Multiplication cancellation property by Peano axioms
$begingroup$
I am trying to prove cancellation property of multiplication of natural numbers, $xy=xz$ implies $y=z$, with Peano axioms and arithmetic but not using or defining order on natural numbers. It can be done for addition. But for proving multiplication cancellation property one uses order. Why is that so?
peano-axioms
asked Aug 8 '18 at 1:00
jnyanjnyan
1,669615
$endgroup$
add a comment |
$begingroup$
I am trying to prove cancellation property of multiplication of natural numbers, $xy=xz$ implies $y=z$, with Peano axioms and arithmetic but not using or defining order on natural numbers. It can be done for addition. But for proving multiplication cancellation property one uses order. Why is that so?
peano-axioms
asked Aug 8 '18 at 1:00
jnyanjnyan
1,669615
$endgroup$
add a comment |
$begingroup$
I am trying to prove cancellation property of multiplication of natural numbers, $xy=xz$ implies $y=z$, with Peano axioms and arithmetic but not using or defining order on natural numbers. It can be done for addition. But for proving multiplication cancellation property one uses order. Why is that so?
peano-axioms
asked Aug 8 '18 at 1:00
jnyanjnyan
1,669615
$endgroup$
I am trying to prove cancellation property of multiplication of natural numbers, $xy=xz$ implies $y=z$, with Peano axioms and arithmetic but not using or defining order on natural numbers. It can be done for addition. But for proving multiplication cancellation property one uses order. Why is that so?
peano-axioms
peano-axioms
asked Aug 8 '18 at 1:00
jnyanjnyan
1,669615
asked Aug 8 '18 at 1:00
jnyanjnyan
1,669615
asked Aug 8 '18 at 1:00
jnyanjnyan
1,669615
asked Aug 8 '18 at 1:00
jnyanjnyan
1,669615
asked Aug 8 '18 at 1:00
jnyanjnyan
1,669615
1,669615
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You do not need to use order. Do it by induction on $y$. Define $M:={y |yx=zxRightarrow y=z }$.
For $y=1$ we have $1cdot x=zcdot x$. If $z=1$ we are done. Suppose $zneq 1$. Then $z=s(p)$ for some $p$ and hence
$$1cdot x=s(p)cdot x= px+x=(pcdot x)+1cdot x$$
and
$$1+(1cdot x)=1+(pcdot x)+(1cdot x).$$
Using cancellation for addition we get
$$1=1+(pcdot x)=(pcdot x)+1=s(pcdot x),$$
that is a contradiction. Therefore $z=1$.
Now, let $y=kin M$, i.e. for any positive integers $z,x$ if $kx=zx$ then $k=z$. Let $t,s in mathbb{N}_+$ be such that $s(k)cdot s=tcdot s$. We show that $s(k)=t$. Obviously $tneq 1$. Hence $t=s(m)$ for some $m$ and therefore we get
$$s(k)s=s(m)s$$
that is equaivalent to
$$ks+s=ms+s$$
from which it follows that $ks=ms$ (cancellation for addition). From the induction hypothesis we have $k=m$, and therefore $s(k)=s(m)=t$.
That means that $n=k+1in M$ and the proof is completed.
answered Jan 9 at 14:31
mikismikis
1,4631821
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2875553%2fmultiplication-cancellation-property-by-peano-axioms%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You do not need to use order. Do it by induction on $y$. Define $M:={y |yx=zxRightarrow y=z }$.
For $y=1$ we have $1cdot x=zcdot x$. If $z=1$ we are done. Suppose $zneq 1$. Then $z=s(p)$ for some $p$ and hence
$$1cdot x=s(p)cdot x= px+x=(pcdot x)+1cdot x$$
and
$$1+(1cdot x)=1+(pcdot x)+(1cdot x).$$
Using cancellation for addition we get
$$1=1+(pcdot x)=(pcdot x)+1=s(pcdot x),$$
that is a contradiction. Therefore $z=1$.
Now, let $y=kin M$, i.e. for any positive integers $z,x$ if $kx=zx$ then $k=z$. Let $t,s in mathbb{N}_+$ be such that $s(k)cdot s=tcdot s$. We show that $s(k)=t$. Obviously $tneq 1$. Hence $t=s(m)$ for some $m$ and therefore we get
$$s(k)s=s(m)s$$
that is equaivalent to
$$ks+s=ms+s$$
from which it follows that $ks=ms$ (cancellation for addition). From the induction hypothesis we have $k=m$, and therefore $s(k)=s(m)=t$.
That means that $n=k+1in M$ and the proof is completed.
answered Jan 9 at 14:31
mikismikis
1,4631821
$endgroup$
add a comment |
$begingroup$
You do not need to use order. Do it by induction on $y$. Define $M:={y |yx=zxRightarrow y=z }$.
For $y=1$ we have $1cdot x=zcdot x$. If $z=1$ we are done. Suppose $zneq 1$. Then $z=s(p)$ for some $p$ and hence
$$1cdot x=s(p)cdot x= px+x=(pcdot x)+1cdot x$$
and
$$1+(1cdot x)=1+(pcdot x)+(1cdot x).$$
Using cancellation for addition we get
$$1=1+(pcdot x)=(pcdot x)+1=s(pcdot x),$$
that is a contradiction. Therefore $z=1$.
Now, let $y=kin M$, i.e. for any positive integers $z,x$ if $kx=zx$ then $k=z$. Let $t,s in mathbb{N}_+$ be such that $s(k)cdot s=tcdot s$. We show that $s(k)=t$. Obviously $tneq 1$. Hence $t=s(m)$ for some $m$ and therefore we get
$$s(k)s=s(m)s$$
that is equaivalent to
$$ks+s=ms+s$$
from which it follows that $ks=ms$ (cancellation for addition). From the induction hypothesis we have $k=m$, and therefore $s(k)=s(m)=t$.
That means that $n=k+1in M$ and the proof is completed.
answered Jan 9 at 14:31
mikismikis
1,4631821
$endgroup$
add a comment |
$begingroup$
You do not need to use order. Do it by induction on $y$. Define $M:={y |yx=zxRightarrow y=z }$.
For $y=1$ we have $1cdot x=zcdot x$. If $z=1$ we are done. Suppose $zneq 1$. Then $z=s(p)$ for some $p$ and hence
$$1cdot x=s(p)cdot x= px+x=(pcdot x)+1cdot x$$
and
$$1+(1cdot x)=1+(pcdot x)+(1cdot x).$$
Using cancellation for addition we get
$$1=1+(pcdot x)=(pcdot x)+1=s(pcdot x),$$
that is a contradiction. Therefore $z=1$.
Now, let $y=kin M$, i.e. for any positive integers $z,x$ if $kx=zx$ then $k=z$. Let $t,s in mathbb{N}_+$ be such that $s(k)cdot s=tcdot s$. We show that $s(k)=t$. Obviously $tneq 1$. Hence $t=s(m)$ for some $m$ and therefore we get
$$s(k)s=s(m)s$$
that is equaivalent to
$$ks+s=ms+s$$
from which it follows that $ks=ms$ (cancellation for addition). From the induction hypothesis we have $k=m$, and therefore $s(k)=s(m)=t$.
That means that $n=k+1in M$ and the proof is completed.
answered Jan 9 at 14:31
mikismikis
1,4631821
$endgroup$
You do not need to use order. Do it by induction on $y$. Define $M:={y |yx=zxRightarrow y=z }$.
For $y=1$ we have $1cdot x=zcdot x$. If $z=1$ we are done. Suppose $zneq 1$. Then $z=s(p)$ for some $p$ and hence
$$1cdot x=s(p)cdot x= px+x=(pcdot x)+1cdot x$$
and
$$1+(1cdot x)=1+(pcdot x)+(1cdot x).$$
Using cancellation for addition we get
$$1=1+(pcdot x)=(pcdot x)+1=s(pcdot x),$$
that is a contradiction. Therefore $z=1$.
Now, let $y=kin M$, i.e. for any positive integers $z,x$ if $kx=zx$ then $k=z$. Let $t,s in mathbb{N}_+$ be such that $s(k)cdot s=tcdot s$. We show that $s(k)=t$. Obviously $tneq 1$. Hence $t=s(m)$ for some $m$ and therefore we get
$$s(k)s=s(m)s$$
that is equaivalent to
$$ks+s=ms+s$$
from which it follows that $ks=ms$ (cancellation for addition). From the induction hypothesis we have $k=m$, and therefore $s(k)=s(m)=t$.
That means that $n=k+1in M$ and the proof is completed.
answered Jan 9 at 14:31
mikismikis
1,4631821
answered Jan 9 at 14:31
mikismikis
1,4631821
answered Jan 9 at 14:31
mikismikis
1,4631821
answered Jan 9 at 14:31
mikismikis
1,4631821
1,4631821
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2875553%2fmultiplication-cancellation-property-by-peano-axioms%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
