Bijection from modular multiplication?
$begingroup$
Say we have two integers $n$ and $k$, such that $0 < k < n$ and $gcd(n, k) = 1$.
Why is it that if we take every integer $xin{0,dots, n-1}$ and compute $kcdot x (mod n)$, we get as remainders all numbers from $0$ to $n-1$, with no repeats?
In other words, why is $ f:{0,dots,n-1}rightarrow{0,dots,n-1}$, $ f(x)=kcdot x (mod n)$ a bijection for $k,ninmathbb{N}, gcd(k,n)=1$?
modular-arithmetic
asked Dec 2 '18 at 18:09
SeriousCraneSeriousCrane
111
$endgroup$
add a comment |
$begingroup$
Say we have two integers $n$ and $k$, such that $0 < k < n$ and $gcd(n, k) = 1$.
Why is it that if we take every integer $xin{0,dots, n-1}$ and compute $kcdot x (mod n)$, we get as remainders all numbers from $0$ to $n-1$, with no repeats?
In other words, why is $ f:{0,dots,n-1}rightarrow{0,dots,n-1}$, $ f(x)=kcdot x (mod n)$ a bijection for $k,ninmathbb{N}, gcd(k,n)=1$?
modular-arithmetic
asked Dec 2 '18 at 18:09
SeriousCraneSeriousCrane
111
$endgroup$
add a comment |
$begingroup$
Say we have two integers $n$ and $k$, such that $0 < k < n$ and $gcd(n, k) = 1$.
Why is it that if we take every integer $xin{0,dots, n-1}$ and compute $kcdot x (mod n)$, we get as remainders all numbers from $0$ to $n-1$, with no repeats?
In other words, why is $ f:{0,dots,n-1}rightarrow{0,dots,n-1}$, $ f(x)=kcdot x (mod n)$ a bijection for $k,ninmathbb{N}, gcd(k,n)=1$?
modular-arithmetic
asked Dec 2 '18 at 18:09
SeriousCraneSeriousCrane
111
$endgroup$
Say we have two integers $n$ and $k$, such that $0 < k < n$ and $gcd(n, k) = 1$.
Why is it that if we take every integer $xin{0,dots, n-1}$ and compute $kcdot x (mod n)$, we get as remainders all numbers from $0$ to $n-1$, with no repeats?
In other words, why is $ f:{0,dots,n-1}rightarrow{0,dots,n-1}$, $ f(x)=kcdot x (mod n)$ a bijection for $k,ninmathbb{N}, gcd(k,n)=1$?
modular-arithmetic
modular-arithmetic
asked Dec 2 '18 at 18:09
SeriousCraneSeriousCrane
111
asked Dec 2 '18 at 18:09
SeriousCraneSeriousCrane
111
asked Dec 2 '18 at 18:09
SeriousCraneSeriousCrane
111
asked Dec 2 '18 at 18:09
SeriousCraneSeriousCrane
111
asked Dec 2 '18 at 18:09
SeriousCraneSeriousCrane
111
111
add a comment |
add a comment |
2 Answers
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The assumptions imply that $k$ is invertible $pmod n$. More precisely, Bezout tells us that $gcd(n,k)=1$ implies the existence of integers $a,b$ such that $ak+bn=1$. We note that $akequiv 1 pmod n$ so $a$ is a multiplicative inverse of $kpmod n$.
Your claim follows at once from this. To see that a residue class $r$ is in the image of multiplication by $k$, we note that $ktimes atimes r equiv r pmod n$. Thus the map is surjective (and as it is a map between finite sets of the same size, surjective implies bijective).
As another argument, note that the map is injective since $$ktimes r_1equiv ktimes r_2pmod nimplies n,|,k(r_1-r_2)$$
But, since $gcd(n,k)=1$ this implies $n,|,r_1-r_2$. Again, since this is a map between finite sets of the same size, injective implies bijective.
answered Dec 2 '18 at 18:24
lulululu
39.8k24778
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$begingroup$
See below for a few ways to prove it. $ $ OP is $rm, c,d = k,0,$ below.
Theorem $ $ The following are equivalent for integers $rm:c,d, n.$
$(1)rm gcd(c,n) = 1$
$(2)rm c:$ is invertible $rm,(mod n)$
$(3)rm xto cx+d:$ is $:1$-$1:$ $rm,(mod n)$
$(4)rm xto cx+d:$ is onto $rm,(mod n)$
Proof $ (1Rightarrow 2) $ By Bezout $rm, gcd(c,n)! =! 1Rightarrow cd!+!kn =! 1,$ for $rm,d,kinBbb Z,$ $rmRightarrow cdequiv 1!pmod{! n}$
$(2Rightarrow 3) rm cx!+!d equiv cy!+!d,Rightarrow,c(x!-!y)equiv 0,Rightarrow,x!-!yequiv 0,$ by multiplying by $rm,c^{-1}$
$(3Rightarrow 4) $ Every $1$-$1$ function on a finite set is onto (pigeonhole).
$(4Rightarrow 1) rm xto cx,$ is onto, so $rm,cdequiv 1,,$ some $rm,d,,$ i.e. $rm, cd+kn = 1,,$ some $rm,k,,$ so $rmgcd(c,n)=1$
See here for a conceptual proof of said Bezout identity for the gcd.
answered Dec 2 '18 at 18:43
Bill DubuqueBill Dubuque
209k29191634
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2 Answers
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2 Answers
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$begingroup$
The assumptions imply that $k$ is invertible $pmod n$. More precisely, Bezout tells us that $gcd(n,k)=1$ implies the existence of integers $a,b$ such that $ak+bn=1$. We note that $akequiv 1 pmod n$ so $a$ is a multiplicative inverse of $kpmod n$.
Your claim follows at once from this. To see that a residue class $r$ is in the image of multiplication by $k$, we note that $ktimes atimes r equiv r pmod n$. Thus the map is surjective (and as it is a map between finite sets of the same size, surjective implies bijective).
As another argument, note that the map is injective since $$ktimes r_1equiv ktimes r_2pmod nimplies n,|,k(r_1-r_2)$$
But, since $gcd(n,k)=1$ this implies $n,|,r_1-r_2$. Again, since this is a map between finite sets of the same size, injective implies bijective.
answered Dec 2 '18 at 18:24
lulululu
39.8k24778
$endgroup$
add a comment |
$begingroup$
The assumptions imply that $k$ is invertible $pmod n$. More precisely, Bezout tells us that $gcd(n,k)=1$ implies the existence of integers $a,b$ such that $ak+bn=1$. We note that $akequiv 1 pmod n$ so $a$ is a multiplicative inverse of $kpmod n$.
Your claim follows at once from this. To see that a residue class $r$ is in the image of multiplication by $k$, we note that $ktimes atimes r equiv r pmod n$. Thus the map is surjective (and as it is a map between finite sets of the same size, surjective implies bijective).
As another argument, note that the map is injective since $$ktimes r_1equiv ktimes r_2pmod nimplies n,|,k(r_1-r_2)$$
But, since $gcd(n,k)=1$ this implies $n,|,r_1-r_2$. Again, since this is a map between finite sets of the same size, injective implies bijective.
answered Dec 2 '18 at 18:24
lulululu
39.8k24778
$endgroup$
add a comment |
$begingroup$
The assumptions imply that $k$ is invertible $pmod n$. More precisely, Bezout tells us that $gcd(n,k)=1$ implies the existence of integers $a,b$ such that $ak+bn=1$. We note that $akequiv 1 pmod n$ so $a$ is a multiplicative inverse of $kpmod n$.
Your claim follows at once from this. To see that a residue class $r$ is in the image of multiplication by $k$, we note that $ktimes atimes r equiv r pmod n$. Thus the map is surjective (and as it is a map between finite sets of the same size, surjective implies bijective).
As another argument, note that the map is injective since $$ktimes r_1equiv ktimes r_2pmod nimplies n,|,k(r_1-r_2)$$
But, since $gcd(n,k)=1$ this implies $n,|,r_1-r_2$. Again, since this is a map between finite sets of the same size, injective implies bijective.
answered Dec 2 '18 at 18:24
lulululu
39.8k24778
$endgroup$
The assumptions imply that $k$ is invertible $pmod n$. More precisely, Bezout tells us that $gcd(n,k)=1$ implies the existence of integers $a,b$ such that $ak+bn=1$. We note that $akequiv 1 pmod n$ so $a$ is a multiplicative inverse of $kpmod n$.
Your claim follows at once from this. To see that a residue class $r$ is in the image of multiplication by $k$, we note that $ktimes atimes r equiv r pmod n$. Thus the map is surjective (and as it is a map between finite sets of the same size, surjective implies bijective).
As another argument, note that the map is injective since $$ktimes r_1equiv ktimes r_2pmod nimplies n,|,k(r_1-r_2)$$
But, since $gcd(n,k)=1$ this implies $n,|,r_1-r_2$. Again, since this is a map between finite sets of the same size, injective implies bijective.
answered Dec 2 '18 at 18:24
lulululu
39.8k24778
answered Dec 2 '18 at 18:24
lulululu
39.8k24778
answered Dec 2 '18 at 18:24
lulululu
39.8k24778
answered Dec 2 '18 at 18:24
lulululu
39.8k24778
39.8k24778
add a comment |
add a comment |
$begingroup$
See below for a few ways to prove it. $ $ OP is $rm, c,d = k,0,$ below.
Theorem $ $ The following are equivalent for integers $rm:c,d, n.$
$(1)rm gcd(c,n) = 1$
$(2)rm c:$ is invertible $rm,(mod n)$
$(3)rm xto cx+d:$ is $:1$-$1:$ $rm,(mod n)$
$(4)rm xto cx+d:$ is onto $rm,(mod n)$
Proof $ (1Rightarrow 2) $ By Bezout $rm, gcd(c,n)! =! 1Rightarrow cd!+!kn =! 1,$ for $rm,d,kinBbb Z,$ $rmRightarrow cdequiv 1!pmod{! n}$
$(2Rightarrow 3) rm cx!+!d equiv cy!+!d,Rightarrow,c(x!-!y)equiv 0,Rightarrow,x!-!yequiv 0,$ by multiplying by $rm,c^{-1}$
$(3Rightarrow 4) $ Every $1$-$1$ function on a finite set is onto (pigeonhole).
$(4Rightarrow 1) rm xto cx,$ is onto, so $rm,cdequiv 1,,$ some $rm,d,,$ i.e. $rm, cd+kn = 1,,$ some $rm,k,,$ so $rmgcd(c,n)=1$
See here for a conceptual proof of said Bezout identity for the gcd.
answered Dec 2 '18 at 18:43
Bill DubuqueBill Dubuque
209k29191634
$endgroup$
add a comment |
$begingroup$
See below for a few ways to prove it. $ $ OP is $rm, c,d = k,0,$ below.
Theorem $ $ The following are equivalent for integers $rm:c,d, n.$
$(1)rm gcd(c,n) = 1$
$(2)rm c:$ is invertible $rm,(mod n)$
$(3)rm xto cx+d:$ is $:1$-$1:$ $rm,(mod n)$
$(4)rm xto cx+d:$ is onto $rm,(mod n)$
Proof $ (1Rightarrow 2) $ By Bezout $rm, gcd(c,n)! =! 1Rightarrow cd!+!kn =! 1,$ for $rm,d,kinBbb Z,$ $rmRightarrow cdequiv 1!pmod{! n}$
$(2Rightarrow 3) rm cx!+!d equiv cy!+!d,Rightarrow,c(x!-!y)equiv 0,Rightarrow,x!-!yequiv 0,$ by multiplying by $rm,c^{-1}$
$(3Rightarrow 4) $ Every $1$-$1$ function on a finite set is onto (pigeonhole).
$(4Rightarrow 1) rm xto cx,$ is onto, so $rm,cdequiv 1,,$ some $rm,d,,$ i.e. $rm, cd+kn = 1,,$ some $rm,k,,$ so $rmgcd(c,n)=1$
See here for a conceptual proof of said Bezout identity for the gcd.
answered Dec 2 '18 at 18:43
Bill DubuqueBill Dubuque
209k29191634
$endgroup$
add a comment |
$begingroup$
See below for a few ways to prove it. $ $ OP is $rm, c,d = k,0,$ below.
Theorem $ $ The following are equivalent for integers $rm:c,d, n.$
$(1)rm gcd(c,n) = 1$
$(2)rm c:$ is invertible $rm,(mod n)$
$(3)rm xto cx+d:$ is $:1$-$1:$ $rm,(mod n)$
$(4)rm xto cx+d:$ is onto $rm,(mod n)$
Proof $ (1Rightarrow 2) $ By Bezout $rm, gcd(c,n)! =! 1Rightarrow cd!+!kn =! 1,$ for $rm,d,kinBbb Z,$ $rmRightarrow cdequiv 1!pmod{! n}$
$(2Rightarrow 3) rm cx!+!d equiv cy!+!d,Rightarrow,c(x!-!y)equiv 0,Rightarrow,x!-!yequiv 0,$ by multiplying by $rm,c^{-1}$
$(3Rightarrow 4) $ Every $1$-$1$ function on a finite set is onto (pigeonhole).
$(4Rightarrow 1) rm xto cx,$ is onto, so $rm,cdequiv 1,,$ some $rm,d,,$ i.e. $rm, cd+kn = 1,,$ some $rm,k,,$ so $rmgcd(c,n)=1$
See here for a conceptual proof of said Bezout identity for the gcd.
answered Dec 2 '18 at 18:43
Bill DubuqueBill Dubuque
209k29191634
$endgroup$
See below for a few ways to prove it. $ $ OP is $rm, c,d = k,0,$ below.
Theorem $ $ The following are equivalent for integers $rm:c,d, n.$
$(1)rm gcd(c,n) = 1$
$(2)rm c:$ is invertible $rm,(mod n)$
$(3)rm xto cx+d:$ is $:1$-$1:$ $rm,(mod n)$
$(4)rm xto cx+d:$ is onto $rm,(mod n)$
Proof $ (1Rightarrow 2) $ By Bezout $rm, gcd(c,n)! =! 1Rightarrow cd!+!kn =! 1,$ for $rm,d,kinBbb Z,$ $rmRightarrow cdequiv 1!pmod{! n}$
$(2Rightarrow 3) rm cx!+!d equiv cy!+!d,Rightarrow,c(x!-!y)equiv 0,Rightarrow,x!-!yequiv 0,$ by multiplying by $rm,c^{-1}$
$(3Rightarrow 4) $ Every $1$-$1$ function on a finite set is onto (pigeonhole).
$(4Rightarrow 1) rm xto cx,$ is onto, so $rm,cdequiv 1,,$ some $rm,d,,$ i.e. $rm, cd+kn = 1,,$ some $rm,k,,$ so $rmgcd(c,n)=1$
See here for a conceptual proof of said Bezout identity for the gcd.
answered Dec 2 '18 at 18:43
Bill DubuqueBill Dubuque
209k29191634
answered Dec 2 '18 at 18:43
Bill DubuqueBill Dubuque
209k29191634
answered Dec 2 '18 at 18:43
Bill DubuqueBill Dubuque
209k29191634
answered Dec 2 '18 at 18:43
Bill DubuqueBill Dubuque
209k29191634
209k29191634
add a comment |
add a comment |
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