Mixing
Show $aBbb Z+bBbb Z = gcd(a,b)Bbb Z$
$begingroup$
I have the following problem:
Let $a, b inmathbb{Z}$. Show that $,{ ax + by : x, y in mathbb{Z}} = { n gcd(a,b) : nin mathbb{Z} }$
I understand that the Bezout's lemma says that $gcd(a,b) = ax +by$, so Im not really how you would go about proving the above, it doesn't really make sense to me. Any help is appreciated1
elementary-number-theory proof-writing divisibility
edited Jan 11 at 0:16
Bill Dubuque
210k29192641
asked Mar 19 '14 at 3:58
Alex ChavezAlex Chavez
438720
$endgroup$
add a comment |
$begingroup$
I have the following problem:
Let $a, b inmathbb{Z}$. Show that $,{ ax + by : x, y in mathbb{Z}} = { n gcd(a,b) : nin mathbb{Z} }$
I understand that the Bezout's lemma says that $gcd(a,b) = ax +by$, so Im not really how you would go about proving the above, it doesn't really make sense to me. Any help is appreciated1
elementary-number-theory proof-writing divisibility
edited Jan 11 at 0:16
Bill Dubuque
210k29192641
asked Mar 19 '14 at 3:58
Alex ChavezAlex Chavez
438720
$endgroup$
$begingroup$
do you have any other conditions on $x$ and $y$ ?
$endgroup$
– wanderer
Mar 19 '14 at 4:11
$begingroup$
nope, theyre just integers
$endgroup$
– Alex Chavez
Mar 19 '14 at 4:15
$begingroup$
This is closely related to Proof of Extended Euclidean Algorithm?
$endgroup$
– robjohn♦
Apr 4 '15 at 14:08
add a comment |
$begingroup$
I have the following problem:
Let $a, b inmathbb{Z}$. Show that $,{ ax + by : x, y in mathbb{Z}} = { n gcd(a,b) : nin mathbb{Z} }$
I understand that the Bezout's lemma says that $gcd(a,b) = ax +by$, so Im not really how you would go about proving the above, it doesn't really make sense to me. Any help is appreciated1
elementary-number-theory proof-writing divisibility
edited Jan 11 at 0:16
Bill Dubuque
210k29192641
asked Mar 19 '14 at 3:58
Alex ChavezAlex Chavez
438720
$endgroup$
I have the following problem:
Let $a, b inmathbb{Z}$. Show that $,{ ax + by : x, y in mathbb{Z}} = { n gcd(a,b) : nin mathbb{Z} }$
I understand that the Bezout's lemma says that $gcd(a,b) = ax +by$, so Im not really how you would go about proving the above, it doesn't really make sense to me. Any help is appreciated1
elementary-number-theory proof-writing divisibility
elementary-number-theory proof-writing divisibility
edited Jan 11 at 0:16
Bill Dubuque
210k29192641
asked Mar 19 '14 at 3:58
Alex ChavezAlex Chavez
438720
edited Jan 11 at 0:16
Bill Dubuque
210k29192641
asked Mar 19 '14 at 3:58
Alex ChavezAlex Chavez
438720
edited Jan 11 at 0:16
Bill Dubuque
210k29192641
edited Jan 11 at 0:16
Bill Dubuque
210k29192641
edited Jan 11 at 0:16
Bill Dubuque
210k29192641
210k29192641
asked Mar 19 '14 at 3:58
Alex ChavezAlex Chavez
438720
asked Mar 19 '14 at 3:58
Alex ChavezAlex Chavez
438720
asked Mar 19 '14 at 3:58
Alex ChavezAlex Chavez
438720
438720
$begingroup$
do you have any other conditions on $x$ and $y$ ?
$endgroup$
– wanderer
Mar 19 '14 at 4:11
$begingroup$
nope, theyre just integers
$endgroup$
– Alex Chavez
Mar 19 '14 at 4:15
$begingroup$
This is closely related to Proof of Extended Euclidean Algorithm?
$endgroup$
– robjohn♦
Apr 4 '15 at 14:08
add a comment |
$begingroup$
do you have any other conditions on $x$ and $y$ ?
$endgroup$
– wanderer
Mar 19 '14 at 4:11
$begingroup$
nope, theyre just integers
$endgroup$
– Alex Chavez
Mar 19 '14 at 4:15
$begingroup$
This is closely related to Proof of Extended Euclidean Algorithm?
$endgroup$
– robjohn♦
Apr 4 '15 at 14:08
$begingroup$
do you have any other conditions on $x$ and $y$ ?
$endgroup$
– wanderer
Mar 19 '14 at 4:11
$begingroup$
do you have any other conditions on $x$ and $y$ ?
$endgroup$
– wanderer
Mar 19 '14 at 4:11
$begingroup$
nope, theyre just integers
$endgroup$
– Alex Chavez
Mar 19 '14 at 4:15
$begingroup$
nope, theyre just integers
$endgroup$
– Alex Chavez
Mar 19 '14 at 4:15
$begingroup$
This is closely related to Proof of Extended Euclidean Algorithm?
$endgroup$
– robjohn♦
Apr 4 '15 at 14:08
$begingroup$
This is closely related to Proof of Extended Euclidean Algorithm?
$endgroup$
– robjohn♦
Apr 4 '15 at 14:08
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
By Bezout, $ ngcd(a,b) = n(aj+bk),$ $Rightarrow$ $,gcd(a,b),Bbb Zsubseteq aBbb Z+bBbb Z., $ Conversely,
$,gcd(a,b)mid a,b,Rightarrow,gcd(a,b)mid ax!+!by,,$ so $,ax!+!by = ngcd(a,b),,$ so $,a,Bbb Z+b,Bbb Zsubseteq gcd(a,b)Bbb Z$
Or we can induct. $ $ wlog $,a,b> 0,$ by $,-aBbb Z = aBbb Z,, (pm a,pm b) = (a,b),,$ and it is true if $,a,$ or $,b=0.$
Proof by induction on $,color{#90f}{{rm size} := a+b}.,$ True if $,a = b!: aBbb Z + aBbb Z = aBbb Z.,$ Else $,aneq b.,$ By symmetry, wlog $,a>b.,$ $,aBbb Z+bBbb Z = color{#0a0}{(a!-!b)Bbb Z+bBbb Z} = (a!-!b,b)Bbb Z = (a,b)Bbb Z,$ because the $,color{#0a0}{rm green},$ instance has smaller $,color{#90f}{{rm size}} = (a!-!b)+b = a < color{#90f}{a+b},,$ so $rmcolor{}{induction},$ applies. $ $ QED
answered Mar 19 '14 at 4:17
Bill DubuqueBill Dubuque
210k29192641
$endgroup$
add a comment |
$begingroup$
Bezout's lemma says that there is a solution for every pair of natural numbers $(a,b)$ to the linear Diophantine equation: $$gcd(a,b)=ax+by$$
Now define $k=ax+by$
And let $(x,y)$ run over the integers.
We have $gcd(a,b)mid k$ because it divides the right hand side.
Thus $k$ must be an integer multiple of $gcd(a,b)$.
But by Bezout's lemma we know there is a solution $(x_0,y_0)=(nx,yx)$ to $$ngcd(a,b)=ax_0+by_0$$
Thus we have shown all values of $k$ must be of the form $ngcd(a,b)$ and that for every $ngcd(a,b)$ with $nin mathbb{Z}$ there exists integers $(x_0,y_0)$ satisfying the linear equation. Which completes the proof.
answered Mar 19 '14 at 4:20
EthanEthan
6,85512164
$endgroup$
add a comment |
$begingroup$
To prove equality of two sets, you need to show that each is contained in the other.
${ax+by | x,y in Bbb Z} subset {n gcd(a,b) | n in Bbb Z}$, because for a given element of the former $ax+by$, we can take $n=frac{a}{gcd(a,b)}x+frac{b}{gcd(a,b)}y$ to see that it is an element of the latter set.
Now as you pointed out there exists a pair of integers $x,y$ such that $ax+by = gcd(a,b)$. Can you use this to show that ${n gcd(a,b) | n in Bbb Z} subset {ax+by | x,y in Bbb Z}$?
answered Mar 19 '14 at 4:15
Mike MillerMike Miller
37.2k472139
$endgroup$
add a comment |
$begingroup$
Let $k = text{gcd}(a, b)$
Then $ax + by = k(frac{a}{k} x + frac{b}{k} y)$
Then since $x, y, frac{a}{k}, frac{b}{k}$ are all integers, so is the entire expression, which is $n$.
answered Mar 19 '14 at 4:16
MCTMCT
14.4k42667
$endgroup$
add a comment |
$begingroup$
Trivially $(a,b)mid ax+by$ (since $(a,b)mid a$ and $(a,b)mid b$).
This shows ${ax+bymid x,yinmathbb Z}subseteq {n(a,b)mid ninmathbb Z}$.
Now prove $exists x,yinmathbb Z$ such that $n(a,b)=ax+by$.$ (2)$
Euclid's algorithm shows that $exists x_0,y_0inmathbb Z ((a,b)=ax_0+by_0)$. Use this to show $(2)$.
This concludes that ${n(a,b)mid ninmathbb Z}subseteq{ax+bymid x,yinmathbb Z}$.
answered Apr 4 '15 at 14:12
user26486user26486
9,29021948
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add a comment |
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5 Answers
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active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By Bezout, $ ngcd(a,b) = n(aj+bk),$ $Rightarrow$ $,gcd(a,b),Bbb Zsubseteq aBbb Z+bBbb Z., $ Conversely,
$,gcd(a,b)mid a,b,Rightarrow,gcd(a,b)mid ax!+!by,,$ so $,ax!+!by = ngcd(a,b),,$ so $,a,Bbb Z+b,Bbb Zsubseteq gcd(a,b)Bbb Z$
Or we can induct. $ $ wlog $,a,b> 0,$ by $,-aBbb Z = aBbb Z,, (pm a,pm b) = (a,b),,$ and it is true if $,a,$ or $,b=0.$
Proof by induction on $,color{#90f}{{rm size} := a+b}.,$ True if $,a = b!: aBbb Z + aBbb Z = aBbb Z.,$ Else $,aneq b.,$ By symmetry, wlog $,a>b.,$ $,aBbb Z+bBbb Z = color{#0a0}{(a!-!b)Bbb Z+bBbb Z} = (a!-!b,b)Bbb Z = (a,b)Bbb Z,$ because the $,color{#0a0}{rm green},$ instance has smaller $,color{#90f}{{rm size}} = (a!-!b)+b = a < color{#90f}{a+b},,$ so $rmcolor{}{induction},$ applies. $ $ QED
answered Mar 19 '14 at 4:17
Bill DubuqueBill Dubuque
210k29192641
$endgroup$
add a comment |
$begingroup$
By Bezout, $ ngcd(a,b) = n(aj+bk),$ $Rightarrow$ $,gcd(a,b),Bbb Zsubseteq aBbb Z+bBbb Z., $ Conversely,
$,gcd(a,b)mid a,b,Rightarrow,gcd(a,b)mid ax!+!by,,$ so $,ax!+!by = ngcd(a,b),,$ so $,a,Bbb Z+b,Bbb Zsubseteq gcd(a,b)Bbb Z$
Or we can induct. $ $ wlog $,a,b> 0,$ by $,-aBbb Z = aBbb Z,, (pm a,pm b) = (a,b),,$ and it is true if $,a,$ or $,b=0.$
Proof by induction on $,color{#90f}{{rm size} := a+b}.,$ True if $,a = b!: aBbb Z + aBbb Z = aBbb Z.,$ Else $,aneq b.,$ By symmetry, wlog $,a>b.,$ $,aBbb Z+bBbb Z = color{#0a0}{(a!-!b)Bbb Z+bBbb Z} = (a!-!b,b)Bbb Z = (a,b)Bbb Z,$ because the $,color{#0a0}{rm green},$ instance has smaller $,color{#90f}{{rm size}} = (a!-!b)+b = a < color{#90f}{a+b},,$ so $rmcolor{}{induction},$ applies. $ $ QED
answered Mar 19 '14 at 4:17
Bill DubuqueBill Dubuque
210k29192641
$endgroup$
add a comment |
$begingroup$
By Bezout, $ ngcd(a,b) = n(aj+bk),$ $Rightarrow$ $,gcd(a,b),Bbb Zsubseteq aBbb Z+bBbb Z., $ Conversely,
$,gcd(a,b)mid a,b,Rightarrow,gcd(a,b)mid ax!+!by,,$ so $,ax!+!by = ngcd(a,b),,$ so $,a,Bbb Z+b,Bbb Zsubseteq gcd(a,b)Bbb Z$
Or we can induct. $ $ wlog $,a,b> 0,$ by $,-aBbb Z = aBbb Z,, (pm a,pm b) = (a,b),,$ and it is true if $,a,$ or $,b=0.$
Proof by induction on $,color{#90f}{{rm size} := a+b}.,$ True if $,a = b!: aBbb Z + aBbb Z = aBbb Z.,$ Else $,aneq b.,$ By symmetry, wlog $,a>b.,$ $,aBbb Z+bBbb Z = color{#0a0}{(a!-!b)Bbb Z+bBbb Z} = (a!-!b,b)Bbb Z = (a,b)Bbb Z,$ because the $,color{#0a0}{rm green},$ instance has smaller $,color{#90f}{{rm size}} = (a!-!b)+b = a < color{#90f}{a+b},,$ so $rmcolor{}{induction},$ applies. $ $ QED
answered Mar 19 '14 at 4:17
Bill DubuqueBill Dubuque
210k29192641
$endgroup$
By Bezout, $ ngcd(a,b) = n(aj+bk),$ $Rightarrow$ $,gcd(a,b),Bbb Zsubseteq aBbb Z+bBbb Z., $ Conversely,
$,gcd(a,b)mid a,b,Rightarrow,gcd(a,b)mid ax!+!by,,$ so $,ax!+!by = ngcd(a,b),,$ so $,a,Bbb Z+b,Bbb Zsubseteq gcd(a,b)Bbb Z$
Or we can induct. $ $ wlog $,a,b> 0,$ by $,-aBbb Z = aBbb Z,, (pm a,pm b) = (a,b),,$ and it is true if $,a,$ or $,b=0.$
Proof by induction on $,color{#90f}{{rm size} := a+b}.,$ True if $,a = b!: aBbb Z + aBbb Z = aBbb Z.,$ Else $,aneq b.,$ By symmetry, wlog $,a>b.,$ $,aBbb Z+bBbb Z = color{#0a0}{(a!-!b)Bbb Z+bBbb Z} = (a!-!b,b)Bbb Z = (a,b)Bbb Z,$ because the $,color{#0a0}{rm green},$ instance has smaller $,color{#90f}{{rm size}} = (a!-!b)+b = a < color{#90f}{a+b},,$ so $rmcolor{}{induction},$ applies. $ $ QED
answered Mar 19 '14 at 4:17
Bill DubuqueBill Dubuque
210k29192641
edited Apr 4 '15 at 21:57
answered Mar 19 '14 at 4:17
Bill DubuqueBill Dubuque
210k29192641
answered Mar 19 '14 at 4:17
Bill DubuqueBill Dubuque
210k29192641
answered Mar 19 '14 at 4:17
Bill DubuqueBill Dubuque
210k29192641
210k29192641
add a comment |
add a comment |
$begingroup$
Bezout's lemma says that there is a solution for every pair of natural numbers $(a,b)$ to the linear Diophantine equation: $$gcd(a,b)=ax+by$$
Now define $k=ax+by$
And let $(x,y)$ run over the integers.
We have $gcd(a,b)mid k$ because it divides the right hand side.
Thus $k$ must be an integer multiple of $gcd(a,b)$.
But by Bezout's lemma we know there is a solution $(x_0,y_0)=(nx,yx)$ to $$ngcd(a,b)=ax_0+by_0$$
Thus we have shown all values of $k$ must be of the form $ngcd(a,b)$ and that for every $ngcd(a,b)$ with $nin mathbb{Z}$ there exists integers $(x_0,y_0)$ satisfying the linear equation. Which completes the proof.
answered Mar 19 '14 at 4:20
EthanEthan
6,85512164
$endgroup$
add a comment |
$begingroup$
Bezout's lemma says that there is a solution for every pair of natural numbers $(a,b)$ to the linear Diophantine equation: $$gcd(a,b)=ax+by$$
Now define $k=ax+by$
And let $(x,y)$ run over the integers.
We have $gcd(a,b)mid k$ because it divides the right hand side.
Thus $k$ must be an integer multiple of $gcd(a,b)$.
But by Bezout's lemma we know there is a solution $(x_0,y_0)=(nx,yx)$ to $$ngcd(a,b)=ax_0+by_0$$
Thus we have shown all values of $k$ must be of the form $ngcd(a,b)$ and that for every $ngcd(a,b)$ with $nin mathbb{Z}$ there exists integers $(x_0,y_0)$ satisfying the linear equation. Which completes the proof.
answered Mar 19 '14 at 4:20
EthanEthan
6,85512164
$endgroup$
add a comment |
$begingroup$
Bezout's lemma says that there is a solution for every pair of natural numbers $(a,b)$ to the linear Diophantine equation: $$gcd(a,b)=ax+by$$
Now define $k=ax+by$
And let $(x,y)$ run over the integers.
We have $gcd(a,b)mid k$ because it divides the right hand side.
Thus $k$ must be an integer multiple of $gcd(a,b)$.
But by Bezout's lemma we know there is a solution $(x_0,y_0)=(nx,yx)$ to $$ngcd(a,b)=ax_0+by_0$$
Thus we have shown all values of $k$ must be of the form $ngcd(a,b)$ and that for every $ngcd(a,b)$ with $nin mathbb{Z}$ there exists integers $(x_0,y_0)$ satisfying the linear equation. Which completes the proof.
answered Mar 19 '14 at 4:20
EthanEthan
6,85512164
$endgroup$
Bezout's lemma says that there is a solution for every pair of natural numbers $(a,b)$ to the linear Diophantine equation: $$gcd(a,b)=ax+by$$
Now define $k=ax+by$
And let $(x,y)$ run over the integers.
We have $gcd(a,b)mid k$ because it divides the right hand side.
Thus $k$ must be an integer multiple of $gcd(a,b)$.
But by Bezout's lemma we know there is a solution $(x_0,y_0)=(nx,yx)$ to $$ngcd(a,b)=ax_0+by_0$$
Thus we have shown all values of $k$ must be of the form $ngcd(a,b)$ and that for every $ngcd(a,b)$ with $nin mathbb{Z}$ there exists integers $(x_0,y_0)$ satisfying the linear equation. Which completes the proof.
answered Mar 19 '14 at 4:20
EthanEthan
6,85512164
answered Mar 19 '14 at 4:20
EthanEthan
6,85512164
answered Mar 19 '14 at 4:20
EthanEthan
6,85512164
answered Mar 19 '14 at 4:20
EthanEthan
6,85512164
6,85512164
add a comment |
add a comment |
$begingroup$
To prove equality of two sets, you need to show that each is contained in the other.
${ax+by | x,y in Bbb Z} subset {n gcd(a,b) | n in Bbb Z}$, because for a given element of the former $ax+by$, we can take $n=frac{a}{gcd(a,b)}x+frac{b}{gcd(a,b)}y$ to see that it is an element of the latter set.
Now as you pointed out there exists a pair of integers $x,y$ such that $ax+by = gcd(a,b)$. Can you use this to show that ${n gcd(a,b) | n in Bbb Z} subset {ax+by | x,y in Bbb Z}$?
answered Mar 19 '14 at 4:15
Mike MillerMike Miller
37.2k472139
$endgroup$
add a comment |
$begingroup$
To prove equality of two sets, you need to show that each is contained in the other.
${ax+by | x,y in Bbb Z} subset {n gcd(a,b) | n in Bbb Z}$, because for a given element of the former $ax+by$, we can take $n=frac{a}{gcd(a,b)}x+frac{b}{gcd(a,b)}y$ to see that it is an element of the latter set.
Now as you pointed out there exists a pair of integers $x,y$ such that $ax+by = gcd(a,b)$. Can you use this to show that ${n gcd(a,b) | n in Bbb Z} subset {ax+by | x,y in Bbb Z}$?
answered Mar 19 '14 at 4:15
Mike MillerMike Miller
37.2k472139
$endgroup$
add a comment |
$begingroup$
To prove equality of two sets, you need to show that each is contained in the other.
${ax+by | x,y in Bbb Z} subset {n gcd(a,b) | n in Bbb Z}$, because for a given element of the former $ax+by$, we can take $n=frac{a}{gcd(a,b)}x+frac{b}{gcd(a,b)}y$ to see that it is an element of the latter set.
Now as you pointed out there exists a pair of integers $x,y$ such that $ax+by = gcd(a,b)$. Can you use this to show that ${n gcd(a,b) | n in Bbb Z} subset {ax+by | x,y in Bbb Z}$?
answered Mar 19 '14 at 4:15
Mike MillerMike Miller
37.2k472139
$endgroup$
To prove equality of two sets, you need to show that each is contained in the other.
${ax+by | x,y in Bbb Z} subset {n gcd(a,b) | n in Bbb Z}$, because for a given element of the former $ax+by$, we can take $n=frac{a}{gcd(a,b)}x+frac{b}{gcd(a,b)}y$ to see that it is an element of the latter set.
Now as you pointed out there exists a pair of integers $x,y$ such that $ax+by = gcd(a,b)$. Can you use this to show that ${n gcd(a,b) | n in Bbb Z} subset {ax+by | x,y in Bbb Z}$?
answered Mar 19 '14 at 4:15
Mike MillerMike Miller
37.2k472139
answered Mar 19 '14 at 4:15
Mike MillerMike Miller
37.2k472139
answered Mar 19 '14 at 4:15
Mike MillerMike Miller
37.2k472139
answered Mar 19 '14 at 4:15
Mike MillerMike Miller
37.2k472139
37.2k472139
add a comment |
add a comment |
$begingroup$
Let $k = text{gcd}(a, b)$
Then $ax + by = k(frac{a}{k} x + frac{b}{k} y)$
Then since $x, y, frac{a}{k}, frac{b}{k}$ are all integers, so is the entire expression, which is $n$.
answered Mar 19 '14 at 4:16
MCTMCT
14.4k42667
$endgroup$
add a comment |
$begingroup$
Let $k = text{gcd}(a, b)$
Then $ax + by = k(frac{a}{k} x + frac{b}{k} y)$
Then since $x, y, frac{a}{k}, frac{b}{k}$ are all integers, so is the entire expression, which is $n$.
answered Mar 19 '14 at 4:16
MCTMCT
14.4k42667
$endgroup$
add a comment |
$begingroup$
Let $k = text{gcd}(a, b)$
Then $ax + by = k(frac{a}{k} x + frac{b}{k} y)$
Then since $x, y, frac{a}{k}, frac{b}{k}$ are all integers, so is the entire expression, which is $n$.
answered Mar 19 '14 at 4:16
MCTMCT
14.4k42667
$endgroup$
Let $k = text{gcd}(a, b)$
Then $ax + by = k(frac{a}{k} x + frac{b}{k} y)$
Then since $x, y, frac{a}{k}, frac{b}{k}$ are all integers, so is the entire expression, which is $n$.
answered Mar 19 '14 at 4:16
MCTMCT
14.4k42667
answered Mar 19 '14 at 4:16
MCTMCT
14.4k42667
answered Mar 19 '14 at 4:16
MCTMCT
14.4k42667
answered Mar 19 '14 at 4:16
MCTMCT
14.4k42667
14.4k42667
add a comment |
add a comment |
$begingroup$
Trivially $(a,b)mid ax+by$ (since $(a,b)mid a$ and $(a,b)mid b$).
This shows ${ax+bymid x,yinmathbb Z}subseteq {n(a,b)mid ninmathbb Z}$.
Now prove $exists x,yinmathbb Z$ such that $n(a,b)=ax+by$.$ (2)$
Euclid's algorithm shows that $exists x_0,y_0inmathbb Z ((a,b)=ax_0+by_0)$. Use this to show $(2)$.
This concludes that ${n(a,b)mid ninmathbb Z}subseteq{ax+bymid x,yinmathbb Z}$.
answered Apr 4 '15 at 14:12
user26486user26486
9,29021948
$endgroup$
add a comment |
$begingroup$
Trivially $(a,b)mid ax+by$ (since $(a,b)mid a$ and $(a,b)mid b$).
This shows ${ax+bymid x,yinmathbb Z}subseteq {n(a,b)mid ninmathbb Z}$.
Now prove $exists x,yinmathbb Z$ such that $n(a,b)=ax+by$.$ (2)$
Euclid's algorithm shows that $exists x_0,y_0inmathbb Z ((a,b)=ax_0+by_0)$. Use this to show $(2)$.
This concludes that ${n(a,b)mid ninmathbb Z}subseteq{ax+bymid x,yinmathbb Z}$.
answered Apr 4 '15 at 14:12
user26486user26486
9,29021948
$endgroup$
add a comment |
$begingroup$
Trivially $(a,b)mid ax+by$ (since $(a,b)mid a$ and $(a,b)mid b$).
This shows ${ax+bymid x,yinmathbb Z}subseteq {n(a,b)mid ninmathbb Z}$.
Now prove $exists x,yinmathbb Z$ such that $n(a,b)=ax+by$.$ (2)$
Euclid's algorithm shows that $exists x_0,y_0inmathbb Z ((a,b)=ax_0+by_0)$. Use this to show $(2)$.
This concludes that ${n(a,b)mid ninmathbb Z}subseteq{ax+bymid x,yinmathbb Z}$.
answered Apr 4 '15 at 14:12
user26486user26486
9,29021948
$endgroup$
Trivially $(a,b)mid ax+by$ (since $(a,b)mid a$ and $(a,b)mid b$).
This shows ${ax+bymid x,yinmathbb Z}subseteq {n(a,b)mid ninmathbb Z}$.
Now prove $exists x,yinmathbb Z$ such that $n(a,b)=ax+by$.$ (2)$
Euclid's algorithm shows that $exists x_0,y_0inmathbb Z ((a,b)=ax_0+by_0)$. Use this to show $(2)$.
This concludes that ${n(a,b)mid ninmathbb Z}subseteq{ax+bymid x,yinmathbb Z}$.
answered Apr 4 '15 at 14:12
user26486user26486
9,29021948
answered Apr 4 '15 at 14:12
user26486user26486
9,29021948
answered Apr 4 '15 at 14:12
user26486user26486
9,29021948
answered Apr 4 '15 at 14:12
user26486user26486
9,29021948
9,29021948
add a comment |
add a comment |
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$begingroup$
do you have any other conditions on $x$ and $y$ ?
$endgroup$
– wanderer
Mar 19 '14 at 4:11
$begingroup$
nope, theyre just integers
$endgroup$
– Alex Chavez
Mar 19 '14 at 4:15
$begingroup$
This is closely related to Proof of Extended Euclidean Algorithm?
$endgroup$
– robjohn♦
Apr 4 '15 at 14:08