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Theorem 16.39. If \( D \) is a PID, then every non-zero, non-unit element of \( D \) can be expressed as a product of irreducibles in \( D \) . | Proof. Let \( c \in D, c \neq 0 \), and \( c \) not a unit. If \( c \) is irreducible, we are done. Otherwise, we can write \( c = {ab} \), where neither \( a \) nor \( b \) are units. As ideals, we have \( {cD} \varsubsetneq {aD} \) and \( {cD} \varsubsetneq {bD} \) . If we continue this process recursively, building ... | No |
Theorem 16.40. Let \( D \) be a PID. For all \( a, b \in D \), there exists a greatest common divisor \( d \) of \( a \) and \( b \), and moreover, \( {aD} + {bD} = {dD} \) . | Proof. Exercise. | No |
Theorem 16.41. Let \( D \) be a PID. For all \( a, b, c \in D \) such that \( c \mid {ab} \) and \( a \) and \( c \) are relatively prime, we have \( c \mid b \) . | Proof. Exercise. | No |
Theorem 16.42. Let \( D \) be a PID. Let \( p \in D \) be irreducible, and let \( a, b \in D \) . Then \( p \mid {ab} \) implies that \( p \mid a \) or \( p \mid b \) . | Proof. Exercise. | No |
Theorem 16.44. Let \( D \) be a UFD. Every non-zero, non-unit element of \( D\left\lbrack X\right\rbrack \) can be expressed as a product of irreducibles in \( D\left\lbrack X\right\rbrack \) . | Proof. Let \( f \) be a non-zero, non-unit polynomial in \( D\left\lbrack X\right\rbrack \) . If \( f \) is a constant, then because \( D \) is a UFD, \( f \) factors into irreducibles in \( D \) . So assume \( f \) is not constant. If \( f \) is not primitive, we can write \( f = c{f}^{\prime } \), where \( c \) is a ... | Yes |
Theorem 16.45. Let \( D \) be a UFD, let \( p \) be an irreducible in \( D \), and let \( g, h \in D\left\lbrack X\right\rbrack \) . Then \( p \mid {gh} \) implies \( p \mid g \) or \( p \mid h \) . | Proof. Consider the quotient ring \( D/{pD} \), which is an integral domain (because \( D \) is a UFD), and the corresponding ring of polynomials \( \left( {D/{pD}}\right) \left\lbrack X\right\rbrack \), which is also an integral domain. Also consider the natural map that sends \( a \in D \) to \( \bar{a} \mathrel{\tex... | Yes |
Theorem 16.46. Let \( D \) be a UFD. The product of two primitive polynomials in \( D\left\lbrack X\right\rbrack \) is also primitive. | Proof. Let \( g, h \in D\left\lbrack X\right\rbrack \) be primitive polynomials, and let \( f \mathrel{\text{:=}} {gh} \) . If \( f \) is not primitive, then \( c \mid f \) for some non-zero, non-unit \( c \in D \), and as \( D \) is a UFD, there is some irreducible element \( p \in D \) that divides \( c \), and there... | Yes |
Theorem 16.47. Let \( D \) be a UFD and let \( F \) be its field of fractions. Suppose that \( f, g \in D\left\lbrack X\right\rbrack \) and \( h \in F\left\lbrack X\right\rbrack \) are non-zero polynomials such that \( f = {gh} \) and \( g \) is primitive. Then \( h \in D\left\lbrack X\right\rbrack \) . | Proof. Write \( h = \left( {c/d}\right) {h}^{\prime } \), where \( c, d \in D \) and \( {h}^{\prime } \in D\left\lbrack X\right\rbrack \) is primitive. Let us assume that \( c \) and \( d \) are relatively prime. Then we have\n\n\[ d \cdot f = c \cdot g{h}^{\prime }.\]\n\n(16.9)\n\nWe claim that \( d \in {D}^{ * } \) .... | Yes |
Theorem 16.48. Let \( D \) be a UFD and \( F \) its field of fractions. If \( f \in D\left\lbrack X\right\rbrack \) with \( \deg \left( f\right) > 0 \) is irreducible, then \( f \) is also irreducible in \( F\left\lbrack X\right\rbrack \) . | Proof. Suppose that \( f \) is not irreducible in \( F\left\lbrack X\right\rbrack \), so that \( f = {gh} \) for non-constant polynomials \( g, h \in F\left\lbrack X\right\rbrack \), both of degree strictly less than that of \( f \) . We may write \( g = \left( {c/d}\right) {g}^{\prime } \), where \( c, d \in D \) and ... | Yes |
Theorem 16.49. Let \( D \) be a UFD. Let \( f \in D\left\lbrack X\right\rbrack \) with \( \deg \left( f\right) > 0 \) be irreducible, and let \( g, h \in D\left\lbrack X\right\rbrack \) . If \( f \) divides \( {gh} \) in \( D\left\lbrack X\right\rbrack \), then \( f \) divides either \( g \) or \( h \) in \( D\left\lbr... | Proof. Suppose that \( f \in D\left\lbrack X\right\rbrack \) with \( \deg \left( f\right) > 0 \) is irreducible. This implies that \( f \) is a primitive polynomial. By Theorem 16.48, \( f \) is irreducible in \( F\left\lbrack X\right\rbrack \), where \( F \) is the field of fractions of \( D \) . Suppose \( f \) divid... | Yes |
Theorem 16.50 (Eisenstein's criterion). Let \( D \) be a UFD and \( F \) its field of fractions. Let \( f = {c}_{n}{X}^{n} + {c}_{n - 1}{X}^{n - 1} + \cdots + {c}_{0} \in D\left\lbrack X\right\rbrack \) . If there exists an irreducible \( p \in D \) such that\n\n\[ p \nmid {c}_{n}, p \mid {c}_{n - 1},\cdots, p \mid {c}... | Proof. Let \( f \) be as above, and suppose it were not irreducible in \( F\left\lbrack X\right\rbrack \) . Then by Theorem 16.48, we could write \( f = {gh} \), where \( g, h \in D\left\lbrack X\right\rbrack \), both of degree strictly less than that of \( f \) . Let us write\n\n\[ g = {a}_{k}{X}^{k} + \cdots + {a}_{0... | Yes |
Theorem 16.51. For every prime number \( q \), the \( q \) th cyclotomic polynomial\n\n\[{\Phi }_{q} \mathrel{\text{:=}} \frac{{X}^{q} - 1}{X - 1} = {X}^{q - 1} + {X}^{q - 2} + \cdots + 1\]\n\nis irreducible over \( \mathbb{Q} \) . | Proof. Let\n\n\[f \mathrel{\text{:=}} {\Phi }_{q}\left( {X + 1}\right) = \frac{{\left( X + 1\right) }^{q} - 1}{\left( {X + 1}\right) - 1}.\n\]\n\nIt is easy to see that\n\n\[f = \mathop{\sum }\limits_{{i = 0}}^{{q - 1}}{c}_{i}{X}^{i},\text{ where }{c}_{i} = \left( \begin{matrix} q \\ i + 1 \end{matrix}\right) \left( {i... | Yes |
Theorem 17.2. Let \( g, h \in F\left\lbrack X\right\rbrack \), with \( \deg \left( g\right) \geq \deg \left( h\right) \) and \( g \neq 0 \) . Define the polynomials \( {r}_{0},{r}_{1},\ldots ,{r}_{\lambda + 1} \in F\left\lbrack X\right\rbrack \) and \( {q}_{1},\ldots ,{q}_{\lambda } \in F\left\lbrack X\right\rbrack \),... | Proof. Arguing as in the proof of Theorem 4.1, one sees that\n\n\[ \gcd \left( {g, h}\right) = \gcd \left( {{r}_{0},{r}_{1}}\right) = \cdots = \gcd \left( {{r}_{\lambda },{r}_{\lambda + 1}}\right) = \gcd \left( {{r}_{\lambda },0}\right) = {r}_{\lambda }/\operatorname{lc}\left( {r}_{\lambda }\right) . \]\n\nThat proves ... | Yes |
Theorem 17.3. Euclid’s algorithm for polynomials performs \( O\left( {\operatorname{len}\left( g\right) \operatorname{len}\left( h\right) }\right) \) operations in \( F \) . | Proof. The proof is almost identical to that of Theorem 4.2. Details are left to the reader. | No |
Theorem 17.4. Let \( g, h,{r}_{0},\ldots ,{r}_{\lambda + 1} \) and \( {q}_{1},\ldots ,{q}_{\lambda } \) be as in Theorem 17.2. Define polynomials \( {s}_{0},\ldots ,{s}_{\lambda + 1} \in F\left\lbrack X\right\rbrack \) and \( {t}_{0},\ldots ,{t}_{\lambda + 1} \in F\left\lbrack X\right\rbrack \) as follows:\n\n\[ \n{s}_... | Proof. (i), (ii), and (iii) are proved just as in the corresponding parts of Theorem 4.3.\n\nFor (iv), the proof will hinge on the following facts:\n\n- For \( i = 1,\ldots ,\lambda \), we have \( \deg \left( {r}_{i - 1}\right) \geq \deg \left( {r}_{i}\right) \), and since \( {q}_{i} \) is the quotient in dividing \( {... | Yes |
Theorem 17.5. The extended Euclidean algorithm for polynomials performs \( O\left( {\operatorname{len}\left( g\right) \operatorname{len}\left( h\right) }\right) \) operations in \( F \) . | Proof. Exercise. | No |
Theorem 17.6. Suppose we are given polynomials \( f, h \in F\left\lbrack X\right\rbrack \), where \( \deg \left( h\right) < \) \( \deg \left( f\right) \) . Then using \( O\left( {\operatorname{len}{\left( f\right) }^{2}}\right) \) operations in \( F \), we can determine if \( h \) is relatively prime to \( f \), and if... | Proof. We may assume \( \deg \left( f\right) > 0 \), since \( \deg \left( f\right) = 0 \) implies \( h = 0 = {h}^{-1}{\;\operatorname{mod}\;f} \) . We run the extended Euclidean algorithm on input \( f, h \), obtaining polynomials \( d, s, t \) such that \( d = \gcd \left( {f, h}\right) \) and \( {fs} + {ht} = d \) . I... | Yes |
Theorem 17.7 (Effective Chinese remainder theorem). Suppose we are given polynomials \( {f}_{1},\ldots ,{f}_{k} \in F\left\lbrack X\right\rbrack \) and \( {g}_{1},\ldots ,{g}_{k} \in F\left\lbrack X\right\rbrack \), where the family \( {\left\{ {f}_{i}\right\} }_{i = 1}^{k} \) is pairwise relatively prime, and where \(... | Proof. Exercise (just use the formulas given after Theorem 16.19). | No |
Theorem 17.8 (Rational function reconstruction). Let \( f, h \in F\left\lbrack X\right\rbrack \) be polynomials, and let \( {r}^{ * },{t}^{ * } \) be non-negative integers, such that\n\n\[ \deg \left( h\right) < \deg \left( f\right) \text{ and }{r}^{ * } + {t}^{ * } \leq \deg \left( f\right) .\n\]\n\nFurther, let \( \o... | Proof. Since \( \deg \left( {r}_{0}\right) = \deg \left( f\right) \geq {r}^{ * } > - \infty = \deg \left( {r}_{\lambda + 1}\right) \), the value of \( j \) is well defined, and moreover, \( j \geq 1,\deg \left( {r}_{j - 1}\right) \geq {r}^{ * } \), and \( {t}_{j} \neq 0 \) .\n\nFrom the equalities \( {r}_{j} = f{s}_{j}... | Yes |
Theorem 18.1. The set \( G\left( \Psi \right) \) is an ideal of \( F\left\lbrack X\right\rbrack \) . | Proof. First, note that for all \( g, h \in F\left\lbrack X\right\rbrack \), we have \( \left( {g + h}\right) \star \Psi = \left( {g \star \Psi }\right) + \left( {h \star \Psi }\right) - \) this is clear from the definitions. It is also clear that for all \( c \in F \) and \( g \in F\left\lbrack X\right\rbrack \) , we ... | Yes |
One can always define a linearly generated sequence by simply choosing an initial segment \( {\alpha }_{0},{\alpha }_{1},\ldots ,{\alpha }_{k - 1} \), along with scalars \( {c}_{0},\ldots ,{c}_{k - 1} \in F \) defining the recurrence relation. | One can enumerate as many elements of the sequence as one wants by using storage for \( k \) elements of \( V \), along with storage for the scalars \( {c}_{0},\ldots ,{c}_{k - 1} \), as follows:\n\n\( \left( {{\beta }_{0},\ldots ,{\beta }_{k - 1}}\right) \leftarrow \left( {{\alpha }_{0},\ldots ,{\alpha }_{k - 1}}\righ... | Yes |
Let \( V \) be a vector space over \( F \) of dimension \( \ell > 0 \), and let \( \tau : V \rightarrow V \) be an \( F \)-linear map. Let \( \beta \in V \), and consider the sequence \( \Psi \mathrel{\text{:=}} {\left\{ {\alpha }_{i}\right\} }_{i = 0}^{\infty } \), where \( {\alpha }_{i} = {\tau }^{i}\left( \beta \rig... | \[ g \star \Psi = \mathop{\sum }\limits_{{j = 0}}^{k}{a}_{j}{\tau }^{j}\left( \beta \right) \] and for every \( i \geq 0 \), we have \[ \left( {{X}^{i}g}\right) \star \Psi = \mathop{\sum }\limits_{{j = 0}}^{k}{a}_{j}{\tau }^{i + j}\left( \beta \right) = {\tau }^{i}\left( {\mathop{\sum }\limits_{{j = 0}}^{k}{a}_{j}{\tau... | Yes |
Theorem 18.2. Let \( \Psi = {\left\{ {z}_{i}\right\} }_{i = 0}^{\infty } \) be a sequence of elements of \( F \), and define the reversed Laurent series\n\n\[ z \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 0}}^{\infty }{z}_{i}{X}^{-\left( {i + 1}\right) } \in F\left( \left( {X}^{-1}\right) \right) ,\]\n\nwhose coeff... | Proof. Observe that for every polynomial \( g \in F\left\lbrack X\right\rbrack \) and every integer \( i \geq 0 \) , the coefficient of \( {X}^{-\left( {i + 1}\right) } \) in the product \( {gz} \) is equal to \( {X}^{i}g \star \Psi \) -just look at the formulas defining these expressions! It follows that \( g \) is a ... | Yes |
Theorem 18.3. Let \( \Psi = {\left\{ {\alpha }_{i}\right\} }_{i = 0}^{\infty } \) be a linearly generated sequence over the field \( F \), where the \( {\alpha }_{i} \) ’s are elements of a vector space \( V \) of finite dimension \( \ell > 0 \) . Let \( \phi \) be the minimal polynomial of \( \Psi \) over \( F \), let... | Proof. While the statement of this theorem looks a bit complicated, its proof is quite straightforward, given our characterization of linearly generated sequences in Theorem 18.2 in terms of rational functions. We build the linear map \( \sigma \) as the composition of two linear maps, \( {\sigma }_{0} \) and \( {\sigm... | Yes |
Theorem 18.4. If \( F \) is a finite field of cardinality \( q \), and \( m \) and \( s \) are positive integers, then we have\n\n\[ \n{\Lambda }_{F}^{m}\left( s\right) = 1 - 1/{q}^{s - 1} + \left( {q - 1}\right) /{q}^{sm}.\n\] | Proof. For each positive integer \( n \), let \( {U}_{n} \) denote the set of all tuples of polynomials \( \left( {{f}_{1},\ldots ,{f}_{s}}\right) \in F{\left\lbrack X\right\rbrack }_{ < n}^{{ \times }_{S}} \) with \( \gcd \left( {{f}_{1},\ldots ,{f}_{s}}\right) = 1 \), and let \( {u}_{n} \mathrel{\text{:=}} \left| {U}... | Yes |
Theorem 18.6. For all \( \tau ,{\tau }^{\prime },{\tau }^{\prime \prime } \in {\mathcal{L}}_{F}\left( V\right) \), and for all \( c \in F \), we have:\n\n(i) \( \tau \circ \left( {{\tau }^{\prime } + {\tau }^{\prime \prime }}\right) = \tau \circ {\tau }^{\prime } + \tau \circ {\tau }^{\prime \prime } \) ;\n\n(ii) \( \l... | Proof. Exercise. | No |
Theorem 18.7. For all \( \tau \in {\mathcal{L}}_{F}\left( V\right) \), for all \( c \in F \), and for all \( g, h \in F\left\lbrack X\right\rbrack \), we have:\n\n(i) \( g\left( \tau \right) + h\left( \tau \right) = \left( {g + h}\right) \left( \tau \right) \) ;\n\n(ii) \( c \cdot g\left( \tau \right) = \left( {cg}\rig... | Proof. Exercise. | No |
Theorem 18.8. The scalar multiplication \( \odot \), together with the usual addition operation on \( V \), makes \( V \) into an \( F\left\lbrack X\right\rbrack \) -module; that is, for all \( g, h \in F\left\lbrack X\right\rbrack \) and \( \alpha ,\beta \in V \), we have\n\n\[ g \odot \left( {h \odot \alpha }\right) ... | Proof. Exercise. | No |
Theorem 18.12. Let \( \tau \in {\mathcal{L}}_{F}\left( V\right) \), and suppose that \( \tau \) has non-zero minimal polynomial \( \phi \) . Then there exists \( \beta \in V \) such that the minimal polynomial of \( \beta \) under \( \tau \) is \( \phi \) . | Proof. Let \( \odot \) be the scalar multiplication associated with \( \tau \) . Let \( \phi = {\phi }_{1}^{{e}_{1}}\cdots {\phi }_{r}^{{e}_{r}} \) be the factorization of \( \phi \) into monic irreducible polynomials in \( F\left\lbrack X\right\rbrack \) . First, we claim that for each \( i = 1,\ldots, r \), there exi... | Yes |
Theorem 19.1. If \( F \) is a field, and \( f \in F\left\lbrack X\right\rbrack \) with \( \gcd \left( {f,\mathbf{D}\left( f\right) }\right) = 1 \), then \( f \) is square-free. | Proof. Suppose \( f \) is not square-free, and write \( f = {g}^{2}h \), for \( g, h \in F\left\lbrack X\right\rbrack \) with \( \deg \left( g\right) > 0 \) . Taking formal derivatives, we have\n\n\[ \mathbf{D}\left( f\right) = {2g}\mathbf{D}\left( g\right) h + {g}^{2}\mathbf{D}\left( h\right) ,\]\n\nand so clearly, \(... | Yes |
Theorem 19.2. Let \( F \) be a field, and let \( k,\ell \) be positive integers. Then \( {X}^{k} - 1 \) divides \( {X}^{\ell } - 1 \) in \( F\left\lbrack X\right\rbrack \) if and only if \( k \) divides \( \ell \) . | Proof. Let \( \ell = {kq} + r \), with \( 0 \leq r < k \) . We have\n\n\[ \n{X}^{\ell } \equiv {X}^{kq}{X}^{r} \equiv {X}^{r}\left( {{\;\operatorname{mod}\;{X}^{k}} - 1}\right) ,\n\]\n\nand \( {X}^{r} \equiv 1\left( {{\;\operatorname{mod}\;{X}^{k}} - 1}\right) \) if and only if \( r = 0 \) . | Yes |
Theorem 19.3. Let \( a \geq 2 \) be an integer and let \( k,\ell \) be positive integers. Then \( {a}^{k} - 1 \) divides \( {a}^{\ell } - 1 \) if and only if \( k \) divides \( \ell \) . | Proof. The proof is analogous to that of Theorem 19.2. We leave the details to the reader. | No |
Theorem 19.4. Let \( a \geq 2 \) be an integer, \( k,\ell \) be positive integers, and \( F \) a field. Then \( {X}^{{a}^{k}} - X \) divides \( {X}^{{a}^{\ell }} - X \) in \( F\left\lbrack X\right\rbrack \) if and only if \( k \) divides \( \ell \) . | Proof. Now, \( {X}^{{a}^{k}} - X \) divides \( {X}^{{a}^{\ell }} - X \) if and only if \( {X}^{{a}^{k} - 1} - 1 \) divides \( {X}^{{a}^{\ell } - 1} - 1 \) . By Theorem 19.2, this happens if and only if \( {a}^{k} - 1 \) divides \( {a}^{\ell } - 1 \) . By Theorem 19.3, this happens if and only if \( k \) divides \( \ell... | Yes |
Theorem 19.6. We have\n\n\[ \n{X}^{q} - X = \mathop{\prod }\limits_{{a \in F}}\left( {X - a}\right) \n\] | Proof. Since each \( a \in F \) is a root of \( {X}^{q} - X \), by Theorem 7.13, the polynomial \( \mathop{\prod }\limits_{{a \in F}}\left( {X - a}\right) \) divides the polynomial \( {X}^{q} - X \) . Since the degrees and leading coefficients of these two polynomials are the same, the two polynomials must be equal. \(... | Yes |
Theorem 19.7. Let \( E \) be an \( F \) -algebra. Then the map \( \sigma : E \rightarrow E \) that sends \( \alpha \in E \) to \( {\alpha }^{q} \) is an \( F \) -algebra homomorphism. | Proof. By Theorem 16.3, either \( E \) is trivial or contains an isomorphic copy of \( F \) as a subring. In the former case, there is nothing to prove. So assume that \( E \) contains an isomorphic copy of \( F \) as a subring. It follows that \( E \) must have characteristic \( p \) .\n\nSince \( q = {p}^{w} \), we s... | Yes |
Theorem 19.8. Let \( E \) be a finite extension of \( F \), and let \( \sigma \) be the Frobenius map on \( E \) over \( F \) . Then \( \sigma \) is an \( F \) -algebra automorphism on \( E \) . Moreover, for all \( \alpha \in E \), we have \( \sigma \left( \alpha \right) = \alpha \) if and only if \( \alpha \in F \) . | Proof. The fact that \( \sigma \) is an \( F \) -algebra homomorphism follows from the previous theorem. Any ring homomorphism from a field into a field is injective (see Exercise 7.47). Surjectivity follows from injectivity and finiteness.\n\nFor the second statement, observe that \( \sigma \left( \alpha \right) = \al... | Yes |
Theorem 19.9. Let \( E \) be a extension of degree \( \ell \) over \( F \), and let \( \sigma \) be the Frobenius map on \( E \) over \( F \) . Then for all integers \( i \) and \( j \), we have \( {\sigma }^{i} = {\sigma }^{j} \) if and only if \( i \equiv j\left( {\;\operatorname{mod}\;\ell }\right) \) . | Proof. We may assume \( i \geq j \) . We have\n\n\[ \n{\sigma }^{i} = {\sigma }^{j} \Leftrightarrow {\sigma }^{i - j} = {\sigma }^{0} \Leftrightarrow {\alpha }^{{q}^{i - j}} - \alpha = 0\text{ for all }\alpha \in E \n\] \n\n\[ \n\Leftrightarrow \left( {\mathop{\prod }\limits_{{\alpha \in E}}\left( {X - \alpha }\right) ... | Yes |
For \( k \geq 1 \), let \( {P}_{k} \) denote the product of all the monic irreducible polynomials in \( F\left\lbrack X\right\rbrack \) of degree \( k \) . For all positive integers \( \ell \), we have\n\n\[ \n{X}^{{q}^{\ell }} - X = \mathop{\prod }\limits_{{k \mid \ell }}{P}_{k} \n\]\n\nwhere the product is over all p... | Proof. First, we claim that the polynomial \( {X}^{{q}^{\ell }} - X \) is square-free. This follows immediately from Theorem 19.1, since \( \mathbf{D}\left( {{X}^{{q}^{\ell }} - X}\right) = {q}^{\ell }{X}^{{q}^{\ell } - 1} - 1 = - 1 \) .\n\nThus, we have reduced the proof to showing that if \( f \) is a monic irreducib... | Yes |
Theorem 19.11. For all \( \ell \geq 1 \), we have\n\n\[ \n{q}^{\ell } = \mathop{\sum }\limits_{{k \mid \ell }}k{\Pi }_{F}\left( k\right) \n\] | Proof. Just equate the degrees of both sides of the identity in Theorem 19.10. | No |
Theorem 19.12. For all \( \ell \geq 1 \), we have\n\n\[ \frac{{q}^{\ell }}{2\ell } \leq {\Pi }_{F}\left( \ell \right) \leq \frac{{q}^{\ell }}{\ell } \]\n\n(19.2)\n\nand\n\n\[ {\Pi }_{F}\left( \ell \right) = \frac{{q}^{\ell }}{\ell } + O\left( \frac{{q}^{\ell /2}}{\ell }\right) . \]\n\n(19.3) | Proof. First, since all the terms in the sum on the right hand side of (19.1) are non-negative, and \( \ell {\Pi }_{F}\left( \ell \right) \) is one of these terms, we may deduce that \( \ell {\Pi }_{F}\left( \ell \right) \leq {q}^{\ell } \) , which proves the second inequality in (19.2). Since this holds for all \( \el... | Yes |
Theorem 19.13. Let \( E \) be an extension of degree \( \ell \) over a finite field \( F \) . Let \( \sigma \) be the Frobenius map on \( E \) over \( F \) . Then the intermediate fields \( K \), with \( F \subseteq K \subseteq E \) , are in one-to-one correspondence with the divisors \( k \) of \( \ell \), where the d... | Proof. Let \( q \) be the cardinality of \( F \) .\n\nSuppose \( k \) is a divisor of \( \ell \) . By Theorem 19.6 (applied to \( E \) ), the polynomial \( {X}^{{q}^{\ell }} - X \) splits into distinct monic linear factors over \( E \) . By Theorem 19.4, the polynomial \( {X}^{{q}^{k}} - X \) divides \( {X}^{{q}^{\ell ... | Yes |
Theorem 19.14. Let \( E \) and \( {E}^{\prime } \) be finite extensions of the same degree over a finite field \( F \) . Then \( E \) and \( {E}^{\prime } \) are isomorphic as \( F \) -algebras. | Proof. Let \( q \) be the cardinality of \( F \), and let \( \ell \) be the degree of the extensions. As we have argued before, we have \( {E}^{\prime } = F\left\lbrack {\alpha }^{\prime }\right\rbrack \) for some \( {\alpha }^{\prime } \in {E}^{\prime } \), and so \( {E}^{\prime } \) is isomorphic as an \( F \) -algeb... | Yes |
Theorem 19.16. If \( \alpha \in {E}^{ * } \) has multiplicative order \( r \), then the degree of \( \alpha \) over \( F \) is equal to the multiplicative order of \( q \) modulo \( r \) . | For \( \alpha \in E \), define the polynomial\n\n\[ \chi \mathrel{\text{:=}} \mathop{\prod }\limits_{{i = 0}}^{{\ell - 1}}\left( {X - {\sigma }^{i}\left( \alpha \right) }\right) \]\n\nIt is easy to see, using the same type of argument as was used to prove Theorem 19.15, that \( \chi \in F\left\lbrack X\right\rbrack \),... | No |
Theorem 19.17. The function \( {\mathbf{N}}_{E/F} \), restricted to \( {E}^{ * } \), is a group homomorphism from \( {E}^{ * } \) onto \( {F}^{ * } \) . | Proof. We have\n\n\[ \n{\mathbf{N}}_{E/F}\left( \alpha \right) = \mathop{\prod }\limits_{{i = 0}}^{{\ell - 1}}{\alpha }^{{q}^{i}} = {\alpha }^{\mathop{\sum }\limits_{{i = 0}}^{{\ell - 1}}{q}^{i}} = {\alpha }^{\left( {{q}^{\ell } - 1}\right) /\left( {q - 1}\right) }. \n\]\n\nSince \( {E}^{ * } \) is a cyclic group of or... | Yes |
As an application of some of the above theory, let us investigate the factorization of the polynomial \( {X}^{r} - 1 \) over \( F \), a finite field of cardinality \( q \) . Let us assume that \( r > 0 \) and is relatively prime to \( q \) . Let \( E \) be a splitting field of \( {X}^{r} - 1 \) (see Theorem 16.25), so ... | \[ {X}^{r} - 1 = \mathop{\prod }\limits_{{i = 1}}^{r}\left( {X - {\alpha }_{i}}\right) \] We claim that the roots \( {\alpha }_{i} \) of \( {X}^{r} - 1 \) are distinct-this follows from the Theorem 19.1 and the fact that \( \gcd \left( {{X}^{r} - 1, r{X}^{r - 1}}\right) = 1 \) . Next, observe that the \( r \) roots of ... | Yes |
Theorem 20.1. Algorithm IPT uses \( O\left( {{\ell }^{3}\operatorname{len}\left( q\right) }\right) \) operations in \( F \) . | Proof. Consider an execution of a single iteration of the main loop. The cost of the \( q \) th-powering step (using a standard repeated-squaring algorithm) is \( O\left( {\operatorname{len}\left( q\right) }\right) \) multiplications modulo \( f \), and so \( O\left( {{\ell }^{2}\operatorname{len}\left( q\right) }\righ... | Yes |
Theorem 20.2. Algorithm RIP uses an expected number of \( O\left( {{\ell }^{4}\operatorname{len}\left( q\right) }\right) \) operations in \( F \), and its output is uniformly distributed over all monic irreducibles of degree \( \ell \) . | Proof. This is a simple application of the generate-and-test paradigm (see Theorem 9.3, and Example 9.10 in particular). Because of Theorem 19.12, the expected number of loop iterations of the above algorithm is \( O\left( \ell \right) \) . Since Algorithm IPT uses \( O\left( {{\ell }^{3}\operatorname{len}\left( q\righ... | Yes |
Theorem 20.3. Suppose that \( f \in F\left\lbrack X\right\rbrack \) is a monic polynomial of degree \( \ell > 0 \) , and that \( \gcd \left( {f,\mathbf{D}\left( f\right) }\right) = f \) . Then \( f = g\left( {X}^{p}\right) \) for some \( g \in F\left\lbrack X\right\rbrack \) . Moreover, if \( g = \mathop{\sum }\limits_... | Proof. Since \( \deg \left( {\mathbf{D}\left( f\right) }\right) < \deg \left( f\right) \) and \( \gcd \left( {f,\mathbf{D}\left( f\right) }\right) = f \), we must have \( \mathbf{D}\left( f\right) = 0 \) . If \( f = \mathop{\sum }\limits_{i}{c}_{i}{X}^{i} \), then \( \mathbf{D}\left( f\right) = \mathop{\sum }\limits_{i... | Yes |
Theorem 20.4. Let \( f \in F\left\lbrack X\right\rbrack \) be a monic polynomial of degree \( \ell > 0 \) . Suppose that the factorization of \( f \) into irreducibles is \( f = {f}_{1}^{{e}_{1}}\cdots {f}_{r}^{{e}_{r}} \) . Then\n\n\[ \frac{f}{\gcd \left( {f,\mathbf{D}\left( f\right) }\right) } = \mathop{\prod }\limit... | Proof. The theorem can be restated in terms of the following claim: for each \( i = 1,\ldots, r \), we have\n\n- \( {f}_{i}^{{e}_{i}} \mid \mathbf{D}\left( f\right) \) if \( {e}_{i} \equiv 0\left( {\;\operatorname{mod}\;p}\right) \), and\n\n- \( {f}_{i}^{{e}_{i} - 1} \mid \mathbf{D}\left( f\right) \) but \( {f}_{i}^{{e... | Yes |
Theorem 20.6. Algorithm DDF uses \( O\left( {{\ell }^{3}\operatorname{len}\left( q\right) }\right) \) operations in \( F \) . | Proof. Note that the body of the main loop is executed at most \( \ell \) times, since after \( \ell \) iterations, we will have removed all the factors of \( f \) . Thus, we perform at most \( \ell q \) th-powering steps, each of which takes \( O\left( {{\ell }^{2}\operatorname{len}\left( q\right) }\right) \) operatio... | Yes |
Theorem 20.7. In the case \( p = 2 \), Algorithm EDF uses an expected number of \( O\left( {k{\ell }^{2}\operatorname{len}\left( q\right) }\right) \) operations in \( F \) . | Proof. We may assume \( r \geq 2 \) . Let \( L \) be the random variable that represents the number of iterations of the main loop of the algorithm. For \( n \geq 1 \), let \( {H}_{n} \) be the random variable that represents the value of \( H \) at the beginning of the \( n \) th loop iteration. For \( i, j = 1,\ldots... | Yes |
Theorem 20.8. In the case \( p > 2 \), Algorithm EDF uses an expected number of \( O\left( {k{\ell }^{2}\operatorname{len}\left( q\right) }\right) \) operations in \( F \) . | Proof. The analysis is essentially the same as in the case \( p = 2 \), except that now the probability that we fail to split a given pair of irreducible factors is at most \( 5/9 \) , rather than equal to \( 1/2 \) . The details are left as an exercise for the reader. | No |
Theorem 20.9. The Cantor-Zassenhaus factoring algorithm uses an expected number of \( O\left( {{\ell }^{3}\operatorname{len}\left( q\right) }\right) \) operations in \( F \) . | This bound is tight, since in the worst case, when the input is irreducible, the algorithm really does do this much work. Also, we have assumed the input to the Cantor-Zassenhaus is a square-free polynomial. However, we may use Algorithm SFD as a preprocessing step to ensure that this is the case. Even if we include th... | Yes |
Theorem 20.10. Algorithm B1 uses \( O\left( {{\ell }^{3} + {\ell }^{2}\operatorname{len}\left( q\right) }\right) \) operations in \( F \) . | Proof. This is just a matter of counting. The computation of \( \alpha \) takes \( O\left( {\operatorname{len}\left( q\right) }\right) \) operations in \( E \) using repeated squaring, and hence \( O\left( {{\ell }^{2}\operatorname{len}\left( q\right) }\right) \) operations in \( F \) . To build the matrix \( Q \), we ... | Yes |
Theorem 20.11. Algorithm B2 uses an expected number of\n\n\[ O\left( {\operatorname{len}\left( r\right) {\ell }^{2}\operatorname{len}\left( q\right) }\right) \]\n\noperations in \( F \) . | Proof. The proof follows the same line of reasoning as the analysis of Algorithm EDF. Indeed, using the same argument as was used there, the expected number of iterations of the main loop is \( O\left( {\operatorname{len}\left( r\right) }\right) \) . As discussed in the paragraph above this theorem, the cost per loop i... | Yes |
Theorem 20.12. Berlekamp's factoring algorithm uses an expected number of \( O\left( {{\ell }^{3} + {\ell }^{2}\operatorname{len}\left( \ell \right) \operatorname{len}\left( q\right) }\right) \) operations in \( F \) . | We have assumed the input to Berlekamp's algorithm is a square-free polynomial. However, we may use Algorithm SFD as a preprocessing step to ensure that this is the case. Even if we include the cost of this preprocessing step, the running time estimate in Theorem 20.12 remains valid. | No |
Theorem 21.1. Let \( n > 1 \) be an integer. If \( n \) is prime, then for all \( a \in {\mathbb{Z}}_{n} \), we have the following identity in the ring \( {\mathbb{Z}}_{n}\left\lbrack X\right\rbrack \) :\n\n\[ \n{\left( X + a\right) }^{n} = {X}^{n} + a.\n\]\n\nConversely, if \( n \) is composite, then for all \( a \in ... | Proof. Note that\n\n\[ \n{\left( X + a\right) }^{n} = {X}^{n} + {a}^{n} + \mathop{\sum }\limits_{{i = 1}}^{{n - 1}}\left( \begin{matrix} n \\ i \end{matrix}\right) {a}^{i}{X}^{n - i}.\n\]\n\nIf \( n \) is prime, then by Fermat’s little theorem (Theorem 2.14), we have \( {a}^{n} = a \) , and by Exercise 1.14, all of the... | Yes |
Theorem 21.2. For integers \( n > 1 \) and \( m \geq 1 \), the least prime \( r \) such that \( r \nmid n \) and the multiplicative order of \( {\left\lbrack n\right\rbrack }_{r} \in {\mathbb{Z}}_{r}^{ * } \) is greater than \( m \) is \( O\left( {{m}^{2}\operatorname{len}\left( n\right) }\right) \) . | Proof. Call a prime \( r \) \ | No |
Theorem 21.3. Algorithm AKS can be implemented so that its running time is \( O\left( {\operatorname{len}{\left( n\right) }^{16.5}}\right) \) . | Proof. As discussed above, the value of \( r \) determined in step 2 will be \( O\left( {\operatorname{len}{\left( n\right) }^{5}}\right) \) . It is fairly straightforward to see that the running time of the algorithm is dominated by the running time of step 5 . Here, we have to perform \( O\left( {{r}^{1/2}\operatorna... | Yes |
Theorem 21.4. If the input to Algorithm AKS is prime, then the output is true. | Proof. Assume that the input \( n \) is prime. The test in step 1 will certainly fail. If the algorithm does not return true in step 3, then certainly the test in step 4 will fail as well. If the algorithm reaches step 5 , then all of the tests in the loop in step 5 will fail-this follows from Theorem 21.1. | Yes |
For all \( k \in {\mathbb{Z}}^{\left( r\right) } \), the kernel of \( {\widehat{\sigma }}_{k} \) is \( \left( {{X}^{r} - 1}\right) \), and the image of \( {\widehat{\sigma }}_{k} \) is \( E \) . | Proof. Let \( J \mathrel{\text{:=}} \operatorname{Ker}{\widehat{\sigma }}_{k} \), which is an ideal of \( {\mathbb{Z}}_{p}\left\lbrack X\right\rbrack \) . Let \( {k}^{\prime } \) be a positive integer such that \( k{k}^{\prime } \equiv 1\left( {\;\operatorname{mod}\;r}\right) \), which exists because \( \gcd \left( {r,... | Yes |
Lemma 21.7. For every \( \alpha \in E \), if \( k \in C\left( \alpha \right) \) and \( {k}^{\prime } \in C\left( \alpha \right) \), then \( k{k}^{\prime } \in C\left( \alpha \right) \) . | Proof. If \( {\sigma }_{k}\left( \alpha \right) = {\alpha }^{k} \) and \( {\sigma }_{{k}^{\prime }}\left( \alpha \right) = {\alpha }^{{k}^{\prime }} \), then\n\n\[ \n{\sigma }_{k{k}^{\prime }}\left( \alpha \right) = {\sigma }_{k}\left( {{\sigma }_{{k}^{\prime }}\left( \alpha \right) }\right) = {\sigma }_{k}\left( {\alp... | Yes |
Lemma 21.8. For every \( k \in {\mathbb{Z}}^{\left( r\right) } \), if \( \alpha \in D\left( k\right) \) and \( \beta \in D\left( k\right) \), then \( {\alpha \beta } \in D\left( k\right) \) . | Proof. If \( {\sigma }_{k}\left( \alpha \right) = {\alpha }^{k} \) and \( {\sigma }_{k}\left( \beta \right) = {\beta }^{k} \), then\n\n\[ \n{\sigma }_{k}\left( {\alpha \beta }\right) = {\sigma }_{k}\left( \alpha \right) {\sigma }_{k}\left( \beta \right) = {\alpha }^{k}{\beta }^{k} = {\left( \alpha \beta \right) }^{k}, ... | Yes |
Lemma 21.11. Under assumptions (A4) and (A5), we have\n\n\[ \n{2}^{\min \left( {t,\ell }\right) } - 1 > {n}^{2\left\lfloor {t}^{1/2}\right\rfloor } \n\] | Proof. Observe that \( {\log }_{2}n \leq \operatorname{len}\left( n\right) \), and so it suffices to show that\n\n\[ \n{2}^{\min \left( {t,\ell }\right) } - 1 > {2}^{2\operatorname{len}\left( n\right) \left\lfloor {t}^{1/2}\right\rfloor } \n\]\n\nand for this, it suffices to show that\n\n\[ \n\min \left( {t,\ell }\righ... | Yes |
For which real numbers \( x \) do you have \( \left| x\right| = 3 \) ? | Since \( \left| 3\right| = 3 \) and \( \left| {-3}\right| = 3 \), we see that there are two solutions, \( x = 3 \) or \( x = - 3 \) . The solution set is \( S = \{ - 3,3\} \). | Yes |
Example 1.6. Solve for \( x : \left| x\right| = - 7 \) . | Solution. Note that \( \left| {-7}\right| = 7 \) and \( \left| 7\right| = 7 \) so that these cannot give any solutions. Indeed, there are no solutions, since the absolute value is always non-negative. The solution set is the empty set \( S = \{ \} \) . | Yes |
Example 1.7. Solve for \( x : \left| x\right| = 0 \) . | Solution. Since \( - 0 = 0 \), there is only one solution, \( x = 0 \) . Thus, \( S = \{ 0\} \) . | Yes |
Solve for \( x : \left| {x + 2}\right| = 6 \) . | Since the absolute value of \( x + 2 \) is 6, we see that \( x + 2 \) has to be either 6 or -6 . We evaluate each case,\n\n\[ \n\begin{array}{ll} \text{ either }x + 2 = 6, & \text{ or }x + 2 = - 6, \\ \Rightarrow x = 6 - 2, & \Rightarrow x = - 6 - 2, \\ \Rightarrow x = 4; & \Rightarrow x = - 8. \end{array} \n\]\n\nThe ... | Yes |
Example 1.9. Solve for \( x : \left| {{3x} - 4}\right| = 5 \) | Solution.\n\n\[ \n\begin{array}{ll} \text{ Either }{3x} - 4 = 5 & \text{ or }{3x} - 4 = - 5 \\ \Rightarrow {3x} = 9 & \Rightarrow {3x} = - 1 \\ \Rightarrow x = 3 & \Rightarrow x = - \frac{1}{3} \end{array} \n\] \n\nThe solution set is \( S = \left\{ {-\frac{1}{3},3}\right\} \) . | Yes |
Solve for \( x : - 2 \cdot \left| {{12} + {3x}}\right| = - {18} \) | Solution. Dividing both sides by -2 gives \( \left| {{12} + {3x}}\right| = 9 \) . With this, we have the two cases\n\n\[ \begin{array}{ll} \text{ Either }{12} + {3x} = 9 & \text{ or }{12} + {3x} = - 9 \\ \Rightarrow {3x} = - 3 & \Rightarrow {3x} = - {21} \\ \Rightarrow x = - 1 & \Rightarrow x = - 7 \end{array} \]\n\nTh... | Yes |
Example 1.14. Graph the the inequality \( \pi < x \leq 5 \) on the number line and write it in interval notation. | Solution.\n\nOn the number line:\n\n\n\nInterval notation: | No |
Example 1.15. Write the following interval as an inequality and in interval notation: | Solution.\n\n\[ \text{Inequality notation:}\; - 3 \leq x \]\n\n\[ \text{Interval notation:}\;\lbrack - 3,\infty ) \] | No |
Example 1.16. Write the following interval as an inequality and in interval notation: | Solution.\n\n\[ \text{Inequality notation:}\;x < 2 \]\n\n\[ \text{Interval notation:}\;\left( {-\infty ,2}\right) \] | Yes |
Solve for \( x \) : a) \( \left| {x + 7}\right| < 2 \) | Solution. a) We follow the three steps described above. In step 1, we solve the corresponding equality, \( \left| {x + 7}\right| = 2 \). \n\n\[ \n\begin{array}{l} x + 7 = 2 \\ \Rightarrow x = - 5 \end{array}\left| {\;\begin{array}{l} x + 7 = - 2 \\ \Rightarrow x = - 9 \end{array}}\right. \n\] \n\nThe solutions \( x = -... | Yes |
Solve for \( x : \;\left| {{12} - {5x}}\right| \leq 1 \) | Note that \( \left| {{12} - {5x}}\right| \leq 1 \) implies that\n\n\[- 1 \leq {12} - {5x} \leq 1\]\n\nso that\n\n\[- {13} \leq - {5x} \leq - {11}\]\n\nand by dividing by -5 (remembering to switch the direction of the inequalities when multiplying or dividing by a negative number) we see that\n\n\[\frac{13}{5} \geq x \g... | Yes |
Example 1.22. Solve for \( x \) : a) \( \left| {x - 6}\right| = 4 \) | Solution. a) Consider the distance between \( x \) and 6 to be 4 on a number line:\n\n\n\nThere are two solutions, \( x = 2 \) or \( x = {10} \) . That is, the distance between 2 and 6 is 4 and the distance between 10 ... | Yes |
Example 2.2. Graph the line \( y = {2x} + 3 \) . | Solution. We calculate \( y \) for various values of \( x \) . For example, when \( x \) is \( - 2, - 1,0,1,2 \), or 3, we calculate\n\n<table><tr><td>\( x \)</td><td>\( - 2 \)</td><td>\( - 1 \)</td><td>0</td><td>1</td><td>2</td><td>3</td></tr><tr><td>\( y \)</td><td>\( - 1 \)</td><td>1</td><td>3</td><td>5</td><td>7</t... | Yes |
Example 2.3. Find the equation of the line in slope-intercept form. | Solution. The \( y \) -intercept can be read off the graph giving us that \( b = 2 \) . As for the slope, we use formula (2.1) and the two points on the line \( {P}_{1}\left( {0,2}\right) \) and \( {P}_{2}\left( {4,0}\right) \) . We obtain\n\n\[ m = \frac{0 - 2}{4 - 0} = \frac{-2}{4} = - \frac{1}{2}. \]\n\nThus, the li... | Yes |
Example 2.4. Find the equation of the line in slope-intercept form. | Solution. The \( y \) -intercept is \( b = - 4 \) . To obtain the slope we can again use the \( y \) -intercept \( {P}_{1}\left( {0, - 4}\right) \) . To use (2.1), we need another point \( {P}_{2} \) on the line. We may pick any second point on the line, for example, \( {P}_{2}\left( {3, - 3}\right) \) . With this,\n\n... | Yes |
Find the equation of the line in point-slope form (2.2). | Solution. We need to identify one point \( \left( {{x}_{1},{y}_{1}}\right) \) on the line together with the slope \( m \) of the line so that we can write the line in point-slope form: \( y - {y}_{1} = m\left( {x - {x}_{1}}\right) \) . By direct inspection, we identify the two points \( {P}_{1}\left( {5,1}\right) \) an... | Yes |
Find the slope, find the \( y \) -intercept, and graph the line\n\n\[ {4x} + {2y} - 2 = 0. \] | Solution. We first rewrite the equation in slope-intercept form.\n\n\[ {4x} + {2y} - 2 = 0\overset{\left( -4x + 2\right) }{ \Rightarrow }{2y} = - {4x} + 2 \]\n\n\[ \overset{\text{ (divide 2) }}{ \Rightarrow }y = - {2x} + 1 \]\n\nWe see that the slope is -2 and the \( y \) -intercept is \( \left( {0,1}\right) \).\n\nWe ... | Yes |
Find the slope, \( y \) -intercept, and graph the line \( {5y} + {2x} = - {10} \) . | Solution. Again, we first rewrite the equation in slope-intercept form.\n\n\[ \n{5y} + {2x} = - {10}\;\overset{\left( \text{subtract }2x\right) }{ \Rightarrow }\;{5y} = - {2x} - {10} \n\]\n\n\[ \n\overset{\left( \text{divide }5\right) }{ \Rightarrow }\;y = \frac{-{2x} - {10}}{5} \n\]\n\n\[ \n\Rightarrow \;y = - \frac{2... | Yes |
Define the assignment \( f \) by the following table\n\n<table><tr><td>\( x \)</td><td>2</td><td>5</td><td>\( - 3 \)</td><td>0</td><td>7</td><td>4</td></tr><tr><td>\( y \)</td><td>6</td><td>8</td><td>6</td><td>4</td><td>\( - 1 \)</td><td>8</td></tr></table>\n\nThe assignment \( f \) assigns to the input 2 the output 6,... | The domain \( D \) is the set of all inputs. The domain is therefore\n\n\[ D = \{ - 3,0,2,4,5,7\} \]\n\nThe range \( R \) is the set of all outputs. The range is therefore\n\n\[ R = \{ - 1,4,6,8\} \]\n\nThe assignment \( f \) is indeed a function since each element of the domain gets assigned exactly one element in the... | Yes |
Example 2.12. Consider the assignment \( f \) that is given by the following table.\n\n<table><tr><td>\( x \)</td><td>2</td><td>5</td><td>\( - 3 \)</td><td>0</td><td>5</td><td>4</td></tr><tr><td>\( y \)</td><td>6</td><td>8</td><td>6</td><td>4</td><td>\( - 1 \)</td><td>8</td></tr></table>\n\nThis assignment does not def... | Consider the input value 5 . What does \( f \) assign to the input 5 ? The third column states that \( f \) assigns to 5 the output 8, whereas the sixth column states that \( f \) assigns to 5 the output -1,\n\n\[ f\left( 5\right) = 8,\;f\left( 5\right) = - 1.\]\n\nHowever, by the definition of a function, to each inpu... | Yes |
A university creates a mentoring program, which matches each freshman student with a senior student as his or her mentor. Within this program it is guaranteed that each freshman gets precisely one mentor, however two freshmen may receive the same mentor. Does the assignment of freshmen to mentor, or mentor to freshmen ... | Since a senior may mentor several freshman, we cannot take a mentor as an \ | No |
a) Is the rainfall a function of the month?\nb) Is the month a function of the rainfall? | a) Each month has exactly one amount of rainfall associated to it. Therefore, the assignment that associates to a month its rainfall (in inches) is a function.\nb) If we take a certain rainfall amount as our input data, can we associate a unique month to it? For example, February and March have the same amount of rainf... | Yes |
Consider the function \( y = {5x} + 4 \) with domain all real numbers and range all real numbers. Note that for each input \( x \), we obtain an exactly one induced output \( y \) . | For example, for the input \( x = 3 \) we get the output \( y = 5 \cdot 3 + 4 = {19} \), etc. | Yes |
For each real number \( x \), denote by \( \lfloor x\rfloor \) the greatest integer that is less or equal to \( x \). We call \( \lfloor x\rfloor \) the floor of \( x \). | For example, to calculate \( \lfloor {4.37}\rfloor \), note that all integers \( 4,3,2,\ldots \) are less or equal to 4.37 :\n\n\[ \ldots , - 3, - 2, - 1,0,1,2,3,4\; \leq \;{4.37} \]\n\nThe greatest of these integers is 4, so that \( \lfloor {4.37}\rfloor = 4 \) . We define the floor function as \( f\left( x\right) = \... | Yes |
Let \( A \) be the area of an isosceles right triangle with base side length \( x \) . Express \( A \) as a function of \( x \) . | Being an isosceles right triangle means that two side lengths are \( x \), and the angles are \( {45}^{ \circ },{45}^{ \circ } \), and \( {90}^{ \circ } \) (or in radian measure \( \frac{\pi }{4},\frac{\pi }{4} \), and \( \left. \frac{\pi }{2}\right) \) :\n\n . This equation associates to each input number \( a \) exactly one output number \( b = {a}^{2} + 3 \) . Therefore, the equation defines a function. | The domain \( D \) is all real numbers, \( D = \mathbb{R} \) . Since \( {x}^{2} \) is always \( \geq 0 \), we see that \( {x}^{2} + 3 \geq 3 \), and vice versa every number \( y \geq 3 \) can be written as \( y = {x}^{2} + 3 \) . (To see this, note that the input \( x = \sqrt{y - 3} \) for \( y \geq 3 \) gives the outp... | Yes |
Consider the equation \( {x}^{2} + {y}^{2} = {25} \) . Does this equation define \( y \) as a function of \( x \) ? That is, does this equation assign to each input \( x \) exactly one output \( y \) ? | An input number \( x \) gets assigned to \( y \) with \( {x}^{2} + {y}^{2} = {25} \) . Solving this for \( y \), we obtain\n\n\[ \n{y}^{2} = {25} - {x}^{2}\; \Rightarrow \;y = \pm \sqrt{{25} - {x}^{2}}. \n\]\n\nTherefore, there are two possible outputs associated to the input \( x\left( { \neq 5}\right) \) :\n\n\[ \n\t... | Yes |
For the given function \( f \), calculate the outputs \( f\left( 2\right), f\left( {-3}\right) \) , and \( f\left( {-1}\right) \) . | Solution. We substitute the input values into the function and simplify.\n\n\[ \text{a)}\;f\left( 2\right) = 3 \cdot 2 + 4 = 6 + 4 = {10}\text{,}\]\n\n\[ f\left( {-3}\right) = 3 \cdot \left( {-3}\right) + 4 = - 9 + 4 = - 5,\]\n\n\[ f\left( {-1}\right) = 3 \cdot \left( {-1}\right) + 4 = - 3 + 4 = 1.\] | Yes |
Example 3.2. Let \( f \) be the function given by \( f\left( x\right) = {x}^{2} + {2x} - 3 \) . Find the following function values.\n\na) \( f\left( 5\right) \) b) \( f\left( 2\right) \) c) \( f\left( {-2}\right) \) d) \( f\left( 0\right) \)\n\ne) \( f\left( \sqrt{5}\right) \) f) \( f\left( {\sqrt{3} + 1}\right) \) g) ... | Solution. The first four function values ((a)-(d)) can be calculated directly:\n\n\[ f\left( 5\right) = {5}^{2} + 2 \cdot 5 - 3 = {25} + {10} - 3 = {32}, \]\n\n\[ f\left( 2\right) = {2}^{2} + 2 \cdot 2 - 3 = 4 + 4 - 3 = 5, \]\n\n\[ f\left( {-2}\right) = {\left( -2\right) }^{2} + 2 \cdot \left( {-2}\right) - 3 = 4 + - 4... | Yes |
Calculate the difference quotient \( \frac{f\left( {x + h}\right) - f\left( x\right) }{h} \) for a) \( f\left( x\right) = {x}^{3} + 2 \) | Solution. We calculate first the difference quotient step by step.\n\n\[ \text{a)}f\left( {x + h}\right) = {\left( x + h\right) }^{3} + 2 = \left( {x + h}\right) \cdot \left( {x + h}\right) \cdot \left( {x + h}\right) + 2 \]\n\n\[ = \left( {{x}^{2} + {2xh} + {h}^{2}}\right) \cdot \left( {x + h}\right) + 2 \]\n\n\[ = {x... | Yes |
Example 3.5. Find the domain of each of the following functions. | Solution.\n\na) There is no problem taking a real number \( x \) to any (positive) power. Therefore, \( f \) is defined for all real numbers \( x \), and the domain is written as \( D = \mathbb{R} \) .\n\nb) Again, we can take the absolute value for any real number \( x \) . The domain is all real numbers, \( D = \math... | Yes |
Let \( y = {x}^{2} \) with domain \( D = \mathbb{R} \) being the set of all real numbers. We can graph this after calculating a table as follows: | <table><tr><td>\( x \)</td><td>\( - 3 \)</td><td>\( - 2 \)</td><td>\( - 1 \)</td><td>0</td><td>1</td><td>2</td><td>3</td></tr><tr><td>\( y \)</td><td>9</td><td>4</td><td>1</td><td>0</td><td>1</td><td>4</td><td>9</td></tr></table> | Yes |
Let \( f \) be the function given by the following graph. | Here, the dashed lines show, that the input \( x = 3 \) gives an output of \( y = 2 \) . Similarly, we can obtain other output values from the graph:\n\n\[ f\left( 2\right) = 4,\;f\left( 3\right) = 2,\;f\left( 5\right) = 2,\;f\left( 7\right) = 4. \]\n\nNote, that in the above graph, a closed point means that the point ... | Yes |
Let \( f \) be the function given by the following graph. | Here are some function values that can be read from the graph:\n\n\[ f\left( {-5}\right) = 2,\;f\left( {-4}\right) = 3,\;f\left( {-3}\right) \text{and}f\left( {-2}\right) \text{are undefined,}\]\n\n\[ f\left( {-1}\right) = 2,\;f\left( 0\right) = 1,\;f\left( 1\right) = 2,\;f\left( 2\right) = - 1,\;f\left( 4\right) = 0,\... | Yes |
Consider the input \( x = 4 \) . There are several outputs that we get for \( x = 4 \) from this graph:\n\n\[ f\left( 4\right) = 1,\;f\left( 4\right) = 2,\;f\left( 4\right) = 3. \]\n\nHowever, in a function, it is not allowed to obtain more than one output for one input! Therefore, this graph is not the graph of a func... | The reason why the previous example is not a function is due to some input having more than one output: \( f\left( 4\right) = 1, f\left( 4\right) = 2, f\left( 4\right) = 3 \) .\n\n\n\nIn other words, there is a vertica... | Yes |
Consider the graph of the equation \( x = {y}^{2} \) | This does not pass the vertical line test so \( y \) is not a function of \( x \) . However, \( x \) is a function of \( y \) since, if you consider \( y \) to be the input, each input has exactly one output (it passes the 'horizontal line' test). | Yes |
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