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Proof that integers (except $1$ and $-1$) don't have an integer multiplicative inverse
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This is a very elementary question, so please forgive me, but precisely because of this reason I haven't found an answer elsewhere:
Is the statement "$frac1ainmathbbZquad textif and only if quad (a=1)lor(a=-1);$" an axiom? Or can it be proven from other, antecedent axioms?
(If we had to prove that every integer $a$ different from $pm 1$ doesn't have a multiplicative inverse, we would have to "test" that "$forall binmathbbZ: abne1$")
elementary-number-theory discrete-mathematics predicate-logic axioms
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user21820
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Anakhand
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This is a very elementary question, so please forgive me, but precisely because of this reason I haven't found an answer elsewhere:
Is the statement "$frac1ainmathbbZquad textif and only if quad (a=1)lor(a=-1);$" an axiom? Or can it be proven from other, antecedent axioms?
(If we had to prove that every integer $a$ different from $pm 1$ doesn't have a multiplicative inverse, we would have to "test" that "$forall binmathbbZ: abne1$")
elementary-number-theory discrete-mathematics predicate-logic axioms
edited 4 secs ago
user21820
36.3k440141
asked 32 mins ago
Anakhand
658
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is a very elementary question, so please forgive me, but precisely because of this reason I haven't found an answer elsewhere:
Is the statement "$frac1ainmathbbZquad textif and only if quad (a=1)lor(a=-1);$" an axiom? Or can it be proven from other, antecedent axioms?
(If we had to prove that every integer $a$ different from $pm 1$ doesn't have a multiplicative inverse, we would have to "test" that "$forall binmathbbZ: abne1$")
elementary-number-theory discrete-mathematics predicate-logic axioms
edited 4 secs ago
user21820
36.3k440141
asked 32 mins ago
Anakhand
658
This is a very elementary question, so please forgive me, but precisely because of this reason I haven't found an answer elsewhere:
Is the statement "$frac1ainmathbbZquad textif and only if quad (a=1)lor(a=-1);$" an axiom? Or can it be proven from other, antecedent axioms?
(If we had to prove that every integer $a$ different from $pm 1$ doesn't have a multiplicative inverse, we would have to "test" that "$forall binmathbbZ: abne1$")
elementary-number-theory discrete-mathematics predicate-logic axioms
elementary-number-theory discrete-mathematics predicate-logic axioms
edited 4 secs ago
user21820
36.3k440141
asked 32 mins ago
Anakhand
658
edited 4 secs ago
user21820
36.3k440141
asked 32 mins ago
Anakhand
658
edited 4 secs ago
user21820
36.3k440141
edited 4 secs ago
user21820
36.3k440141
edited 4 secs ago
user21820
36.3k440141
36.3k440141
asked 32 mins ago
Anakhand
658
asked 32 mins ago
Anakhand
658
asked 32 mins ago
Anakhand
658
658
add a comment |Â
add a comment |Â
6 Answers
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This is readily proved once you properly define $Bbb Z$ and its operations. We may want to start from $Bbb N$ with its successor map $S$ and the Peano axioms, then define addition $Bbb Ntimes Bbb NtoBbb N$ recursively as $x+0:=x$, $x+Sy:=S(x+y)$, and then define multiplication recursively as $xcdot 0=0$, $xcdot Sy:=xcdot y+y$.
Now we see that $(a,b)sim (c,d):iff a+d=b+c$ is an equivalence relation on $Bbb N^2$, define $Bbb Z:=Bbb N^2/sim$ as the set of equivalence classes, define $overline(a,b)+overline(c,d):=overline(a+c,b+d)$ and define $overline(a,b)cdot overline(c,d):=overline(acdot c+bcdot d, acdot d+bcdot c)$ (which involves showing that this is well-defined). The claim now boils down to $overline(a,b)cdotoverline(c,d)=overline(1,0)implies a=Sblor b=Sa$. It does take a bit of work, but by going back the way we came can ultimately be reduced to a statement about $Bbb N$ that is provable by induction.
answered 15 mins ago
Hagen von Eitzen
268k21260484
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Well, let's assume $ab=1$ when $a,binmathbbZ$. Of course it means that both $a$ and $b$ are non zero. Because they are integers we can conclude that their absolute values are at least $1$. And then $1=|ab|=|a||b|geq |a|$, and in the same way $1geq |b|$. So both $a$ and $b$ must belong to the set $-1,1$.
answered 24 mins ago
Mark
1,811110
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It follows from the well-ordering principle for the natural numbers.
Suppose $ab=1$ for $a,b in BbbN$ with $b>1$. Then $c=a(b-1)$ is a natural number with $0<c<1$. So $c,c^2,c^3,dots$ is a set of natural numbers which has no least element, contradicting the well-ordering principle.
Now, if $ab=1$ for $a,b in BbbZ$, then either $a,b$ or $-a,-b$ will fit into the proof in the above paragraph...
answered 24 mins ago
Micah
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First of all, of course integers, have multiplicative inverses and of course that statement is obviously false as $a = frac 12, frac 13, frac 14,$ etc. bear out.
What you mean to say is $1$ and $-1$ are the only two integers whose multiplicative inverses are also integers.
I.e. If $a, frac 1a in mathbb Z$ then $a = 1$ or $a = -1$.
That is not an axiom.
If $a in mathbb Z$ then you have five possibilities. $a > 1$ or $a = 1$ or $a = 0$ or $a = -1$ or $a < 1$.
Case 1: $a > 1$. Then if $frac 1a < 0$ we would have $a*frac 1a < a*0$ and $1 < 0$ which is a contradiction. We can't have $frac 1a = 0$ because $0$ has no multiplicative inverses so $frac 1a > 0$. If $frac 1a ge 1$ then $a*frac 1a ge a*1 = a$ or $1 ge a$ which is a contradiction so $0< frac 1a < 1$ and not an integer.
Case 2: $a = 1$. then $frac 1a = 1$.
Case 3: $a = 0$ then $frac 1a$ is undefined.
Case 4: $a = -1$ then $frac 1a = -1$.
Case 5: $a < -1$. that is so similar to case 1: I'll leave it to you.
answered 19 mins ago
fleablood
61.8k22678
1
I don't agree with your first two paragraphs; at least not on a minimally generous reading of the question. Whether a number has a multiplicative inverse depends on which structure you're considering. In the ring of integers, non-units don't have multiplicative inverses. To say that "their inverses aren't integers" doesn't make sense within the structure; it only makes sense within the rationals. Though the question doesn't explicitly say "within the integers", it only talks about integers, and it stands to reason that the intended context is the ring of integers.
– joriki
8 mins ago
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As we are interested in 1/a which is just a symbol being adopted to denote the inverse of an element in Z. Further identity element under multiplication in Z is 1. So if any member say ''a' of Z is invertible then there must be a member in Z denoted as 1/a or a^(-1) such that
a. a^(-1) =1.
Now it's a simple algebra that in Z 1/a is an integer iff a= 1 or -1.
Hope it will help you.
answered 21 mins ago
prakash nainwal
524
Please see this tutorial and reference on how to typeset math on this site.
– joriki
1 min ago
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It depends upon the way that $mathbb Z$ and $times$ are defined, of course, but I doubt that anyone ever created axioms for the integers such that what you stated is an axiom.
Suppos, for instance, that you define:
$mathbb Z$ is the set of equivalence classes of $mathbbN^2$ with respect to the equivalence relation$$(a,b)sim(c,d)iff a+d=b+c;$$
$bigl[(a,b)bigr]timesbigl[(c,d)bigr]=bigl[(ac+bd,ad+bc)bigr]$.
Then your statement becomes a proposition, which can in fact be proved.
answered 21 mins ago
José Carlos Santos
123k17101186
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6 Answers
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6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
This is readily proved once you properly define $Bbb Z$ and its operations. We may want to start from $Bbb N$ with its successor map $S$ and the Peano axioms, then define addition $Bbb Ntimes Bbb NtoBbb N$ recursively as $x+0:=x$, $x+Sy:=S(x+y)$, and then define multiplication recursively as $xcdot 0=0$, $xcdot Sy:=xcdot y+y$.
Now we see that $(a,b)sim (c,d):iff a+d=b+c$ is an equivalence relation on $Bbb N^2$, define $Bbb Z:=Bbb N^2/sim$ as the set of equivalence classes, define $overline(a,b)+overline(c,d):=overline(a+c,b+d)$ and define $overline(a,b)cdot overline(c,d):=overline(acdot c+bcdot d, acdot d+bcdot c)$ (which involves showing that this is well-defined). The claim now boils down to $overline(a,b)cdotoverline(c,d)=overline(1,0)implies a=Sblor b=Sa$. It does take a bit of work, but by going back the way we came can ultimately be reduced to a statement about $Bbb N$ that is provable by induction.
answered 15 mins ago
Hagen von Eitzen
268k21260484
add a comment |Â
up vote
3
down vote
This is readily proved once you properly define $Bbb Z$ and its operations. We may want to start from $Bbb N$ with its successor map $S$ and the Peano axioms, then define addition $Bbb Ntimes Bbb NtoBbb N$ recursively as $x+0:=x$, $x+Sy:=S(x+y)$, and then define multiplication recursively as $xcdot 0=0$, $xcdot Sy:=xcdot y+y$.
Now we see that $(a,b)sim (c,d):iff a+d=b+c$ is an equivalence relation on $Bbb N^2$, define $Bbb Z:=Bbb N^2/sim$ as the set of equivalence classes, define $overline(a,b)+overline(c,d):=overline(a+c,b+d)$ and define $overline(a,b)cdot overline(c,d):=overline(acdot c+bcdot d, acdot d+bcdot c)$ (which involves showing that this is well-defined). The claim now boils down to $overline(a,b)cdotoverline(c,d)=overline(1,0)implies a=Sblor b=Sa$. It does take a bit of work, but by going back the way we came can ultimately be reduced to a statement about $Bbb N$ that is provable by induction.
answered 15 mins ago
Hagen von Eitzen
268k21260484
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This is readily proved once you properly define $Bbb Z$ and its operations. We may want to start from $Bbb N$ with its successor map $S$ and the Peano axioms, then define addition $Bbb Ntimes Bbb NtoBbb N$ recursively as $x+0:=x$, $x+Sy:=S(x+y)$, and then define multiplication recursively as $xcdot 0=0$, $xcdot Sy:=xcdot y+y$.
Now we see that $(a,b)sim (c,d):iff a+d=b+c$ is an equivalence relation on $Bbb N^2$, define $Bbb Z:=Bbb N^2/sim$ as the set of equivalence classes, define $overline(a,b)+overline(c,d):=overline(a+c,b+d)$ and define $overline(a,b)cdot overline(c,d):=overline(acdot c+bcdot d, acdot d+bcdot c)$ (which involves showing that this is well-defined). The claim now boils down to $overline(a,b)cdotoverline(c,d)=overline(1,0)implies a=Sblor b=Sa$. It does take a bit of work, but by going back the way we came can ultimately be reduced to a statement about $Bbb N$ that is provable by induction.
answered 15 mins ago
Hagen von Eitzen
268k21260484
This is readily proved once you properly define $Bbb Z$ and its operations. We may want to start from $Bbb N$ with its successor map $S$ and the Peano axioms, then define addition $Bbb Ntimes Bbb NtoBbb N$ recursively as $x+0:=x$, $x+Sy:=S(x+y)$, and then define multiplication recursively as $xcdot 0=0$, $xcdot Sy:=xcdot y+y$.
Now we see that $(a,b)sim (c,d):iff a+d=b+c$ is an equivalence relation on $Bbb N^2$, define $Bbb Z:=Bbb N^2/sim$ as the set of equivalence classes, define $overline(a,b)+overline(c,d):=overline(a+c,b+d)$ and define $overline(a,b)cdot overline(c,d):=overline(acdot c+bcdot d, acdot d+bcdot c)$ (which involves showing that this is well-defined). The claim now boils down to $overline(a,b)cdotoverline(c,d)=overline(1,0)implies a=Sblor b=Sa$. It does take a bit of work, but by going back the way we came can ultimately be reduced to a statement about $Bbb N$ that is provable by induction.
answered 15 mins ago
Hagen von Eitzen
268k21260484
answered 15 mins ago
Hagen von Eitzen
268k21260484
answered 15 mins ago
Hagen von Eitzen
268k21260484
answered 15 mins ago
Hagen von Eitzen
268k21260484
268k21260484
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Well, let's assume $ab=1$ when $a,binmathbbZ$. Of course it means that both $a$ and $b$ are non zero. Because they are integers we can conclude that their absolute values are at least $1$. And then $1=|ab|=|a||b|geq |a|$, and in the same way $1geq |b|$. So both $a$ and $b$ must belong to the set $-1,1$.
answered 24 mins ago
Mark
1,811110
add a comment |Â
up vote
2
down vote
Well, let's assume $ab=1$ when $a,binmathbbZ$. Of course it means that both $a$ and $b$ are non zero. Because they are integers we can conclude that their absolute values are at least $1$. And then $1=|ab|=|a||b|geq |a|$, and in the same way $1geq |b|$. So both $a$ and $b$ must belong to the set $-1,1$.
answered 24 mins ago
Mark
1,811110
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Well, let's assume $ab=1$ when $a,binmathbbZ$. Of course it means that both $a$ and $b$ are non zero. Because they are integers we can conclude that their absolute values are at least $1$. And then $1=|ab|=|a||b|geq |a|$, and in the same way $1geq |b|$. So both $a$ and $b$ must belong to the set $-1,1$.
answered 24 mins ago
Mark
1,811110
Well, let's assume $ab=1$ when $a,binmathbbZ$. Of course it means that both $a$ and $b$ are non zero. Because they are integers we can conclude that their absolute values are at least $1$. And then $1=|ab|=|a||b|geq |a|$, and in the same way $1geq |b|$. So both $a$ and $b$ must belong to the set $-1,1$.
answered 24 mins ago
Mark
1,811110
answered 24 mins ago
Mark
1,811110
answered 24 mins ago
Mark
1,811110
answered 24 mins ago
Mark
1,811110
1,811110
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It follows from the well-ordering principle for the natural numbers.
Suppose $ab=1$ for $a,b in BbbN$ with $b>1$. Then $c=a(b-1)$ is a natural number with $0<c<1$. So $c,c^2,c^3,dots$ is a set of natural numbers which has no least element, contradicting the well-ordering principle.
Now, if $ab=1$ for $a,b in BbbZ$, then either $a,b$ or $-a,-b$ will fit into the proof in the above paragraph...
answered 24 mins ago
Micah
28.9k1361102
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up vote
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It follows from the well-ordering principle for the natural numbers.
Suppose $ab=1$ for $a,b in BbbN$ with $b>1$. Then $c=a(b-1)$ is a natural number with $0<c<1$. So $c,c^2,c^3,dots$ is a set of natural numbers which has no least element, contradicting the well-ordering principle.
Now, if $ab=1$ for $a,b in BbbZ$, then either $a,b$ or $-a,-b$ will fit into the proof in the above paragraph...
answered 24 mins ago
Micah
28.9k1361102
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It follows from the well-ordering principle for the natural numbers.
Suppose $ab=1$ for $a,b in BbbN$ with $b>1$. Then $c=a(b-1)$ is a natural number with $0<c<1$. So $c,c^2,c^3,dots$ is a set of natural numbers which has no least element, contradicting the well-ordering principle.
Now, if $ab=1$ for $a,b in BbbZ$, then either $a,b$ or $-a,-b$ will fit into the proof in the above paragraph...
answered 24 mins ago
Micah
28.9k1361102
It follows from the well-ordering principle for the natural numbers.
Suppose $ab=1$ for $a,b in BbbN$ with $b>1$. Then $c=a(b-1)$ is a natural number with $0<c<1$. So $c,c^2,c^3,dots$ is a set of natural numbers which has no least element, contradicting the well-ordering principle.
Now, if $ab=1$ for $a,b in BbbZ$, then either $a,b$ or $-a,-b$ will fit into the proof in the above paragraph...
answered 24 mins ago
Micah
28.9k1361102
edited 17 mins ago
answered 24 mins ago
Micah
28.9k1361102
answered 24 mins ago
Micah
28.9k1361102
answered 24 mins ago
Micah
28.9k1361102
28.9k1361102
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up vote
1
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First of all, of course integers, have multiplicative inverses and of course that statement is obviously false as $a = frac 12, frac 13, frac 14,$ etc. bear out.
What you mean to say is $1$ and $-1$ are the only two integers whose multiplicative inverses are also integers.
I.e. If $a, frac 1a in mathbb Z$ then $a = 1$ or $a = -1$.
That is not an axiom.
If $a in mathbb Z$ then you have five possibilities. $a > 1$ or $a = 1$ or $a = 0$ or $a = -1$ or $a < 1$.
Case 1: $a > 1$. Then if $frac 1a < 0$ we would have $a*frac 1a < a*0$ and $1 < 0$ which is a contradiction. We can't have $frac 1a = 0$ because $0$ has no multiplicative inverses so $frac 1a > 0$. If $frac 1a ge 1$ then $a*frac 1a ge a*1 = a$ or $1 ge a$ which is a contradiction so $0< frac 1a < 1$ and not an integer.
Case 2: $a = 1$. then $frac 1a = 1$.
Case 3: $a = 0$ then $frac 1a$ is undefined.
Case 4: $a = -1$ then $frac 1a = -1$.
Case 5: $a < -1$. that is so similar to case 1: I'll leave it to you.
answered 19 mins ago
fleablood
61.8k22678
1
I don't agree with your first two paragraphs; at least not on a minimally generous reading of the question. Whether a number has a multiplicative inverse depends on which structure you're considering. In the ring of integers, non-units don't have multiplicative inverses. To say that "their inverses aren't integers" doesn't make sense within the structure; it only makes sense within the rationals. Though the question doesn't explicitly say "within the integers", it only talks about integers, and it stands to reason that the intended context is the ring of integers.
– joriki
8 mins ago
add a comment |Â
up vote
1
down vote
First of all, of course integers, have multiplicative inverses and of course that statement is obviously false as $a = frac 12, frac 13, frac 14,$ etc. bear out.
What you mean to say is $1$ and $-1$ are the only two integers whose multiplicative inverses are also integers.
I.e. If $a, frac 1a in mathbb Z$ then $a = 1$ or $a = -1$.
That is not an axiom.
If $a in mathbb Z$ then you have five possibilities. $a > 1$ or $a = 1$ or $a = 0$ or $a = -1$ or $a < 1$.
Case 1: $a > 1$. Then if $frac 1a < 0$ we would have $a*frac 1a < a*0$ and $1 < 0$ which is a contradiction. We can't have $frac 1a = 0$ because $0$ has no multiplicative inverses so $frac 1a > 0$. If $frac 1a ge 1$ then $a*frac 1a ge a*1 = a$ or $1 ge a$ which is a contradiction so $0< frac 1a < 1$ and not an integer.
Case 2: $a = 1$. then $frac 1a = 1$.
Case 3: $a = 0$ then $frac 1a$ is undefined.
Case 4: $a = -1$ then $frac 1a = -1$.
Case 5: $a < -1$. that is so similar to case 1: I'll leave it to you.
answered 19 mins ago
fleablood
61.8k22678
1
I don't agree with your first two paragraphs; at least not on a minimally generous reading of the question. Whether a number has a multiplicative inverse depends on which structure you're considering. In the ring of integers, non-units don't have multiplicative inverses. To say that "their inverses aren't integers" doesn't make sense within the structure; it only makes sense within the rationals. Though the question doesn't explicitly say "within the integers", it only talks about integers, and it stands to reason that the intended context is the ring of integers.
– joriki
8 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First of all, of course integers, have multiplicative inverses and of course that statement is obviously false as $a = frac 12, frac 13, frac 14,$ etc. bear out.
What you mean to say is $1$ and $-1$ are the only two integers whose multiplicative inverses are also integers.
I.e. If $a, frac 1a in mathbb Z$ then $a = 1$ or $a = -1$.
That is not an axiom.
If $a in mathbb Z$ then you have five possibilities. $a > 1$ or $a = 1$ or $a = 0$ or $a = -1$ or $a < 1$.
Case 1: $a > 1$. Then if $frac 1a < 0$ we would have $a*frac 1a < a*0$ and $1 < 0$ which is a contradiction. We can't have $frac 1a = 0$ because $0$ has no multiplicative inverses so $frac 1a > 0$. If $frac 1a ge 1$ then $a*frac 1a ge a*1 = a$ or $1 ge a$ which is a contradiction so $0< frac 1a < 1$ and not an integer.
Case 2: $a = 1$. then $frac 1a = 1$.
Case 3: $a = 0$ then $frac 1a$ is undefined.
Case 4: $a = -1$ then $frac 1a = -1$.
Case 5: $a < -1$. that is so similar to case 1: I'll leave it to you.
answered 19 mins ago
fleablood
61.8k22678
First of all, of course integers, have multiplicative inverses and of course that statement is obviously false as $a = frac 12, frac 13, frac 14,$ etc. bear out.
What you mean to say is $1$ and $-1$ are the only two integers whose multiplicative inverses are also integers.
I.e. If $a, frac 1a in mathbb Z$ then $a = 1$ or $a = -1$.
That is not an axiom.
If $a in mathbb Z$ then you have five possibilities. $a > 1$ or $a = 1$ or $a = 0$ or $a = -1$ or $a < 1$.
Case 1: $a > 1$. Then if $frac 1a < 0$ we would have $a*frac 1a < a*0$ and $1 < 0$ which is a contradiction. We can't have $frac 1a = 0$ because $0$ has no multiplicative inverses so $frac 1a > 0$. If $frac 1a ge 1$ then $a*frac 1a ge a*1 = a$ or $1 ge a$ which is a contradiction so $0< frac 1a < 1$ and not an integer.
Case 2: $a = 1$. then $frac 1a = 1$.
Case 3: $a = 0$ then $frac 1a$ is undefined.
Case 4: $a = -1$ then $frac 1a = -1$.
Case 5: $a < -1$. that is so similar to case 1: I'll leave it to you.
answered 19 mins ago
fleablood
61.8k22678
answered 19 mins ago
fleablood
61.8k22678
answered 19 mins ago
fleablood
61.8k22678
answered 19 mins ago
fleablood
61.8k22678
61.8k22678
1
I don't agree with your first two paragraphs; at least not on a minimally generous reading of the question. Whether a number has a multiplicative inverse depends on which structure you're considering. In the ring of integers, non-units don't have multiplicative inverses. To say that "their inverses aren't integers" doesn't make sense within the structure; it only makes sense within the rationals. Though the question doesn't explicitly say "within the integers", it only talks about integers, and it stands to reason that the intended context is the ring of integers.
– joriki
8 mins ago
add a comment |Â
1
I don't agree with your first two paragraphs; at least not on a minimally generous reading of the question. Whether a number has a multiplicative inverse depends on which structure you're considering. In the ring of integers, non-units don't have multiplicative inverses. To say that "their inverses aren't integers" doesn't make sense within the structure; it only makes sense within the rationals. Though the question doesn't explicitly say "within the integers", it only talks about integers, and it stands to reason that the intended context is the ring of integers.
– joriki
8 mins ago
1
1
I don't agree with your first two paragraphs; at least not on a minimally generous reading of the question. Whether a number has a multiplicative inverse depends on which structure you're considering. In the ring of integers, non-units don't have multiplicative inverses. To say that "their inverses aren't integers" doesn't make sense within the structure; it only makes sense within the rationals. Though the question doesn't explicitly say "within the integers", it only talks about integers, and it stands to reason that the intended context is the ring of integers.
– joriki
8 mins ago
I don't agree with your first two paragraphs; at least not on a minimally generous reading of the question. Whether a number has a multiplicative inverse depends on which structure you're considering. In the ring of integers, non-units don't have multiplicative inverses. To say that "their inverses aren't integers" doesn't make sense within the structure; it only makes sense within the rationals. Though the question doesn't explicitly say "within the integers", it only talks about integers, and it stands to reason that the intended context is the ring of integers.
– joriki
8 mins ago
add a comment |Â
up vote
0
down vote
As we are interested in 1/a which is just a symbol being adopted to denote the inverse of an element in Z. Further identity element under multiplication in Z is 1. So if any member say ''a' of Z is invertible then there must be a member in Z denoted as 1/a or a^(-1) such that
a. a^(-1) =1.
Now it's a simple algebra that in Z 1/a is an integer iff a= 1 or -1.
Hope it will help you.
answered 21 mins ago
prakash nainwal
524
Please see this tutorial and reference on how to typeset math on this site.
– joriki
1 min ago
add a comment |Â
up vote
0
down vote
As we are interested in 1/a which is just a symbol being adopted to denote the inverse of an element in Z. Further identity element under multiplication in Z is 1. So if any member say ''a' of Z is invertible then there must be a member in Z denoted as 1/a or a^(-1) such that
a. a^(-1) =1.
Now it's a simple algebra that in Z 1/a is an integer iff a= 1 or -1.
Hope it will help you.
answered 21 mins ago
prakash nainwal
524
Please see this tutorial and reference on how to typeset math on this site.
– joriki
1 min ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As we are interested in 1/a which is just a symbol being adopted to denote the inverse of an element in Z. Further identity element under multiplication in Z is 1. So if any member say ''a' of Z is invertible then there must be a member in Z denoted as 1/a or a^(-1) such that
a. a^(-1) =1.
Now it's a simple algebra that in Z 1/a is an integer iff a= 1 or -1.
Hope it will help you.
answered 21 mins ago
prakash nainwal
524
As we are interested in 1/a which is just a symbol being adopted to denote the inverse of an element in Z. Further identity element under multiplication in Z is 1. So if any member say ''a' of Z is invertible then there must be a member in Z denoted as 1/a or a^(-1) such that
a. a^(-1) =1.
Now it's a simple algebra that in Z 1/a is an integer iff a= 1 or -1.
Hope it will help you.
answered 21 mins ago
prakash nainwal
524
answered 21 mins ago
prakash nainwal
524
answered 21 mins ago
prakash nainwal
524
answered 21 mins ago
prakash nainwal
524
524
Please see this tutorial and reference on how to typeset math on this site.
– joriki
1 min ago
add a comment |Â
Please see this tutorial and reference on how to typeset math on this site.
– joriki
1 min ago
Please see this tutorial and reference on how to typeset math on this site.
– joriki
1 min ago
Please see this tutorial and reference on how to typeset math on this site.
– joriki
1 min ago
add a comment |Â
up vote
0
down vote
It depends upon the way that $mathbb Z$ and $times$ are defined, of course, but I doubt that anyone ever created axioms for the integers such that what you stated is an axiom.
Suppos, for instance, that you define:
$mathbb Z$ is the set of equivalence classes of $mathbbN^2$ with respect to the equivalence relation$$(a,b)sim(c,d)iff a+d=b+c;$$
$bigl[(a,b)bigr]timesbigl[(c,d)bigr]=bigl[(ac+bd,ad+bc)bigr]$.
Then your statement becomes a proposition, which can in fact be proved.
answered 21 mins ago
José Carlos Santos
123k17101186
add a comment |Â
up vote
0
down vote
It depends upon the way that $mathbb Z$ and $times$ are defined, of course, but I doubt that anyone ever created axioms for the integers such that what you stated is an axiom.
Suppos, for instance, that you define:
$mathbb Z$ is the set of equivalence classes of $mathbbN^2$ with respect to the equivalence relation$$(a,b)sim(c,d)iff a+d=b+c;$$
$bigl[(a,b)bigr]timesbigl[(c,d)bigr]=bigl[(ac+bd,ad+bc)bigr]$.
Then your statement becomes a proposition, which can in fact be proved.
answered 21 mins ago
José Carlos Santos
123k17101186
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It depends upon the way that $mathbb Z$ and $times$ are defined, of course, but I doubt that anyone ever created axioms for the integers such that what you stated is an axiom.
Suppos, for instance, that you define:
$mathbb Z$ is the set of equivalence classes of $mathbbN^2$ with respect to the equivalence relation$$(a,b)sim(c,d)iff a+d=b+c;$$
$bigl[(a,b)bigr]timesbigl[(c,d)bigr]=bigl[(ac+bd,ad+bc)bigr]$.
Then your statement becomes a proposition, which can in fact be proved.
answered 21 mins ago
José Carlos Santos
123k17101186
It depends upon the way that $mathbb Z$ and $times$ are defined, of course, but I doubt that anyone ever created axioms for the integers such that what you stated is an axiom.
Suppos, for instance, that you define:
$mathbb Z$ is the set of equivalence classes of $mathbbN^2$ with respect to the equivalence relation$$(a,b)sim(c,d)iff a+d=b+c;$$
$bigl[(a,b)bigr]timesbigl[(c,d)bigr]=bigl[(ac+bd,ad+bc)bigr]$.
Then your statement becomes a proposition, which can in fact be proved.
answered 21 mins ago
José Carlos Santos
123k17101186
answered 21 mins ago
José Carlos Santos
123k17101186
answered 21 mins ago
José Carlos Santos
123k17101186
answered 21 mins ago
José Carlos Santos
123k17101186
123k17101186
add a comment |Â
add a comment |Â
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