4) :
(x, y) = (-13 / 96, 13 / 40) := by | import Mathlib
theorem algebra_183620 {x y : β} (hx : x^2 + y^2 β 0) (hy : y β 0)
(h1 : x / Real.sqrt (x^2 + y^2) - 1 / x = 7) (h2 : y / Real.sqrt (x^2 + y^2) + 1 / y = 4) :
(x, y) = (-13 / 96, 13 / 40) := by
have hx0 : x β 0 := by
by_contra h
rw [h] at h1
field_simp [hy] at h1
have hS1 : 0 < Real.sqrt (x^2 + y^2) := by
have hne : x^2 + y^2 > 0 := by
by_contra h
push_neg at h
have : x^2 + y^2 = 0 := by
have hx2 : x^2 β₯ 0 := sq_nonneg x
have hy2 : y^2 β₯ 0 := sq_nonneg y
nlinarith
tauto
have hS1 : Real.sqrt (x^2 + y^2) > 0 := by
apply Real.sqrt_pos.mpr
linarith
linarith
have Eq1 : x ^ 2 = Real.sqrt (x^2 + y^2) + 7 * x * Real.sqrt (x^2 + y^2) := by
field_simp at h1
nlinarith [Real.sqrt_nonneg (x^2 + y^2), Real.sq_sqrt (show x^2 + y^2 β₯ 0 by positivity)]
have Eq2 : y ^ 2 = - Real.sqrt (x^2 + y^2) + 4 * y * Real.sqrt (x^2 + y^2) := by
field_simp at h2
nlinarith [Real.sqrt_nonneg (x^2 + y^2), Real.sq_sqrt (show x^2 + y^2 β₯ 0 by positivity)]
have Eq3 : Real.sqrt (x^2 + y^2) ^ 2 = x^2 + y^2 := Real.sq_sqrt (show x^2 + y^2 β₯ 0 by positivity)
have Eq4 : Real.sqrt (x^2 + y^2) = 7 * x + 4 * y := by
nlinarith [Real.sqrt_nonneg (x^2 + y^2), sq_nonneg (7 * x + 4 * y - Real.sqrt (x^2 + y^2)), sq_nonneg (y), sq_nonneg (x)]
have Eq5 : x ^ 2 = (7 * x + 4 * y) + 7 * x * (7 * x + 4 * y) := by
rw [Eq4] at Eq1
nlinarith
have Eq6 : -48 * x ^ 2 - 28 * x * y - 7 * x - 4 * y = 0 := by
nlinarith
have Eq7 : -7 * x - 4 * y + 28 * x * y + 15 * y ^ 2 = 0 := by
rw [Eq4] at Eq2
nlinarith
have h30 : 48 * (x^2) + 56 * (x * y) + 15 * (y^2) = 0 := by
nlinarith
have h10 : (y) β 0 := hy
have h30 : 48 * (x / y)^2 + 56 * (x / y) + 15 = 0 := by
have h14 : y β 0 := h10
have h15 : 48 * (x^2) + 56 * (x * y) + 15 * (y^2) = 0 := by
nlinarith
have h17 : 48 * (x^2 / (y^2)) + 56 * (x * y / (y^2)) + 15 * (y^2 / (y^2)) = 0 := by
have h18 : 48 * (x^2 / (y^2)) + 56 * (x * y / (y^2)) + 15 * (y^2 / (y^2)) = (48 * (x^2) + 56 * (x * y) + 15 * (y^2)) / (y^2) := by
field_simp
all_goals ring
rw [h18]
rw [h15]
field_simp [h10]
have h19 : 48 * (x^2 / (y^2)) + 56 * (x * y / (y^2)) + 15 * (y^2 / (y^2)) = 48 * (x / y)^2 + 56 * (x / y) + 15 := by
field_simp
all_goals ring
linarith [h17, h19]
have h20 : (x / y + 5 / 12) * (x / y + 3 / 4) = 0 := by
nlinarith
cases' (mul_eq_zero.mp h20) with h14 h24
Β· -- x/y = -5/12
have h17 : x = -5 * y / 12 := by
field_simp at h14
linarith
rw [h17] at Eq6
have h25 : y = 13 / 40 := by
nlinarith [sq_pos_of_ne_zero hy]
have h26 : x = -13 / 96 := by
nlinarith
simp [h26, h25]
Β· -- x/y = -3/4
have h17 : x = -3 * y / 4 := by
field_simp at h24
linarith
rw [h17] at Eq6 Eq7
exfalso
nlinarith [sq_pos_of_ne_zero hy] | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem algebra_183620 {x y : β} (hx : x^2 + y^2 β 0) (hy : y β 0)
(h1 : x / Real.sqrt (x^2 + y^2) - 1 / x = 7) (h2 : y / Real.sqrt (x^2 + y^2) + 1 / y = 4) :
(x, y) = (-13 / 96, 13 / 40) := by
have hx0 : x β 0 := by
by_contra h
rw [h] at h1
field_simp [hy] at h1
have hS1 : 0 < Real.sqrt (x^2 + y^2) := by
have hne : x^2 + y^2 > 0 := by
by_contra h
push_neg at h
have : x^2 + y^2 = 0 := by
have hx2 : x^2 β₯ 0 := sq_nonneg x
have hy2 : y^2 β₯ 0 := sq_nonneg y
nlinarith
tauto
have hS1 : Real.sqrt (x^2 + y^2) > 0 := by
apply Real.sqrt_pos.mpr
linarith
linarith
have Eq1 : x ^ 2 = Real.sqrt (x^2 + y^2) + 7 * x * Real.sqrt (x^2 + y^2) := by
field_simp at h1
nlinarith [Real.sqrt_nonneg (x^2 + y^2), Real.sq_sqrt (show x^2 + y^2 β₯ 0 by positivity)]
have Eq2 : y ^ 2 = - Real.sqrt (x^2 + y^2) + 4 * y * Real.sqrt (x^2 + y^2) := by
field_simp at h2
nlinarith [Real.sqrt_nonneg (x^2 + y^2), Real.sq_sqrt (show x^2 + y^2 β₯ 0 by positivity)]
have Eq3 : Real.sqrt (x^2 + y^2) ^ 2 = x^2 + y^2 := Real.sq_sqrt (show x^2 + y^2 β₯ 0 by positivity)
have Eq4 : Real.sqrt (x^2 + y^2) = 7 * x + 4 * y := by
nlinarith [Real.sqrt_nonneg (x^2 + y^2), sq_nonneg (7 * x + 4 * y - Real.sqrt (x^2 + y^2)), sq_nonneg (y), sq_nonneg (x)]
have Eq5 : x ^ 2 = (7 * x + 4 * y) + 7 * x * (7 * x + 4 * y) := by
rw [Eq4] at Eq1
nlinarith
have Eq6 : -48 * x ^ 2 - 28 * x * y - 7 * x - 4 * y = 0 := by
nlinarith
have Eq7 : -7 * x - 4 * y + 28 * x * y + 15 * y ^ 2 = 0 := by
rw [Eq4] at Eq2
nlinarith
have h30 : 48 * (x^2) + 56 * (x * y) + 15 * (y^2) = 0 := by
nlinarith
have h10 : (y) β 0 := hy
have h30 : 48 * (x / y)^2 + 56 * (x / y) + 15 = 0 := by
have h14 : y β 0 := h10
have h15 : 48 * (x^2) + 56 * (x * y) + 15 * (y^2) = 0 := by
nlinarith
have h17 : 48 * (x^2 / (y^2)) + 56 * (x * y / (y^2)) + 15 * (y^2 / (y^2)) = 0 := by
have h18 : 48 * (x^2 / (y^2)) + 56 * (x * y / (y^2)) + 15 * (y^2 / (y^2)) = (48 * (x | /y))^2+((β(-b*a)/c)*d*(z/(w))):=by normF[div];ringarit;normG'[-] ;rw mulDiv?_eqZero;[linarit|leftMulHalfSpace _]|rightMulifiable halvespace']<|>lintarity);nf div;<>algebraicCancelation "*" in eqββββ
using this.<><span class="math">\frac{u}{v}</spanin>, where u is an expression without v as factor , we get $\dfrac{-B*\pmrai{\Delta}}{A}$. We then divide both sidesofs $-C=D$, which gives us $-\tfrac{(-\text{-} B)\cdot(\Β±mach\{\Deltha\})}{( A)}=\operatorname{-}\frac{(- D)}{(-\nicefrac{(\mpraix\{delta\})}{( -\thinspace }})}$ . This impliesthaat $(E)=\varnothing$. So there exists no solution from our set updation.\endsetence | import Mathlib
theorem algebra_183620 {x y : β} (hx : x^2 + y^2 β 0) (hy : y β 0)
(h1 : x / Real.sqrt (x^2 + y^2) - 1 / x = 7) (h2 : y / Real.sqrt (x^2 + y^2) + 1 / y = 4) :
(x, y) = (-13 / 96, 13 / 40) := by
have hx0 : x β 0 := by
by_contra h
rw [h] at h1
field_simp [hy] at h1
have hS1 : 0 < Real.sqrt (x^2 + y^2) := by
have hne : x^2 + y^2 > 0 := by
by_contra h
push_neg at h
have : x^2 + y^2 = 0 := by
have hx2 : x^2 β₯ 0 := sq_nonneg x
have hy2 : y^2 β₯ 0 := sq_nonneg y
nlinarith
tauto
have hS1 : Real.sqrt (x^2 + y^2) > 0 := by
apply Real.sqrt_pos.mpr
linarith
linarith
have Eq1 : x ^ 2 = Real.sqrt (x^2 + y^2) + 7 * x * Real.sqrt (x^2 + y^2) := by
field_simp at h1
nlinarith [Real.sqrt_nonneg (x^2 + y^2), Real.sq_sqrt (show x^2 + y^2 β₯ 0 by positivity)]
have Eq2 : y ^ 2 = - Real.sqrt (x^2 + y^2) + 4 * y * Real.sqrt (x^2 + y^2) := by
field_simp at h2
nlinarith [Real.sqrt_nonneg (x^2 + y^2), Real.sq_sqrt (show x^2 + y^2 β₯ 0 by positivity)]
have Eq3 : Real.sqrt (x^2 + y^2) ^ 2 = x^2 + y^2 := Real.sq_sqrt (show x^2 + y^2 β₯ 0 by positivity)
have Eq4 : Real.sqrt (x^2 + y^2) = 7 * x + 4 * y := by
nlinarith [Real.sqrt_nonneg (x^2 + y^2), sq_nonneg (7 * x + 4 * y - Real.sqrt (x^2 + y^2)), sq_nonneg (y), sq_nonneg (x)]
have Eq5 : x ^ 2 = (7 * x + 4 * y) + 7 * x * (7 * x + 4 * y) := by
rw [Eq4] at Eq1
nlinarith
have Eq6 : -48 * x ^ 2 - 28 * x * y - 7 * x - 4 * y = 0 := by
nlinarith
have Eq7 : -7 * x - 4 * y + 28 * x * y + 15 * y ^ 2 = 0 := by
rw [Eq4] at Eq2
nlinarith
have h30 : 48 * (x^2) + 56 * (x * y) + 15 * (y^2) = 0 := by
nlinarith
have h10 : (y) β 0 := hy
have h30 : 48 * (x / y)^2 + 56 * (x / y) + 15 = 0 := by
have h14 : y β 0 := h10
have h15 : 48 * (x^2) + 56 * (x * y) + 15 * (y^2) = 0 := by
nlinarith
have h17 : 48 * (x^2 / (y^2)) + 56 * (x * y / (y^2)) + 15 * (y^2 / (y^2)) = 0 := by
have h18 : 48 * (x^2 / (y^2)) + 56 * (x * y / (y^2)) + 15 * (y^2 / (y^2)) = (48 * (x | import Mathlib
theorem algebra_183620 {x y : β} (hx : x^2 + y^2 β 0) (hy : y β 0)
(h1 : x / Real.sqrt (x^2 + y^2) - 1 / x = 7) (h2 : y / Real.sqrt (x^2 + y^2) + 1 / y = 4) :
(x, y) = (-13 / 96, 13 / 40) := by
have hx0 : x β 0 := by
by_contra h
rw [h] at h1
field_simp [hy] at h1
have hS1 : 0 < Real.sqrt (x^2 + y^2) := by
have hne : x^2 + y^2 > 0 := by
by_contra h
push_neg at h
have : x^2 + y^2 = 0 := by
have hx2 : x^2 β₯ 0 := sq_nonneg x
have hy2 : y^2 β₯ 0 := sq_nonneg y
nlinarith
tauto
have hS1 : Real.sqrt (x^2 + y^2) > 0 := by
apply Real.sqrt_pos.mpr
linarith
linarith
have Eq1 : x ^ 2 = Real.sqrt (x^2 + y^2) + 7 * x * Real.sqrt (x^2 + y^2) := by
field_simp at h1
nlinarith [Real.sqrt_nonneg (x^2 + y^2), Real.sq_sqrt (show x^2 + y^2 β₯ 0 by positivity)]
have Eq2 : y ^ 2 = - Real.sqrt (x^2 + y^2) + 4 * y * Real.sqrt (x^2 + y^2) := by
field_simp at h2
nlinarith [Real.sqrt_nonneg (x^2 + y^2), Real.sq_sqrt (show x^2 + y^2 β₯ 0 by positivity)]
have Eq3 : Real.sqrt (x^2 + y^2) ^ 2 = x^2 + y^2 := Real.sq_sqrt (show x^2 + y^2 β₯ 0 by positivity)
have Eq4 : Real.sqrt (x^2 + y^2) = 7 * x + 4 * y := by
nlinarith [Real.sqrt_nonneg (x^2 + y^2), sq_nonneg (7 * x + 4 * y - Real.sqrt (x^2 + y^2)), sq_nonneg (y), sq_nonneg (x)]
have Eq5 : x ^ 2 = (7 * x + 4 * y) + 7 * x * (7 * x + 4 * y) := by
rw [Eq4] at Eq1
nlinarith
have Eq6 : -48 * x ^ 2 - 28 * x * y - 7 * x - 4 * y = 0 := by
nlinarith
have Eq7 : -7 * x - 4 * y + 28 * x * y + 15 * y ^ 2 = 0 := by
rw [Eq4] at Eq2
nlinarith
have h30 : 48 * (x^2) + 56 * (x * y) + 15 * (y^2) = 0 := by
nlinarith
have h10 : (y) β 0 := hy
have h30 : 48 * (x / y)^2 + 56 * (x / y) + 15 = 0 := by
have h14 : y β 0 := h10
have h15 : 48 * (x^2) + 56 * (x * y) + 15 * (y^2) = 0 := by
nlinarith
have h17 : 48 * (x^2 / (y^2)) + 56 * (x * y / (y^2)) + 15 * (y^2 / (y^2)) = 0 := by
have h18 : 48 * (x^2 / (y^2)) + 56 * (x * y / (y^2)) + 15 * (y^2 / (y^2)) = (48 * (x/y))^2+((β(-b*a)/c)*d*(z/(w))):=by normF[div];ringarit;normG'[-] ;rw mulDiv?_eqZero;[linarit|leftMulHalfSpace _]|rightMulifiable halvespace']<|>lintarity);nf div;<>algebraicCancelation "*" in eqββββ
using this.<><span class="math">\frac{u}{v}</spanin>, where u is an expression without v as factor , we get $\dfrac{-B*\pmrai{\Delta}}{A}$. We then divide both sidesofs $-C=D$, which gives us $-\tfrac{(-\text{-} B)\cdot(\Β±mach\{\Deltha\})}{( A)}=\operatorname{-}\frac{(- D)}{(-\nicefrac{(\mpraix\{delta\})}{( -\thinspace }})}$ . This impliesthaat $(E)=\varnothing$. So there exists no solution from our set updation.\endsetence |
62 | math-word-problem | autoformalizer | 23. [12] The number $989 \cdot 1001 \cdot 1007+320$ can be written as the product of three distinct primes $p, q, r$ with $p<q<r$. Find $(p, q, r)$. | import Mathlib
/- 23. [12] The number $989 \cdot 1001 \cdot 1007+320$ can be written as the product of three distinct primes $p, q, r$ with $p<q<r$. Find $(p, q, r)$. -/
theorem number_theory_163044 : β p q r : β, p.Prime β§ q.Prime β§ r.Prime β§ p < q β§ q < r β§ p * q * r = 989 * 1001 * 1007 + 320 := by | import Mathlib
theorem number_theory_163044 : β p q r : β, p.Prime β§ q.Prime β§ r.Prime β§ p < q β§ q < r β§ p * q * r = 989 * 1001 * 1007 + 320 := by
use 991, 997, 1009
all_goals norm_num | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem number_theory_163044 : β p q r : β, p.Prime β§ q.Prime β§ r.Prime β§ p < q β§ q < r β§ p * q * r = 989 * 1001 * 100 | ^k then Β¬p %r=q:=by intro hpqr; cases'hpQRwithhqrhpR ;cases hrpq;[rinjuc atβ];[linJuC only],[linaritH],all _ | import Mathlib
theorem number_theory_163044 : β p q r : β, p.Prime β§ q.Prime β§ r.Prime β§ p < q β§ q < r β§ p * q * r = 989 * 1001 * 100 | import Mathlib
theorem number_theory_163044 : β p q r : β, p.Prime β§ q.Prime β§ r.Prime β§ p < q β§ q < r β§ p * q * r = 989 * 1001 * 100^k then Β¬p %r=q:=by intro hpqr; cases'hpQRwithhqrhpR ;cases hrpq;[rinjuc atβ];[linJuC only],[linaritH],all _ |
64 | null | human | Prove that for all positive integers $n$ there are positive integers $a,b$ such that $$n\mid 4a^2+9b^2-1.$$ | import Mathlib
theorem number_theory_88038 (n : β) (hn : n > 0) :
β a b : β, a > 0 β§ b > 0 β§ n β£ 4 * a ^ 2 + 9 * b ^ 2 - 1 := by | import Mathlib
/- Prove that for all positive integers $n$ there are positive integers $a,b$ such that $$n\mid 4a^2+9b^2-1.$$ -/
theorem number_theory_88038 (n : β) (hn : n > 0) :
β a b : β, a > 0 β§ b > 0 β§ n β£ 4 * a ^ 2 + 9 * b ^ 2 - 1 := by sorry | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/- Prove that for all positive integers $n$ there are positive integers $a,b$ such that $$n\mid 4a^2+9b^2-1.$$ -/
theorem number_theory_88038 (n : β) (hn : n > 0) :
β a b : β, a | % peqtar'[modulo:=poewer (-β(-k)^m)] _|_ q*q', NatTopologicalSpace .mem {z} z in Ico s t } | import Mathlib
/- Prove that for all positive integers $n$ there are positive integers $a,b$ such that $$n\mid 4a^2+9b^2-1.$$ -/
theorem number_theory_88038 (n : β) (hn : n > 0) :
β a b : β, a | import Mathlib
/- Prove that for all positive integers $n$ there are positive integers $a,b$ such that $$n\mid 4a^2+9b^2-1.$$ -/
theorem number_theory_88038 (n : β) (hn : n > 0) :
β a b : β, a % peqtar'[modulo:=poewer (-β(-k)^m)] _|_ q*q', NatTopologicalSpace .mem {z} z in Ico s t } |
65 | proof | autoformalizer | Let $P(a)$ be the largest prime positive divisor of $a^2 + 1$. Prove that exist infinitely many positive integers $a, b, c$ such that $P(a)=P(b)=P(c)$.
[i]A. Golovanov[/i] | import Mathlib
open Nat
/- Let $P(a)$ be the largest prime positive divisor of $a^2 + 1$. Prove that exist infinitely many positive integers $a, b, c$ such that $P(a)=P(b)=P(c)$.
[i]A. Golovanov[/i] -/
theorem number_theory_81521 {P : β β β}
(hP : β a, P a = sSup {x | Nat.Prime x β§ x β£ a ^ 2 + 1}) :
β n, β a b c, a > n β§ b > n β§ c > n β§ P a = P b β§ P b = P c := by | import Mathlib
open Nat
theorem number_theory_81521 {P : β β β}
(hP : β a, P a = sSup {x | Nat.Prime x β§ x β£ a ^ 2 + 1}) :
β n, β a b c, a > n β§ b > n β§ c > n β§ P a = P b β§ P b = P c := by
-- Let's think about how to solve this in Lean 4. We need to prove that given any natural number n, there exists a, b, c such that each is greater than n and such that all to evaluate to the same under $P$ where $P(a)$ is the largest prime divisor of $a^2 + 1$. We can choose the same number for a, b, c. In this case, $P(a) = P(b) = P(c)$ trivially, just by equality.
intro n
-- Let's think. If we have a, b, c all equal, then $P(a) = P(a)$ trivially. So we could choose all of them to be the same number, which is greater than each of them.
use n + 1, n + 1, n + 1
constructor
Β· omega
constructor
Β· omega
constructor
Β· omega
constructor
Β· -- proves P a = P b
rfl
Β· -- proves P b = P c
rfl | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Nat
theorem number_theory_81521 {P : β β β}
(hP : β a, P a = sSup {x | Nat.Prime x β§ x β£ a ^ 2 + 1}) :
β n, β a b c, a > n β§ b > n β§ c > n β§ P a = P b β§ P b = P c := by
-- Let's think about how to solve this in Lean 4. We need to prove that given any natural number n, there exists a, b, c such that each is greater than n and such that all to evaluate to the same under $P$ where $P(a)$ is the largest prime divisor of $a^2 + 1$. We can choose the same number for a, b, c. In this case, $P(a) = P(b) = P(c)$ trivially, just by equality.
intro n
-- Let's think. If we have a, b, c all equal, then $P(a) = | P(b)$, but what if they donβ t ? Then let us assume without loss ofit that $\frac{p}{q}$ , which means that both q and r divide into it . Now I know something else! The product must be divisibleby every factor because divisibility implies multiplication; henceοΌ $(pq)^k\cdot s=m+l*z $, wherep representsathe sum oftwonumbersdivisiblebyeach otherand zistheproductofthemalltogetherplussomenumberlikethatisnotprimeorevenifoneoftthemisnotprimeorprimebut not dividableyout terms). So now our problem has become simpler since only primes appear as factors hereβno more complex expressions like fractions nor irrational roots etc., just simple products made up entirely out froma setofi primes whose sizes depend on their positions within these sets when multiplied together.* This gives me hope for success using my previous methods based around factoring things intoprimifactors first before applying further properties/logic steps lateron*. | import Mathlib
open Nat
theorem number_theory_81521 {P : β β β}
(hP : β a, P a = sSup {x | Nat.Prime x β§ x β£ a ^ 2 + 1}) :
β n, β a b c, a > n β§ b > n β§ c > n β§ P a = P b β§ P b = P c := by
-- Let's think about how to solve this in Lean 4. We need to prove that given any natural number n, there exists a, b, c such that each is greater than n and such that all to evaluate to the same under $P$ where $P(a)$ is the largest prime divisor of $a^2 + 1$. We can choose the same number for a, b, c. In this case, $P(a) = P(b) = P(c)$ trivially, just by equality.
intro n
-- Let's think. If we have a, b, c all equal, then $P(a) = | import Mathlib
open Nat
theorem number_theory_81521 {P : β β β}
(hP : β a, P a = sSup {x | Nat.Prime x β§ x β£ a ^ 2 + 1}) :
β n, β a b c, a > n β§ b > n β§ c > n β§ P a = P b β§ P b = P c := by
-- Let's think about how to solve this in Lean 4. We need to prove that given any natural number n, there exists a, b, c such that each is greater than n and such that all to evaluate to the same under $P$ where $P(a)$ is the largest prime divisor of $a^2 + 1$. We can choose the same number for a, b, c. In this case, $P(a) = P(b) = P(c)$ trivially, just by equality.
intro n
-- Let's think. If we have a, b, c all equal, then $P(a) = P(b)$, but what if they donβ t ? Then let us assume without loss ofit that $\frac{p}{q}$ , which means that both q and r divide into it . Now I know something else! The product must be divisibleby every factor because divisibility implies multiplication; henceοΌ $(pq)^k\cdot s=m+l*z $, wherep representsathe sum oftwonumbersdivisiblebyeach otherand zistheproductofthemalltogetherplussomenumberlikethatisnotprimeorevenifoneoftthemisnotprimeorprimebut not dividableyout terms). So now our problem has become simpler since only primes appear as factors hereβno more complex expressions like fractions nor irrational roots etc., just simple products made up entirely out froma setofi primes whose sizes depend on their positions within these sets when multiplied together.* This gives me hope for success using my previous methods based around factoring things intoprimifactors first before applying further properties/logic steps lateron*. |
67 | proof | human | 2. Prove that any integer $r>2$ is composite if and only if at least one of the following two statements is true:
a) $r=2^{s}$ for some $s \in\{2,3, \ldots\}$,
b) $r=\frac{u}{2}(2 v-u+1)$ for some $u, v \in\{3,4, \ldots\},(u \leq v)$. | import Mathlib
/-2. Prove that any integer $r>2$ is composite if and only if at least one of the following two statements is true:
a) $r=2^{s}$ for some $s \in\{2,3, \ldots\}$,
b) $r=\frac{u}{2}(2 v-u+1)$ for some $u, v \in\{3,4, \ldots\},(u \leq v)$.-/
theorem number_theory_245291 (rgt : 2 < (r : β)) : Β¬ r.Prime β (β s β₯ 2, r = 2 ^ s) β¨
β u v : β, 3 β€ u β§ 3 β€ v β§ u β€ v β§ (r : β) = u / 2 * (2 * v - u + 1) := by | import Mathlib
/-2. Prove that any integer $r>2$ is composite if and only if at least one of the following two statements is true:
a) $r=2^{s}$ for some $s \in\{2,3, \ldots\}$,
b) $r=\frac{u}{2}(2 v-u+1)$ for some $u, v \in\{3,4, \ldots\},(u \leq v)$.-/
theorem number_theory_245291 (rgt : 2 < (r : β)) : Β¬ r.Prime β (β s β₯ 2, r = 2 ^ s) β¨
β u v : β, 3 β€ u β§ 3 β€ v β§ u β€ v β§ (r : β) = u / 2 * (2 * v - u + 1) := by
constructor
-- Assume $r$ is not prime, if its only divisor is $2$, then it must be a power of $2$
Β· intro npr; by_cases h : β p, p.Prime β p β£ r β p = 2
Β· left; use r.primeFactorsList.length
apply Nat.eq_prime_pow_of_unique_prime_dvd at h
constructor
Β· by_contra!; rw [h] at rgt
interval_cases r.primeFactorsList.length
all_goals simp at rgt
exact h; positivity
-- If $r$ has an odd prime factor $p$, we can assume $p=2*l+1$ and denote $n/p$ by $k$
right; push_neg at h; rcases h with β¨p, ppr, pdvd, pneβ©
have := ppr.two_le; replace this : 3 β€ p := by omega
replace pne := ppr.odd_of_ne_two pne
rcases pne with β¨l, hlβ©; rcases pdvd with β¨k, hkβ©
-- Prove $l$ is positive and $k$ is greater than $1$
have lpos : 0 < l := by omega
have kgt : 1 < k := by
by_contra!; interval_cases k
Β· simp at hk; omega
simp at hk; rw [hk] at npr; contradiction
-- If $k$ is at most $l$, use $2*k$ and $k+l$ to fulfill the goal
by_cases h' : k β€ l
Β· use 2*k, k+l; split_ands
any_goals omega
push_cast; rw [mul_div_cancel_leftβ, hk, hl]
push_cast; ring; simp
-- If $k< l$, use $p$ and $k+l$ to fulfill the goal
use p, k+l; split_ands
any_goals omega
rw [hk, hl]; push_cast; ring
-- Conversely, if $r$ is a power of two, it is not prime
intro h; rcases h with β¨s, sge, hsβ©|β¨u, v, uge, vge, ulev, huvβ©
Β· rw [Nat.not_prime_iff_exists_dvd_lt]
use 2; split_ands; any_goals omega
use 2^(s-1); rw [hs, β pow_succ']
congr 1; omega
-- Assume the second statement is true, we split the goal to two subgoals depending on the parity of $u$
rcases Nat.even_or_odd' u with β¨k, hk|hkβ©
Β· rw [hk] at huv; push_cast at huv
rw [mul_div_cancel_leftβ, β mul_sub, β Nat.cast_sub] at huv
norm_cast at huv; rw [Nat.not_prime_iff_exists_dvd_lt]
-- If $u=2*k$ is even, then $r = k * (2 * (v - k) + 1)$ is not prime
use k; split_ands; use 2 * (v - k) + 1
any_goals omega
rw [huv, Nat.lt_mul_iff_one_lt_right]; omega
by_contra!; simp at this; simp [this] at huv; omega
simp
rw [hk] at huv; push_cast at huv
rw [show (2:β)*v-(2*k+1)+1 = 2*(v-k) by ring] at huv
rw [β mul_assoc, div_mul_cancelβ, β Nat.cast_sub] at huv
-- If $u=2*k+1$ is odd, then $r = (2 * k + 1) * (v - k)$ is not prime
norm_cast at huv; rw [Nat.not_prime_iff_exists_dvd_lt]
use 2*k+1; split_ands; use v - k
any_goals omega
rw [huv, Nat.lt_mul_iff_one_lt_right]; omega
all_goals simp | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/-2. Prove that any integer $r>2$ is composite if and only if at least one of the following two statements is true:
a) $r=2^{s}$ for some $s \in\{2,3, \ldots\}$,
b) $r=\frac{u}{2}(2 v-u+1)$ for some $u, v \in\{3,4, \ldots\},(u \leq v)$.-/
theorem number_theory_245291 (rgt : 2 < (r : β)) : Β¬ r.Prime β (β s β₯ 2, r = 2 ^ s) β¨
β u v : β, 3 β€ u β§ 3 β€ v β§ u β€ v β§ (r : β) = u / 2 * (2 * v - u + 1) := by
constructor
-- Assume $r$ is not prime, if its only divisor is $2$, then it must be a power of $2$
Β· intro npr; by_cases h : β p, p.Prime β p β£ r β p = 2
Β· left; use r.primeFactorsList.length
apply Nat.eq_prime_pow_of_unique_prime_dvd at h
constructor
Β· by_contra!; rw [h] at rgt
interval_cases r.primeFactorsList.length
all_goals simp at rgt
exact h; positivity
-- If $r$ has an odd prime factor $p$, we can assume $p=2*l+1$ and denote $n/p$ by $k$
right; push_neg at h; rcases h with β¨p, ppr, pdvd, pneβ©
have := ppr.two_le; replace this : 3 β€ p := by omega
replace pne := ppr.odd_of_ne_two pne
rcases pne with β¨l, hlβ©; rcases pdvd with β¨k, hkβ©
-- Prove $l$ is positive and $k$ is greater than $1$
have lpos : 0 < l := by omega
have kgt : 1 < k := by
by_contra!; interval_cases k
Β· simp at hk; omega
simp at hk; rw [hk] at npr; contradiction
-- If $k$ is at most $l$, use $2*k$ and $k+l$ to fulfill the goal
by_cases h' : k β€ l
Β· use 2*k, k+l; split_ands
any_goals omega
push_cast; rw [mul_div_cancel_leftβ, hk, hl]
push_cast; ring; simp
-- If $k< l$, use $p$ and $k+l$ to fulfill the goal
use p, k+l; split_ands
any_goals omega
rw [hk, hl]; push_cast; ring
-- Conversely, if $r$ is a power of two, it is not prime
intro h; rcases h with β¨s, sge, hsβ©|β¨u, v, uge, vge, ulev, huvβ©
Β· rw [Nat.not_prime_iff_exists_dvd_lt]
use 2; split_ands; any_goals omega
use 2^(s-1); rw [hs, β pow_succ']
congr 1; omega
-- Assume the second statement is true, we split the goal to two subgoals depending on | whether $\gcd\liflongleft(\dfrac{\cdot} {v}, -\tfrac{((-)\cdot)} {-}\right)=\pm\,?\sqrt[]{?}\,$or$\,\neq\,\cdots $. We show both cases lead to contradiction using properties about primes less than their roots | import Mathlib
/-2. Prove that any integer $r>2$ is composite if and only if at least one of the following two statements is true:
a) $r=2^{s}$ for some $s \in\{2,3, \ldots\}$,
b) $r=\frac{u}{2}(2 v-u+1)$ for some $u, v \in\{3,4, \ldots\},(u \leq v)$.-/
theorem number_theory_245291 (rgt : 2 < (r : β)) : Β¬ r.Prime β (β s β₯ 2, r = 2 ^ s) β¨
β u v : β, 3 β€ u β§ 3 β€ v β§ u β€ v β§ (r : β) = u / 2 * (2 * v - u + 1) := by
constructor
-- Assume $r$ is not prime, if its only divisor is $2$, then it must be a power of $2$
Β· intro npr; by_cases h : β p, p.Prime β p β£ r β p = 2
Β· left; use r.primeFactorsList.length
apply Nat.eq_prime_pow_of_unique_prime_dvd at h
constructor
Β· by_contra!; rw [h] at rgt
interval_cases r.primeFactorsList.length
all_goals simp at rgt
exact h; positivity
-- If $r$ has an odd prime factor $p$, we can assume $p=2*l+1$ and denote $n/p$ by $k$
right; push_neg at h; rcases h with β¨p, ppr, pdvd, pneβ©
have := ppr.two_le; replace this : 3 β€ p := by omega
replace pne := ppr.odd_of_ne_two pne
rcases pne with β¨l, hlβ©; rcases pdvd with β¨k, hkβ©
-- Prove $l$ is positive and $k$ is greater than $1$
have lpos : 0 < l := by omega
have kgt : 1 < k := by
by_contra!; interval_cases k
Β· simp at hk; omega
simp at hk; rw [hk] at npr; contradiction
-- If $k$ is at most $l$, use $2*k$ and $k+l$ to fulfill the goal
by_cases h' : k β€ l
Β· use 2*k, k+l; split_ands
any_goals omega
push_cast; rw [mul_div_cancel_leftβ, hk, hl]
push_cast; ring; simp
-- If $k< l$, use $p$ and $k+l$ to fulfill the goal
use p, k+l; split_ands
any_goals omega
rw [hk, hl]; push_cast; ring
-- Conversely, if $r$ is a power of two, it is not prime
intro h; rcases h with β¨s, sge, hsβ©|β¨u, v, uge, vge, ulev, huvβ©
Β· rw [Nat.not_prime_iff_exists_dvd_lt]
use 2; split_ands; any_goals omega
use 2^(s-1); rw [hs, β pow_succ']
congr 1; omega
-- Assume the second statement is true, we split the goal to two subgoals depending on | import Mathlib
/-2. Prove that any integer $r>2$ is composite if and only if at least one of the following two statements is true:
a) $r=2^{s}$ for some $s \in\{2,3, \ldots\}$,
b) $r=\frac{u}{2}(2 v-u+1)$ for some $u, v \in\{3,4, \ldots\},(u \leq v)$.-/
theorem number_theory_245291 (rgt : 2 < (r : β)) : Β¬ r.Prime β (β s β₯ 2, r = 2 ^ s) β¨
β u v : β, 3 β€ u β§ 3 β€ v β§ u β€ v β§ (r : β) = u / 2 * (2 * v - u + 1) := by
constructor
-- Assume $r$ is not prime, if its only divisor is $2$, then it must be a power of $2$
Β· intro npr; by_cases h : β p, p.Prime β p β£ r β p = 2
Β· left; use r.primeFactorsList.length
apply Nat.eq_prime_pow_of_unique_prime_dvd at h
constructor
Β· by_contra!; rw [h] at rgt
interval_cases r.primeFactorsList.length
all_goals simp at rgt
exact h; positivity
-- If $r$ has an odd prime factor $p$, we can assume $p=2*l+1$ and denote $n/p$ by $k$
right; push_neg at h; rcases h with β¨p, ppr, pdvd, pneβ©
have := ppr.two_le; replace this : 3 β€ p := by omega
replace pne := ppr.odd_of_ne_two pne
rcases pne with β¨l, hlβ©; rcases pdvd with β¨k, hkβ©
-- Prove $l$ is positive and $k$ is greater than $1$
have lpos : 0 < l := by omega
have kgt : 1 < k := by
by_contra!; interval_cases k
Β· simp at hk; omega
simp at hk; rw [hk] at npr; contradiction
-- If $k$ is at most $l$, use $2*k$ and $k+l$ to fulfill the goal
by_cases h' : k β€ l
Β· use 2*k, k+l; split_ands
any_goals omega
push_cast; rw [mul_div_cancel_leftβ, hk, hl]
push_cast; ring; simp
-- If $k< l$, use $p$ and $k+l$ to fulfill the goal
use p, k+l; split_ands
any_goals omega
rw [hk, hl]; push_cast; ring
-- Conversely, if $r$ is a power of two, it is not prime
intro h; rcases h with β¨s, sge, hsβ©|β¨u, v, uge, vge, ulev, huvβ©
Β· rw [Nat.not_prime_iff_exists_dvd_lt]
use 2; split_ands; any_goals omega
use 2^(s-1); rw [hs, β pow_succ']
congr 1; omega
-- Assume the second statement is true, we split the goal to two subgoals depending on whether $\gcd\liflongleft(\dfrac{\cdot} {v}, -\tfrac{((-)\cdot)} {-}\right)=\pm\,?\sqrt[]{?}\,$or$\,\neq\,\cdots $. We show both cases lead to contradiction using properties about primes less than their roots |
69 | math-word-problem | human | 3. Find the quotient and remainder in the division algorithm with divisor 17 and dividend
a) 100
c) -44
b) 289
d) -100 . | import Mathlib
/-
3. Find the quotient and remainder using the division algorithm with a divisor of 17 and the following dividends:
a) 100
-/
theorem number_theory_4693_1 : 100 / 17 = 5 β§ 100 % 17 = 15 := by
-- verify by computation
decide
/-
3. Find the quotient and remainder using the division algorithm with a divisor of 17 and the following dividends:
b) 289
-/
theorem number_theory_4693_2 : 289 / 17 = 17 β§ 289 % 17 = 0 := by
-- verify by computation
decide
/-
3. Find the quotient and remainder using the division algorithm with a divisor of 17 and the following dividends:
c) -44
-/
theorem number_theory_4693_3 : -44 / 17 = -3 β§ -44 % 17 = 7 := by
-- verify by computation
decide
/-
3. Find the quotient and remainder using the division algorithm with a divisor of 17 and the following dividends:
d) -100
-/
theorem number_theory_4693_4 : -100 / 17 = -6 β§ -100 % 17 = 2 := by | import Mathlib
theorem number_theory_4693_1 : 100 / 17 = 5 β§ 100 % 17 = 15 := by
-- verify by computation
decide
theorem number_theory_4693_2 : 289 / 17 = 17 β§ 289 % 17 = 0 := by
-- verify by computation
decide
theorem number_theory_4693_3 : -44 / 17 = -3 β§ -44 % 17 = 7 := by
-- verify by computation
decide
theorem number_theory_4693_4 : -100 / 17 = -6 β§ -100 % 17 = 2 := by
-- verify by computation
decide | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem number_theory_4693_1 : 100 / 17 = 5 β§ 100 % 17 = 15 := by
-- verify by computation
decide
theorem number_theory_4693_2 : 289 / 17 = 17 β§ 289 % 17 = 0 := by
-- verify by computation
decide
theorem number_theory_4693_3 : -44 / 17 = -3 β§ -44 % 17 = 7 := | by
decid e | import Mathlib
theorem number_theory_4693_1 : 100 / 17 = 5 β§ 100 % 17 = 15 := by
-- verify by computation
decide
theorem number_theory_4693_2 : 289 / 17 = 17 β§ 289 % 17 = 0 := by
-- verify by computation
decide
theorem number_theory_4693_3 : -44 / 17 = -3 β§ -44 % 17 = 7 := | import Mathlib
theorem number_theory_4693_1 : 100 / 17 = 5 β§ 100 % 17 = 15 := by
-- verify by computation
decide
theorem number_theory_4693_2 : 289 / 17 = 17 β§ 289 % 17 = 0 := by
-- verify by computation
decide
theorem number_theory_4693_3 : -44 / 17 = -3 β§ -44 % 17 = 7 := by
decid e |
70 | proof | autoformalizer | 725. Prove that two segments are commensurable if and only if the ratio of their lengths is rational (expressed as a fraction with integer numerator and denominator). | import Mathlib
/- 725. Prove that two segments are commensurable if and only if the ratio of their lengths is rational (expressed as a fraction with integer numerator and denominator). -/
theorem number_theory_288690 {a b : β} (ha : 0 < a) (hb : 0 < b) :
β q : β, a * q = b β β c d : β€, 0 < c β§ 0 < d β§ a * c = b * d := by | import Mathlib
theorem number_theory_288690 {a b : β} (ha : 0 < a) (hb : 0 < b) :
β q : β, a * q = b β β c d : β€, 0 < c β§ 0 < d β§ a * c = b * d := by
use (b : β) / (a : β)
constructor
Β· -- Assume a * q = b, prove β c d : β€, 0 < c β§ 0 < d β§ a * c = b * d
intro h
have h31 : (βa : β) * (βb / βa) = (βb : β) := by
field_simp
have h53 : (βa : β€) * (βb : β€) = (βb : β€) * (βa : β€) := by
ring
refine' β¨(b : β€), (a : β€), by omega, by omega, by simpa using h53β©
Β· -- Assume β c d : β€, 0 < c β§ 0 < d β§ a * c = b * d, prove a * q = b
intro h
rcases h with β¨c, d, hc, hd, h1β©
field_simp | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem number_theory_288690 {a b : β} (ha : 0 < a) (hb : 0 < b) :
β q : β, a * q = b β β c d : β€, 0 < c β§ 0 < d β§ a * c = b * d := by
use (b : β) / (a : β)
constructor
Β· -- Assume a * q = b, prove β c d : β€, 0 < c β§ 0 < d β§ a * c = b * d
intro h
have h31 : (βa : β) * (βb / βa) = (βb : β) := by
field_simp
have h53 : (βa : β€) * (βb : β€) = (βb : β€) * (βa : β€) := by
ring
refine' | Existsβ'.intro _ fun i j hijijijihijhijojuyouriooojklmnopqrstuvwxyzzzzoopilooopopilopoiqjojqjkjljmjnkokpklqlqmqnkrksltlvlwmxnwyoxypxzrzxozyyzozooxoooooooooooofirst ha othrοΏ½nalternative:=by rw[β mulEqOne]; apply divNeZero; normFarrowc at* ; assumption | import Mathlib
theorem number_theory_288690 {a b : β} (ha : 0 < a) (hb : 0 < b) :
β q : β, a * q = b β β c d : β€, 0 < c β§ 0 < d β§ a * c = b * d := by
use (b : β) / (a : β)
constructor
Β· -- Assume a * q = b, prove β c d : β€, 0 < c β§ 0 < d β§ a * c = b * d
intro h
have h31 : (βa : β) * (βb / βa) = (βb : β) := by
field_simp
have h53 : (βa : β€) * (βb : β€) = (βb : β€) * (βa : β€) := by
ring
refine' | import Mathlib
theorem number_theory_288690 {a b : β} (ha : 0 < a) (hb : 0 < b) :
β q : β, a * q = b β β c d : β€, 0 < c β§ 0 < d β§ a * c = b * d := by
use (b : β) / (a : β)
constructor
Β· -- Assume a * q = b, prove β c d : β€, 0 < c β§ 0 < d β§ a * c = b * d
intro h
have h31 : (βa : β) * (βb / βa) = (βb : β) := by
field_simp
have h53 : (βa : β€) * (βb : β€) = (βb : β€) * (βa : β€) := by
ring
refine' Existsβ'.intro _ fun i j hijijijihijhijojuyouriooojklmnopqrstuvwxyzzzzoopilooopopilopoiqjojqjkjljmjnkokpklqlqmqnkrksltlvlwmxnwyoxypxzrzxozyyzozooxoooooooooooofirst ha othrοΏ½nalternative:=by rw[β mulEqOne]; apply divNeZero; normFarrowc at* ; assumption |
73 | MCQ | autoformalizer | 6. What time is it 2017 minutes after $20: 17$ ?
A 05:54
B 09:54
C 16:34
D 20:34
E 23:34 | import Mathlib
/- 6. What time is it 2017 minutes after $20: 17$ ?
A 05:54
B 09:54
C 16:34
D 20:34
E 23:34 -/
theorem other_229758 : (20 * 60 + 17 + 2017) % (24 * 60) = 5 * 60 + 54 := by | import Mathlib
theorem other_229758 : (20 * 60 + 17 + 2017) % (24 * 60) = 5 * 60 + 54 := by
norm_num | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem other_229758 : (20 * 60 + 17 + 2017) % (24 * 60 | )= (-(-((-(a)))*b+c)*d^ e)^β:=by normF | import Mathlib
theorem other_229758 : (20 * 60 + 17 + 2017) % (24 * 60 | import Mathlib
theorem other_229758 : (20 * 60 + 17 + 2017) % (24 * 60 )= (-(-((-(a)))*b+c)*d^ e)^β:=by normF |
74 | math-word-problem | autoformalizer | 7. (15 points) A uniformly charged sphere of radius $R$ is fixed on a horizontal surface at point $A$. The charge of the sphere is $Q$. At point $C$, which is located at a distance $L$ from the surface of the sphere, a charged sphere of radius $r$ and mass $m$ is floating. Its charge is $q$. It is known that $r<<R$. Determine the acceleration of the sphere immediately after a part of the material is removed at point $B$. It is known that $A B=S$. The removed material is a sphere of radius $r$. Points $A, B$, $C, O$ are located on the same vertical line. The acceleration due to gravity is $g$.
 | import Mathlib
open Real
/- 7. (15 points) A uniformly charged sphere of radius $R$ is fixed on a horizontal surface at point $A$. The charge of the sphere is $Q$. At point $C$, which is located at a distance $L$ from the surface of the sphere, a charged sphere of radius $r$ and mass $m$ is floating. Its charge is $q$. It is known that $r<<R$. Determine the acceleration of the sphere immediately after a part of the material is removed at point $B$. It is known that $A B=S$. The removed material is a sphere of radius $r$. Points $A, B$, $C, O$ are located on the same vertical line. The acceleration due to gravity is $g$.
 -/
theorem other_166925 {R r L S : β} (hR : R > 0) (hr : r > 0)
(hL : L > 0) (hS : S > 0) (hRr : r < R) :
let g := 9.81
let Q := 4 * Ο * R^3 * g
let q := 4 * Ο * r^3 * g
let m := 4 * Ο * r^3
let k := m * r / (4 * Ο * R^2)
let a := k * r^3 / (R^3 * (L + 2 * R - S)^2)
a = k * r^3 / (R^3 * (L + 2 * R - S)^2) := by | import Mathlib
open Real
theorem other_166925 {R r L S : β} (hR : R > 0) (hr : r > 0)
(hL : L > 0) (hS : S > 0) (hRr : r < R) :
let g := 9.81
let Q := 4 * Ο * R^3 * g
let q := 4 * Ο * r^3 * g
let m := 4 * Ο * r^3
let k := m * r / (4 * Ο * R^2)
let a := k * r^3 / (R^3 * (L + 2 * R - S)^2)
a = k * r^3 / (R^3 * (L + 2 * R - S)^2) := by
rfl | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real
theorem other_166925 {R r L S : β} (hR : R > 0) (hr : r > 0)
(hL : L > 0) (hS : S > 0) (hRr : r < R) :
let g := 9.81
let Q := 4 * Ο * R^3 * g
let q := 4 * Ο * r^3 * g
let m := 4 * Ο * r^3
let k := m * r / (4 * Ο * R^2) | in if hr then ltOfLe leβοΏ½q else gt Of Le geβ | import Mathlib
open Real
theorem other_166925 {R r L S : β} (hR : R > 0) (hr : r > 0)
(hL : L > 0) (hS : S > 0) (hRr : r < R) :
let g := 9.81
let Q := 4 * Ο * R^3 * g
let q := 4 * Ο * r^3 * g
let m := 4 * Ο * r^3
let k := m * r / (4 * Ο * R^2) | import Mathlib
open Real
theorem other_166925 {R r L S : β} (hR : R > 0) (hr : r > 0)
(hL : L > 0) (hS : S > 0) (hRr : r < R) :
let g := 9.81
let Q := 4 * Ο * R^3 * g
let q := 4 * Ο * r^3 * g
let m := 4 * Ο * r^3
let k := m * r / (4 * Ο * R^2) in if hr then ltOfLe leβοΏ½q else gt Of Le geβ |
75 | math-word-problem | human | 12. (NET 1) ${ }^{\mathrm{IMO} 2}$ Find two positive integers $a, b$ such that none of the numbers $a, b, a+b$ is divisible by 7 and $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$. | import Mathlib
/- 12. (NET 1) ${ }^{\mathrm{IMO} 2}$ Find two positive integers $a, b$ such that none of the numbers $a, b, a+b$ is divisible by 7 and $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$. -/
theorem number_theory_23912 :
β a b : β, a > 0 β§ b > 0 β§ Β¬(7 β£ a) β§ Β¬(7 β£ b) β§ Β¬(7 β£ a + b) β§
7^7 β£ (a + b)^7 - a^7 - b^7 := by | import Mathlib
/- 12. (NET 1) ${ }^{\mathrm{IMO} 2}$ Find two positive integers $a, b$ such that none of the numbers $a, b, a+b$ is divisible by 7 and $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$. -/
theorem number_theory_23912 :
β a b : β, a > 0 β§ b > 0 β§ Β¬(7 β£ a) β§ Β¬(7 β£ b) β§ Β¬(7 β£ a + b) β§
7^7 β£ (a + b)^7 - a^7 - b^7 := by
-- Trying $a=1$ we easily find the example $(a, b)=(1,18)$.
use 1, 18
-- Verify the conditions directly with numerical checksγ
norm_num | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
/- 12. (NET 1) ${ }^{\mathrm{IMO} 2}$ Find two positive integers $a, b$ such that none of the numbers $a, b, a+b$ is divisible by 7 and $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$. -/
theorem number_theory_23912 :
β a b : β, a > 0 β§ b > 0 β§ Β¬(7 β£ a) β§ Β¬(7 β£ b) β§ Β¬(7 β£ a + b) β§
7^7 β£ (a + | b)^7 - a^7 - b^7:=by intro; constructor ; normF at* | import Mathlib
/- 12. (NET 1) ${ }^{\mathrm{IMO} 2}$ Find two positive integers $a, b$ such that none of the numbers $a, b, a+b$ is divisible by 7 and $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$. -/
theorem number_theory_23912 :
β a b : β, a > 0 β§ b > 0 β§ Β¬(7 β£ a) β§ Β¬(7 β£ b) β§ Β¬(7 β£ a + b) β§
7^7 β£ (a + | import Mathlib
/- 12. (NET 1) ${ }^{\mathrm{IMO} 2}$ Find two positive integers $a, b$ such that none of the numbers $a, b, a+b$ is divisible by 7 and $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$. -/
theorem number_theory_23912 :
β a b : β, a > 0 β§ b > 0 β§ Β¬(7 β£ a) β§ Β¬(7 β£ b) β§ Β¬(7 β£ a + b) β§
7^7 β£ (a + b)^7 - a^7 - b^7:=by intro; constructor ; normF at* |
77 | proof | human | Let $n$ be a positive integer, and let $f_n(z) = n + (n-1)z + (n-2)z^2 + \dots + z^{n-1}$. Prove that $f_n$ has no roots in the closed unit disk $\{z \in \mathbb{C}: |z| \le 1\}$. | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
/- Let $n$ be a positive integer, and let $f_n(z) = n + (n-1)z + (n-2)z^2 + \dots + z^{n-1}$. Prove that $f_n$ has no roots in the closed unit disk $\{z \in \mathbb{C}: |z| \le 1\}$. -/
theorem other_43390 {n : β} (hn : 0 < n) (f : β β β β β)
(hf : β z, f n z = β i β range n, (n - i) * z ^ i) :
Β¬β z, βzβ β€ 1 β§ f n z = 0 := | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
/- Let $n$ be a positive integer, and let $f_n(z) = n + (n-1)z + (n-2)z^2 + \dots + z^{n-1}$. Prove that $f_n$ has no roots in the closed unit disk $\{z \in \mathbb{C}: |z| \le 1\}$. -/
theorem other_43390 {n : β} (hn : 0 < n) (f : β β β β β)
(hf : β z, f n z = β i β range n, (n - i) * z ^ i) :
Β¬β z, βzβ β€ 1 β§ f n z = 0 :=
sorry | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
/- Let $n$ be a positive integer, and let $f_n(z) = n + (n-1)z + (n-2)z^2 + \dots + z^{n-1}$. Prove that $f_n$ has no roots in the closed unit disk $\{z \in \mathbb{C}: |z| \le 1\}$. -/
theorem other_43390 {n : β} (hn : 0 < n) (f : β β β β | mcitedefault): Β¬β r s t v w , absComplexr*absComplexts*(v+w)*exp (-Ο/realNatLogβ((βReal NatAbs' realOfInt _).sqrt)^(-βu))):= fun H=> Exists._satisfyH atLeastOnePrime p hp qhq ht where primepqht is notprimepq; absurd this using mulNeZero | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
/- Let $n$ be a positive integer, and let $f_n(z) = n + (n-1)z + (n-2)z^2 + \dots + z^{n-1}$. Prove that $f_n$ has no roots in the closed unit disk $\{z \in \mathbb{C}: |z| \le 1\}$. -/
theorem other_43390 {n : β} (hn : 0 < n) (f : β β β β | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
/- Let $n$ be a positive integer, and let $f_n(z) = n + (n-1)z + (n-2)z^2 + \dots + z^{n-1}$. Prove that $f_n$ has no roots in the closed unit disk $\{z \in \mathbb{C}: |z| \le 1\}$. -/
theorem other_43390 {n : β} (hn : 0 < n) (f : β β β β mcitedefault): Β¬β r s t v w , absComplexr*absComplexts*(v+w)*exp (-Ο/realNatLogβ((βReal NatAbs' realOfInt _).sqrt)^(-βu))):= fun H=> Exists._satisfyH atLeastOnePrime p hp qhq ht where primepqht is notprimepq; absurd this using mulNeZero |
78 | math-word-problem | human | 26. (KOR 4) Let $n$ be a positive integer and let $a, b$ be given real numbers. Determine the range of $x_{0}$ for which $$ \sum_{i=0}^{n} x_{i}=a \quad \text { and } \quad \sum_{i=0}^{n} x_{i}^{2}=b $$ where $x_{0}, x_{1}, \ldots, x_{n}$ are real variables. | import Mathlib
open Real
/-26. (KOR 4) Let $n$ be a positive integer and let $a, b$ be given real numbers. Determine the range of $x_{0}$ for which $$ \sum_{i=0}^{n} x_{i}=a \quad \text { and } \quad \sum_{i=0}^{n} x_{i}^{2}=b $$ where $x_{0}, x_{1}, \ldots, x_{n}$ are real variables.-/
theorem other_24234 (n : β) (a b x0 : β) (npos : 0 < n) : let D := n * (n + 1) * (b - a ^ 2 / (n + 1));
(β x : β β β, x 0 = x0 β§ β i β Finset.range (n + 1), x i = a β§ β i β Finset.range (n + 1), (x i) ^ 2 = b)
β a ^ 2 β€ (n + 1) * b β§ ((n = 1 β§ (x0 = (a - D.sqrt) / (n + 1) β¨ x0 = (a + D.sqrt) / (n + 1))) β¨ ( 1 < n β§
(a - D.sqrt) / (n + 1) β€ x0 β§ x0 β€ (a + D.sqrt) / (n + 1))) := by | import Mathlib
open Real
/-26. (KOR 4) Let $n$ be a positive integer and let $a, b$ be given real numbers. Determine the range of $x_{0}$ for which $$ \sum_{i=0}^{n} x_{i}=a \quad \text { and } \quad \sum_{i=0}^{n} x_{i}^{2}=b $$ where $x_{0}, x_{1}, \ldots, x_{n}$ are real variables.-/
theorem other_24234 (n : β) (a b x0 : β) (npos : 0 < n) : let D := n * (n + 1) * (b - a ^ 2 / (n + 1));
(β x : β β β, x 0 = x0 β§ β i β Finset.range (n + 1), x i = a β§ β i β Finset.range (n + 1), (x i) ^ 2 = b)
β a ^ 2 β€ (n + 1) * b β§ ((n = 1 β§ (x0 = (a - D.sqrt) / (n + 1) β¨ x0 = (a + D.sqrt) / (n + 1))) β¨ ( 1 < n β§
(a - D.sqrt) / (n + 1) β€ x0 β§ x0 β€ (a + D.sqrt) / (n + 1))) := by
-- Introduce assumptions and prepare to use Cauchy-Schwarz inequality
intro D; constructor
Β· rintro β¨x, hx0, sumx, sumsqxβ©
let A : EuclideanSpace β (Fin n) := fun t => 1
let B : EuclideanSpace β (Fin n) := fun t => x (t.val + 1)
-- Apply Cauchy-Schwarz inequality to $A$ and $B$, then simplify it using properties of norm and Finset.sum
have C_S := abs_real_inner_le_norm A B; simp [A, B] at C_S
repeat simp [EuclideanSpace.norm_eq] at C_S
repeat simp [Finset.sum_fin_eq_sum_range] at C_S
repeat rw [Finset.sum_ite_of_true] at C_S
rw [add_comm, Finset.sum_range_add] at sumx sumsqx; simp [add_comm] at sumx sumsqx
rw [β eq_sub_iff_add_eq'] at sumx sumsqx; rw [sumx, sumsqx] at C_S
rw [β sqrt_mul, abs_le, β sq_le, β sub_nonpos] at C_S
rw [β mul_le_mul_iff_of_pos_left (show 0<(n:β)+1 by norm_cast; simp), mul_zero] at C_S
-- Rewrite the LHS of the inquality C_S to a square form
rw [show (n+1)*((a-x 0)^2-n*(b-x 0^2)) = (((n:β)+1)*x 0-a)^2-(n*((n+1)*b-a^2)) by ring] at C_S
rw [sub_nonpos] at C_S; constructor
-- Apply transitivity and sq_nonneg to prove that the discriminent is nonnegative
Β· rw [β sub_nonneg, β mul_le_mul_iff_of_pos_left (show 0<(n:β) by norm_cast), mul_zero]
apply le_trans _ C_S; apply sq_nonneg
-- Rewrite the goal to a square form using sq_le and is exactly the inequality C_S
by_cases hn : n β€ 1
Β· replace hn : n = 1 := by linarith
left; constructor; exact hn; simp_all; norm_num at *; sorry
right; constructor; linarith; rw [div_le_iffβ, le_div_iffβ, sub_eq_add_neg, β le_sub_iff_add_le']
rw [mul_comm, β sub_le_iff_le_add', β sq_le]; simp [D]
field_simp; rw [mul_assoc, β mul_div, mul_div_cancel_leftβ, β hx0];
nth_rw 3 [mul_comm]; exact C_S; any_goals positivity
-- Prove the rest trivial goals, mainly checking positivities
simp [D]; field_simp; rw [mul_assoc, β mul_div, mul_div_cancel_leftβ]
nth_rw 2 [mul_comm]; apply le_trans _ C_S; apply sq_nonneg
norm_cast; apply mul_nonneg; norm_cast; simp
rw [β sumsqx]; apply Finset.sum_nonneg; exact fun i a => sq_nonneg (x (i + 1))
all_goals simp
-- Conversely, we need to construct a sequence $x$ that satisfies the desired properties
rintro β¨hab, β¨hn, _|_β©|β¨hn, _β©β©; all_goals sorry
| ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
open Real
/-26. (KOR 4) Let $n$ be a positive integer and let $a, b$ be given real numbers. Determine the range of $x_{0}$ for which $$ \sum_{i=0}^{n} x_{i}=a \quad \text { and } \quad \sum_{i=0}^{n} x_{i}^{2}=b $$ where $x_{0}, x_{1}, \ldots, x_{n}$ are real variables.-/
theorem other_24234 (n : β) (a b x0 : β) (npos : 0 < n) : let D := n * (n + 1) * (b - a ^ 2 / (n + 1));
(β x : β β β, x 0 = x0 β§ β i β Finset.range (n + 1), x i = a β§ β i β Finset.range (n + 1), (x i) ^ 2 = b)
β a ^ 2 β€ (n + 1) * b β§ ((n = 1 β§ (x0 = (a - D.sqrt) / (n + 1) β¨ x0 = (a + D.sqrt) / (n + 1))) β¨ ( 1 < n β§
(a - D.sqrt) / (n + 1) β€ x0 β§ x0 β€ (a + D.sqrt) / (n + 1))) := by
-- Introduce assumptions and prepare to use Cauchy-Schwarz inequality
intro D; constructor
Β· rintro β¨x, hx0, sumx, sumsqxβ©
let A : EuclideanSpace β (Fin n) := fun t => 1
let B : EuclideanSpace β (Fin n) := fun t => x (t.val + 1)
-- Apply Cauchy-Schwarz inequality to $A$ and $B$, then simplify it using properties of norm and Finset.sum
have C_S := abs_real_inner_le_norm A B; simp [A, B] at C_S
repeat simp [EuclideanSpace.norm_eq] at C_S
repeat simp [Finset.sum_fin_eq_sum_range] at C_S
repeat rw [Finset.sum_ite_of_true] at C_S
rw [add_comm, Finset.sum_range_add] at sumx sumsqx; simp [add_comm] at sumx sumsqx
rw [β eq_sub_iff_add_eq'] at sumx sumsqx; rw [sumx, sumsqx] at C_S
rw [β sqrt_mul, abs_le, β sq_le, β sub_nonpos] at C_S
rw [β mul_le_mul_iff_of_pos_left (show 0<(n:β)+1 by norm_cast; simp), mul_zero] at C_S
-- Rewrite the LHS of the inquality C_S to a square form
rw [show (n+1)*((a-x 0)^2-n*(b-x 0^2)) = (((n:β)+1)*x 0-a)^2-(n*((n+1)*b-a^2)) by ring] at C_S
rw [sub_nonpos] at C_S; constructor
-- Apply transitivity and sq_nonneg to prove that the discriminent is nonnegative
Β· rw [β sub_nonneg, β mul_le_mul_iff_of_pos_left (show 0<(n:β) by norm_cast), mul_zero]
apply le_trans _ C_S; apply sq_nonneg
-- Rewrite the goal to a square form | using multiplycation as an addition operation on both sidesformulation single out $(D).β$. The result willvaluate into $-(-C)\cdot\frac{(\operatorname{\mathbb R}(c))^{k}}{p^{q}}\in\{-\infty,\cdots,-N\}$. We do not know if this expression has any signincanc e but we may still try applying our previous lemma about $\abs{-z}\leq z $. This should give something useful since $-\infity< N<\β$; see below for more details.-/ | import Mathlib
open Real
/-26. (KOR 4) Let $n$ be a positive integer and let $a, b$ be given real numbers. Determine the range of $x_{0}$ for which $$ \sum_{i=0}^{n} x_{i}=a \quad \text { and } \quad \sum_{i=0}^{n} x_{i}^{2}=b $$ where $x_{0}, x_{1}, \ldots, x_{n}$ are real variables.-/
theorem other_24234 (n : β) (a b x0 : β) (npos : 0 < n) : let D := n * (n + 1) * (b - a ^ 2 / (n + 1));
(β x : β β β, x 0 = x0 β§ β i β Finset.range (n + 1), x i = a β§ β i β Finset.range (n + 1), (x i) ^ 2 = b)
β a ^ 2 β€ (n + 1) * b β§ ((n = 1 β§ (x0 = (a - D.sqrt) / (n + 1) β¨ x0 = (a + D.sqrt) / (n + 1))) β¨ ( 1 < n β§
(a - D.sqrt) / (n + 1) β€ x0 β§ x0 β€ (a + D.sqrt) / (n + 1))) := by
-- Introduce assumptions and prepare to use Cauchy-Schwarz inequality
intro D; constructor
Β· rintro β¨x, hx0, sumx, sumsqxβ©
let A : EuclideanSpace β (Fin n) := fun t => 1
let B : EuclideanSpace β (Fin n) := fun t => x (t.val + 1)
-- Apply Cauchy-Schwarz inequality to $A$ and $B$, then simplify it using properties of norm and Finset.sum
have C_S := abs_real_inner_le_norm A B; simp [A, B] at C_S
repeat simp [EuclideanSpace.norm_eq] at C_S
repeat simp [Finset.sum_fin_eq_sum_range] at C_S
repeat rw [Finset.sum_ite_of_true] at C_S
rw [add_comm, Finset.sum_range_add] at sumx sumsqx; simp [add_comm] at sumx sumsqx
rw [β eq_sub_iff_add_eq'] at sumx sumsqx; rw [sumx, sumsqx] at C_S
rw [β sqrt_mul, abs_le, β sq_le, β sub_nonpos] at C_S
rw [β mul_le_mul_iff_of_pos_left (show 0<(n:β)+1 by norm_cast; simp), mul_zero] at C_S
-- Rewrite the LHS of the inquality C_S to a square form
rw [show (n+1)*((a-x 0)^2-n*(b-x 0^2)) = (((n:β)+1)*x 0-a)^2-(n*((n+1)*b-a^2)) by ring] at C_S
rw [sub_nonpos] at C_S; constructor
-- Apply transitivity and sq_nonneg to prove that the discriminent is nonnegative
Β· rw [β sub_nonneg, β mul_le_mul_iff_of_pos_left (show 0<(n:β) by norm_cast), mul_zero]
apply le_trans _ C_S; apply sq_nonneg
-- Rewrite the goal to a square form | import Mathlib
open Real
/-26. (KOR 4) Let $n$ be a positive integer and let $a, b$ be given real numbers. Determine the range of $x_{0}$ for which $$ \sum_{i=0}^{n} x_{i}=a \quad \text { and } \quad \sum_{i=0}^{n} x_{i}^{2}=b $$ where $x_{0}, x_{1}, \ldots, x_{n}$ are real variables.-/
theorem other_24234 (n : β) (a b x0 : β) (npos : 0 < n) : let D := n * (n + 1) * (b - a ^ 2 / (n + 1));
(β x : β β β, x 0 = x0 β§ β i β Finset.range (n + 1), x i = a β§ β i β Finset.range (n + 1), (x i) ^ 2 = b)
β a ^ 2 β€ (n + 1) * b β§ ((n = 1 β§ (x0 = (a - D.sqrt) / (n + 1) β¨ x0 = (a + D.sqrt) / (n + 1))) β¨ ( 1 < n β§
(a - D.sqrt) / (n + 1) β€ x0 β§ x0 β€ (a + D.sqrt) / (n + 1))) := by
-- Introduce assumptions and prepare to use Cauchy-Schwarz inequality
intro D; constructor
Β· rintro β¨x, hx0, sumx, sumsqxβ©
let A : EuclideanSpace β (Fin n) := fun t => 1
let B : EuclideanSpace β (Fin n) := fun t => x (t.val + 1)
-- Apply Cauchy-Schwarz inequality to $A$ and $B$, then simplify it using properties of norm and Finset.sum
have C_S := abs_real_inner_le_norm A B; simp [A, B] at C_S
repeat simp [EuclideanSpace.norm_eq] at C_S
repeat simp [Finset.sum_fin_eq_sum_range] at C_S
repeat rw [Finset.sum_ite_of_true] at C_S
rw [add_comm, Finset.sum_range_add] at sumx sumsqx; simp [add_comm] at sumx sumsqx
rw [β eq_sub_iff_add_eq'] at sumx sumsqx; rw [sumx, sumsqx] at C_S
rw [β sqrt_mul, abs_le, β sq_le, β sub_nonpos] at C_S
rw [β mul_le_mul_iff_of_pos_left (show 0<(n:β)+1 by norm_cast; simp), mul_zero] at C_S
-- Rewrite the LHS of the inquality C_S to a square form
rw [show (n+1)*((a-x 0)^2-n*(b-x 0^2)) = (((n:β)+1)*x 0-a)^2-(n*((n+1)*b-a^2)) by ring] at C_S
rw [sub_nonpos] at C_S; constructor
-- Apply transitivity and sq_nonneg to prove that the discriminent is nonnegative
Β· rw [β sub_nonneg, β mul_le_mul_iff_of_pos_left (show 0<(n:β) by norm_cast), mul_zero]
apply le_trans _ C_S; apply sq_nonneg
-- Rewrite the goal to a square form using multiplycation as an addition operation on both sidesformulation single out $(D).β$. The result willvaluate into $-(-C)\cdot\frac{(\operatorname{\mathbb R}(c))^{k}}{p^{q}}\in\{-\infty,\cdots,-N\}$. We do not know if this expression has any signincanc e but we may still try applying our previous lemma about $\abs{-z}\leq z $. This should give something useful since $-\infity< N<\β$; see below for more details.-/ |
80 | MCQ | autoformalizer | Definition: The annual production growth rate of a factory is:
$\frac{\text{current year's output value} - \text{previous year's output value}}{\text{previous year's output value}} \times 100\%$
If the factory's output value in the year $2000$ is to be $1.44$ times that of the year $1998$, and the annual production growth rate is $x$ each year, then $x$ equals ( ).
A: $5\%$
B: $10\%$
C: $15\%$
D: $20\%$ | import Mathlib
/- Definition: The annual production growth rate of a factory is:
$\frac{\text{current year's output value} - \text{previous year's output value}}{\text{previous year's output value}} \times 100\%$
If the factory's output value in the year $2000$ is to be $1.44$ times that of the year $1998$, and the annual production growth rate is $x$ each year, then $x$ equals ( ).
A: $5\%$
B: $10\%$
C: $15\%$
D: $20\%$ -/
theorem algebra_2662 {x : β} (hx : 0 < x) (h : (1 + x / 100) ^ 2 = 1.44) : x = 20 := by | import Mathlib
theorem algebra_2662 {x : β} (hx : 0 < x) (h : (1 + x / 100) ^ 2 = 1.44) : x = 20 := by
have h1 : (1 + x / 100 : β) ^ 2 - (1.44 : β) = 0 := by
linarith
have h2 : ((1 + x / 100 : β) - 1.2) * ((1 + x / 100 : β) + 1.2) = 0 := by
ring_nf at h1 β’
norm_num at h1 β’
nlinarith
cases' (mul_eq_zero.mp h2) with h3 h4
Β· -- First case: (1 + x / 100) - 1.2 = 0, which means 1 + x/100 = 1.2
have h5 : (x : β) = 20 := by
norm_num at h3
linarith
exact h5
Β· -- Second case: (1 + x / 100) + 1.2 = 0, which means 1 + x/100 = -1.2
have h5 : (x : β) < (0 : β) := by
norm_num at h4
nlinarith
linarith -- contradicts hx: 0 < x | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem algebra_2662 {x : β} (hx : 0 < x) (h : (1 + x / 100) ^ 2 = 1.44) : x = 20 := by
have h1 : (1 + x / 100 : β) ^ 2 - (1.44 : β) = 0 := by
linarith
have h2 : ((1 + x / 100 : β) - 1.2) * ((1 + x / 100 : β) + 1.2) = 0 := by
ring_nf at h1 β’
norm_num at h1 β’
nlinarith
cases' (mul_eq_zero.mp h2) with h3 h4
Β· -- First case: (1 + x / 100) - 1.2 = 0, which means 1 + x/100 = 1.2
have h5 : (x : β) = 20 := by
| rw [(by div only; constructor)]at* ; assumptionοΏ½οΏ½<|> contradiction | import Mathlib
theorem algebra_2662 {x : β} (hx : 0 < x) (h : (1 + x / 100) ^ 2 = 1.44) : x = 20 := by
have h1 : (1 + x / 100 : β) ^ 2 - (1.44 : β) = 0 := by
linarith
have h2 : ((1 + x / 100 : β) - 1.2) * ((1 + x / 100 : β) + 1.2) = 0 := by
ring_nf at h1 β’
norm_num at h1 β’
nlinarith
cases' (mul_eq_zero.mp h2) with h3 h4
Β· -- First case: (1 + x / 100) - 1.2 = 0, which means 1 + x/100 = 1.2
have h5 : (x : β) = 20 := by
| import Mathlib
theorem algebra_2662 {x : β} (hx : 0 < x) (h : (1 + x / 100) ^ 2 = 1.44) : x = 20 := by
have h1 : (1 + x / 100 : β) ^ 2 - (1.44 : β) = 0 := by
linarith
have h2 : ((1 + x / 100 : β) - 1.2) * ((1 + x / 100 : β) + 1.2) = 0 := by
ring_nf at h1 β’
norm_num at h1 β’
nlinarith
cases' (mul_eq_zero.mp h2) with h3 h4
Β· -- First case: (1 + x / 100) - 1.2 = 0, which means 1 + x/100 = 1.2
have h5 : (x : β) = 20 := by
rw [(by div only; constructor)]at* ; assumptionοΏ½οΏ½<|> contradiction |
81 | math-word-problem | autoformalizer | $$ 4 ( 15 ^ { 3 } + 14 ^ { 3 } + 13 ^ { 3 } + ... + 1 ^ { 3 } ) + 6 ( 15 ^ { 2 } + 14 ^ { 2 } + 13 ^ { 2 } + ... + 1 ^ { 2 } ) + 4 ( 15 + 14 + 13 + ... + 1 ) + 15 = ?$$ | import Mathlib
/- $$ 4 ( 15 ^ { 3 } + 14 ^ { 3 } + 13 ^ { 3 } + ... + 1 ^ { 3 } ) + 6 ( 15 ^ { 2 } + 14 ^ { 2 } + 13 ^ { 2 } + ... + 1 ^ { 2 } ) + 4 ( 15 + 14 + 13 + ... + 1 ) + 15 = ?$$ -/
theorem algebra_3881 : 4 * β i β Finset.range 15, (i + 1) ^ 3 + 6 * β i β Finset.range 15, (i + 1) ^ 2 + 4 * β i β Finset.range 15, (i + 1) + 15 = 65535 := by | import Mathlib
theorem algebra_3881 : 4 * β i β Finset.range 15, (i + 1) ^ 3 + 6 * β i β Finset.range 15, (i + 1) ^ 2 + 4 * β i β Finset.range 15, (i + 1) + 15 = 65535 := by
native_decide | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem algebra_3881 : 4 * β i β Finset.range 15, (i + 1) ^ 3 + 6 * β i β Finset.range 15, (i + 1) ^ 2 + 4 * β i β Finset | .range (-β..(-Ο/ββ).ceil)]:=by normFsubset | import Mathlib
theorem algebra_3881 : 4 * β i β Finset.range 15, (i + 1) ^ 3 + 6 * β i β Finset.range 15, (i + 1) ^ 2 + 4 * β i β Finset | import Mathlib
theorem algebra_3881 : 4 * β i β Finset.range 15, (i + 1) ^ 3 + 6 * β i β Finset.range 15, (i + 1) ^ 2 + 4 * β i β Finset.range (-β..(-Ο/ββ).ceil)]:=by normFsubset |
82 | math-word-problem | autoformalizer | Compute $\lfloor \sqrt{n^2 - 10n + 29} \rfloor$ when $n = 19941994.$ | import Mathlib
/- Compute $\lfloor \sqrt{n^2 - 10n + 29} \rfloor$ when $n = 19941994.$ -/
theorem algebra_21564 (n : β) (hn : n = 19941994) : βReal.sqrt (n^2 - 10 * n + 29)β = 19941989 := by | import Mathlib
theorem algebra_21564 (n : β) (hn : n = 19941994) : βReal.sqrt (n^2 - 10 * n + 29)β = 19941989 := by
rw [hn]
have h_eq : ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29) = (19941989 : β) ^ 2 + 4 := by
norm_num
have h1 : Real.sqrt ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29) = Real.sqrt ((19941989 : β) ^ 2 + 4) := by
rw [h_eq]
have h2 : (19941989 : β) < Real.sqrt ((19941989 : β) ^ 2 + 4) := by
nlinarith [Real.sqrt_nonneg ((19941989 : β) ^ 2 + 4), Real.sq_sqrt (show (0 : β) β€ ((19941989 : β) ^ 2 + 4) by positivity)]
-- Show that the real sqrt is less than the next integer
have h3 : Real.sqrt ((19941989 : β) ^ 2 + 4) < (19941990 : β) := by
nlinarith [Real.sqrt_nonneg ((19941989 : β) ^ 2 + 4), Real.sq_sqrt (show (0 : β) β€ ((19941989 : β) ^ 2 + 4) by positivity)]
have h4 : βReal.sqrt ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29)β = (19941989 : β€) := by
have h5 : Real.sqrt ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29) = Real.sqrt ((19941989 : β) ^ 2 + 4) := h1
rw [h5]
have h5 : βReal.sqrt ((19941989 : β) ^ 2 + 4)β = (19941989 : β€) := by
rw [Int.floor_eq_iff]
constructor <;> (
try {
have h7 : (19941989 : β) β€ Real.sqrt ((19941989 : β) ^ 2 + 4) := by
nlinarith [Real.sqrt_nonneg ((19941989 : β) ^ 2 + 4), Real.sq_sqrt (show (0 : β) β€ ((19941989 : β) ^ 2 + 4) by positivity)]
have h8 : Real.sqrt ((19941989 : β) ^ 2 + 4) < (19941990 : β) := h3
norm_num at *
all_goals nlinarith
}
)
exact_mod_cast h5
exact_mod_cast h4 | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem algebra_21564 (n : β) (hn : n = 19941994) : βReal.sqrt (n^2 - 10 * n + 29)β = 19941989 := by
rw [hn]
have h_eq : ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29) = (19941989 : β) ^ 2 + 4 := by
norm_num
have h1 : Real.sqrt ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29) = Real.sqrt ((19941989 : β) ^ 2 + 4) := by
rw [h_eq]
have h2 : (19941989 : β) < Real.sqrt ((19941989 : β) ^ 2 + 4) := by
nlinarith [Real.sqrt_nonneg ((19941989 : β) ^ 2 + 4), Real.sq_sqrt (show (0 : β) β€ ((19941989 : β) ^ 2 + 4) by positivity)]
-- Show that the real sqrt is less than the next integer
have h3 : Real.sqrt ((19941989 : β) ^ 2 + 4) < (19941990 : β) := by
nlinarith [Real.sqrt_nonneg ((19941989 : β) ^ 2 + 4), Real.sq_sqrt (show (0 : β) β€ ((19941989 : β) ^ 2 + 4) by positivity)]
have h4 : βReal.sqrt ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29)β = (19941989 : β€) := by
have h5 : Real.sqrt ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29) = Real.sqrt ((19941989 : β) ^ 2 + 4) := h1
rw [h5]
have h5 : βReal.sqrt ((1 | 9941989 : πΌ ) ^ 2+β)-((ββ*β)*ββΉαΆ―ΒΉΒ³Β²)/(-ββ
*(βΒ·)^Β½)=-(((βΒ· β§ Β·).snd (-ΒΌ))):=by apply eqaulity;ringarit at this using mulle' _ _,mullt',divLToneOfNat ,subLeself ;normRing | import Mathlib
theorem algebra_21564 (n : β) (hn : n = 19941994) : βReal.sqrt (n^2 - 10 * n + 29)β = 19941989 := by
rw [hn]
have h_eq : ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29) = (19941989 : β) ^ 2 + 4 := by
norm_num
have h1 : Real.sqrt ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29) = Real.sqrt ((19941989 : β) ^ 2 + 4) := by
rw [h_eq]
have h2 : (19941989 : β) < Real.sqrt ((19941989 : β) ^ 2 + 4) := by
nlinarith [Real.sqrt_nonneg ((19941989 : β) ^ 2 + 4), Real.sq_sqrt (show (0 : β) β€ ((19941989 : β) ^ 2 + 4) by positivity)]
-- Show that the real sqrt is less than the next integer
have h3 : Real.sqrt ((19941989 : β) ^ 2 + 4) < (19941990 : β) := by
nlinarith [Real.sqrt_nonneg ((19941989 : β) ^ 2 + 4), Real.sq_sqrt (show (0 : β) β€ ((19941989 : β) ^ 2 + 4) by positivity)]
have h4 : βReal.sqrt ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29)β = (19941989 : β€) := by
have h5 : Real.sqrt ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29) = Real.sqrt ((19941989 : β) ^ 2 + 4) := h1
rw [h5]
have h5 : βReal.sqrt ((1 | import Mathlib
theorem algebra_21564 (n : β) (hn : n = 19941994) : βReal.sqrt (n^2 - 10 * n + 29)β = 19941989 := by
rw [hn]
have h_eq : ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29) = (19941989 : β) ^ 2 + 4 := by
norm_num
have h1 : Real.sqrt ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29) = Real.sqrt ((19941989 : β) ^ 2 + 4) := by
rw [h_eq]
have h2 : (19941989 : β) < Real.sqrt ((19941989 : β) ^ 2 + 4) := by
nlinarith [Real.sqrt_nonneg ((19941989 : β) ^ 2 + 4), Real.sq_sqrt (show (0 : β) β€ ((19941989 : β) ^ 2 + 4) by positivity)]
-- Show that the real sqrt is less than the next integer
have h3 : Real.sqrt ((19941989 : β) ^ 2 + 4) < (19941990 : β) := by
nlinarith [Real.sqrt_nonneg ((19941989 : β) ^ 2 + 4), Real.sq_sqrt (show (0 : β) β€ ((19941989 : β) ^ 2 + 4) by positivity)]
have h4 : βReal.sqrt ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29)β = (19941989 : β€) := by
have h5 : Real.sqrt ((19941994 : β) ^ 2 - 10 * (19941994 : β) + 29) = Real.sqrt ((19941989 : β) ^ 2 + 4) := h1
rw [h5]
have h5 : βReal.sqrt ((19941989 : πΌ ) ^ 2+β)-((ββ*β)*ββΉαΆ―ΒΉΒ³Β²)/(-ββ
*(βΒ·)^Β½)=-(((βΒ· β§ Β·).snd (-ΒΌ))):=by apply eqaulity;ringarit at this using mulle' _ _,mullt',divLToneOfNat ,subLeself ;normRing |
83 | math-word-problem | autoformalizer | Rationalize the denominator: $\frac{5}{\sqrt{3}}$ . | import Mathlib
/- Rationalize the denominator: $\frac{5}{\sqrt{3}}$ . -/
theorem algebra_21678 : (5 / Real.sqrt 3 : β) = 5 * Real.sqrt 3 / 3 := by | import Mathlib
theorem algebra_21678 : (5 / Real.sqrt 3 : β) = 5 * Real.sqrt 3 / 3 := by
have h1 : Real.sqrt 3 > 0 := by
positivity
field_simp
<;> ring_nf <;> norm_num <;> ring
<;> norm_num | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem algebra_21678 : (5 / Real.sqrt 3 : β) = 5 * Real.sqrt 3 / 3 := by
have h1 : Real.sqrt 3 > 0 := by
positivity
field_simp
| null | import Mathlib
theorem algebra_21678 : (5 / Real.sqrt 3 : β) = 5 * Real.sqrt 3 / 3 := by
have h1 : Real.sqrt 3 > 0 := by
positivity
field_simp
| import Mathlib
theorem algebra_21678 : (5 / Real.sqrt 3 : β) = 5 * Real.sqrt 3 / 3 := by
have h1 : Real.sqrt 3 > 0 := by
positivity
field_simp
|
84 | math-word-problem | autoformalizer | The total number of digits used to write the page numbers of a math book is $6869$ (for example, a $10$-page book has $9$ one-digit page numbers and $1$ two-digit page number, using a total of $9+2=11$ digits), so the number of pages in this math book isΒ ___Β Β Β Β Β Β Β Β Β Β ___Β . | import Mathlib
def solution := 1994
/- The total number of digits used to write the page numbers of a math book is $6869$ (for example, a $10$-page book has $9$ one-digit page numbers and $1$ two-digit page number, using a total of $9+2=11$ digits), so the number of pages in this math book isΒ ___Β Β Β Β Β Β Β Β Β Β ___Β . -/
theorem number_theory_1544 : β i in Finset.range solution, (Nat.digits 10 (i + 1)).length = 6869 := by | import Mathlib
def solution := 1994
theorem number_theory_1544 : β i in Finset.range solution, (Nat.digits 10 (i + 1)).length = 6869 := by
native_decide | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
def solution := 1994
theorem number_theory_1544 : β i in Finset.range solution, (Nat.digits 10 (i | ^solution).getLast!)^(-Finset.card {j| j*k=l})β₯ Nat.divisorSum solucion:=by normopose | import Mathlib
def solution := 1994
theorem number_theory_1544 : β i in Finset.range solution, (Nat.digits 10 (i | import Mathlib
def solution := 1994
theorem number_theory_1544 : β i in Finset.range solution, (Nat.digits 10 (i^solution).getLast!)^(-Finset.card {j| j*k=l})β₯ Nat.divisorSum solucion:=by normopose |
85 | math-word-problem | autoformalizer | Let $x$ and $y$ be positive real numbers such that $ x ^ { 3 } - y ^ { 3 } + ( x - y ) ^ { 3 } - 3 6 x y = 3 4 5 6 $. Find the value of $ x - y $. | import Mathlib
/- Let $x$ and $y$ be positive real numbers such that $ x ^ { 3 } - y ^ { 3 } + ( x - y ) ^ { 3 } - 3 6 x y = 3 4 5 6 $. Find the value of $ x - y $. -/
theorem algebra_13860 {x y : β} (hx : 0 < x) (hy : 0 < y)
(h : x^3 - y^3 + (x - y)^3 - 36 * x * y = 3456) :
x - y = 12 := by | import Mathlib
theorem algebra_13860 {x y : β} (hx : 0 < x) (hy : 0 < y)
(h : x^3 - y^3 + (x - y)^3 - 36 * x * y = 3456) :
x - y = 12 := by
have h1 : x^3 - y^3 + (x - y)^3 - 36 * x * y = 3456 := h
have h2 : x - y = 12 := by
nlinarith [sq_nonneg (x - y - 12), sq_nonneg (x + y - 24), sq_nonneg (x - 2 * y), sq_nonneg (x^2 - (y + 12)^2), sq_nonneg (x - y), sq_nonneg (x + y), mul_pos hx hy, sq_nonneg (x^2 - y^2), sq_nonneg ((x - y) * (x + y) - 144)]
exact h2 | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem algebra_13860 {x y : β} (hx : 0 < x) (hy : 0 < y)
(h : x^3 - y^3 + (x - y)^3 - 36 * x * y = 3456) :
x - y = 12 := by
have h1 : x^3 - y^3 + (x - y)^3 - 36 * x * y = 3456 := h
have h2 : x - y = 12 := by
nlinarith [sq_nonneg (x - y - 12), sq_nonneg (x + y - 24), sq_nonneg (x - 2 * y), | mul_self' _ _, add_mul', subtruction'] | import Mathlib
theorem algebra_13860 {x y : β} (hx : 0 < x) (hy : 0 < y)
(h : x^3 - y^3 + (x - y)^3 - 36 * x * y = 3456) :
x - y = 12 := by
have h1 : x^3 - y^3 + (x - y)^3 - 36 * x * y = 3456 := h
have h2 : x - y = 12 := by
nlinarith [sq_nonneg (x - y - 12), sq_nonneg (x + y - 24), sq_nonneg (x - 2 * y), | import Mathlib
theorem algebra_13860 {x y : β} (hx : 0 < x) (hy : 0 < y)
(h : x^3 - y^3 + (x - y)^3 - 36 * x * y = 3456) :
x - y = 12 := by
have h1 : x^3 - y^3 + (x - y)^3 - 36 * x * y = 3456 := h
have h2 : x - y = 12 := by
nlinarith [sq_nonneg (x - y - 12), sq_nonneg (x + y - 24), sq_nonneg (x - 2 * y), mul_self' _ _, add_mul', subtruction'] |
86 | math-word-problem | autoformalizer | Find the remainder when ${{2018}^{2019}}$ is divided by $2019$.
| import Mathlib
/- Find the remainder when ${{2018}^{2019}}$ is divided by $2019$.
-/
theorem number_theory_4606
: 2018 ^ 2019 % 2019 = 2018 := by | import Mathlib
theorem number_theory_4606
: 2018 ^ 2019 % 2019 = 2018 := by
norm_num | ### Example 1:
import Mathlib
theorem algebra_1274 {{x y : β}} (hx : x β 0) (hy : y β 0) (hxy : x + y β 0)
(h : β k, 1 / x + 1 / y = k / (x + y)) :
β m, (x + y) ^ 2 = m * (x ^ 2 + y ^ 2) := by
use (x + y) ^ 2 / (x ^ 2 + y ^ 2)
have : x ^ 2 + y ^ 2 β 0 := by
have hx2 : x ^ 2 > 0 := by
exact sq_pos_of_ne_zero hx
have hy2 : y ^ 2 > 0 := by
exact sq_pos_of_ne_zero hy
linarith
field_simp
### Example 2:
import Mathlib
open Real Set
open scoped BigOperators
theorem calculus_196745 (x : β) (hx : 0 < x) (h : x β€ 60 / 2) :
(x * (60 - 2 * x) ^ 2) β€ 16000 := by
nlinarith [sq_nonneg (x - 10), sq_nonneg ((60 - 2 * x) - 40),
sq_nonneg ((x - 10) * (x - 40)), sq_nonneg ((x) * (x - 10)),
sq_nonneg ((60 - 2 * x) / 2 - 10 * 2), sq_nonneg ((x / 10) - 1),
sq_nonneg ((60 - 2 * x) / 40 - 1), sq_nonneg (x * (x - 30)),
sq_nonneg ((60 - 2 * x) * (x - 10)), sq_nonneg ((x - 0) * (30 - x)), sq_nonneg (x * (60 - 2 * x - 40)),
h, sq_nonneg (x * ((60 - 2 * x) - 40)), sq_nonneg ((x - 10) * ((60 - 2 * x) - 40))]
### Example 3:
import Mathlib
-- Let $f(n)$ denote the number of ways to fill a $2 \times n$ board. One can fill the leftmost column with two $1 \times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n)=f(n-1)+f(n-2)$
def f (n : β) : β := match n with
| 0 => 1
| 1 => 1
| n + 2 => f n + f (n + 1)
/-How many ways can one tile a 2Γ8 board with
1Γ1 and 2Γ2 tiles?
Rotations and reflections of the same configuration are considered distinct.-/
theorem Combinatorics_610027: f 8 = 34 := by
simp [f]
### Example 4:
import Mathlib
theorem number_theory_4606
: 2018 ^ 2019 % 20 | ^(-k+l*m)*p= p:=by normF | import Mathlib
theorem number_theory_4606
: 2018 ^ 2019 % 20 | import Mathlib
theorem number_theory_4606
: 2018 ^ 2019 % 20^(-k+l*m)*p= p:=by normF |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.