Determining whether or not a relation involving absolute value is transitive
$begingroup$
I needed help doing a relation problem the specific problem is
$|x+y|$ = $|x|$ + $|y|$
I first tried thinking of some counter-examples but I couldn't think of any so i tried showing it.
So to show that a relation is transitive
you are given that $|x+y| = |x| +|y|$ and $|y+z| = |y| + |z|$ and need to show that $|x+z| = |x| + |z|$.
I tried isolating the second given for $|y|$ so I have $|y| = |y+z| - |z|$ and then using this I plugged it in the first given to get
$|x+y| = |x| + |y+z| - |z|$
Now this is where the problem rises since these have a absolute value around them I cant just cancel out the variables so I get stuck here and I am not sure what to do.
proof-verification relations
asked Dec 6 '18 at 23:29
RufyiRufyi
6618
$endgroup$
add a comment |
$begingroup$
I needed help doing a relation problem the specific problem is
$|x+y|$ = $|x|$ + $|y|$
I first tried thinking of some counter-examples but I couldn't think of any so i tried showing it.
So to show that a relation is transitive
you are given that $|x+y| = |x| +|y|$ and $|y+z| = |y| + |z|$ and need to show that $|x+z| = |x| + |z|$.
I tried isolating the second given for $|y|$ so I have $|y| = |y+z| - |z|$ and then using this I plugged it in the first given to get
$|x+y| = |x| + |y+z| - |z|$
Now this is where the problem rises since these have a absolute value around them I cant just cancel out the variables so I get stuck here and I am not sure what to do.
proof-verification relations
asked Dec 6 '18 at 23:29
RufyiRufyi
6618
$endgroup$
add a comment |
$begingroup$
I needed help doing a relation problem the specific problem is
$|x+y|$ = $|x|$ + $|y|$
I first tried thinking of some counter-examples but I couldn't think of any so i tried showing it.
So to show that a relation is transitive
you are given that $|x+y| = |x| +|y|$ and $|y+z| = |y| + |z|$ and need to show that $|x+z| = |x| + |z|$.
I tried isolating the second given for $|y|$ so I have $|y| = |y+z| - |z|$ and then using this I plugged it in the first given to get
$|x+y| = |x| + |y+z| - |z|$
Now this is where the problem rises since these have a absolute value around them I cant just cancel out the variables so I get stuck here and I am not sure what to do.
proof-verification relations
asked Dec 6 '18 at 23:29
RufyiRufyi
6618
$endgroup$
I needed help doing a relation problem the specific problem is
$|x+y|$ = $|x|$ + $|y|$
I first tried thinking of some counter-examples but I couldn't think of any so i tried showing it.
So to show that a relation is transitive
you are given that $|x+y| = |x| +|y|$ and $|y+z| = |y| + |z|$ and need to show that $|x+z| = |x| + |z|$.
I tried isolating the second given for $|y|$ so I have $|y| = |y+z| - |z|$ and then using this I plugged it in the first given to get
$|x+y| = |x| + |y+z| - |z|$
Now this is where the problem rises since these have a absolute value around them I cant just cancel out the variables so I get stuck here and I am not sure what to do.
proof-verification relations
proof-verification relations
asked Dec 6 '18 at 23:29
RufyiRufyi
6618
asked Dec 6 '18 at 23:29
RufyiRufyi
6618
asked Dec 6 '18 at 23:29
RufyiRufyi
6618
asked Dec 6 '18 at 23:29
RufyiRufyi
6618
asked Dec 6 '18 at 23:29
RufyiRufyi
6618
6618
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $|x+y|=|x|+|y|$, by squaring both sides we get $xy=|xy|$.
Thus either one of the numbers is $0$ or they are both nonzero and have the same sign.
What happens with $x=1$, $y=0$ and $z=-1$?
answered Dec 6 '18 at 23:53
egregegreg
181k1485203
$endgroup$
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%2f3029211%2fdetermining-whether-or-not-a-relation-involving-absolute-value-is-transitive%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
$begingroup$
If $|x+y|=|x|+|y|$, by squaring both sides we get $xy=|xy|$.
Thus either one of the numbers is $0$ or they are both nonzero and have the same sign.
What happens with $x=1$, $y=0$ and $z=-1$?
answered Dec 6 '18 at 23:53
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
If $|x+y|=|x|+|y|$, by squaring both sides we get $xy=|xy|$.
Thus either one of the numbers is $0$ or they are both nonzero and have the same sign.
What happens with $x=1$, $y=0$ and $z=-1$?
answered Dec 6 '18 at 23:53
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
If $|x+y|=|x|+|y|$, by squaring both sides we get $xy=|xy|$.
Thus either one of the numbers is $0$ or they are both nonzero and have the same sign.
What happens with $x=1$, $y=0$ and $z=-1$?
answered Dec 6 '18 at 23:53
egregegreg
181k1485203
$endgroup$
If $|x+y|=|x|+|y|$, by squaring both sides we get $xy=|xy|$.
Thus either one of the numbers is $0$ or they are both nonzero and have the same sign.
What happens with $x=1$, $y=0$ and $z=-1$?
answered Dec 6 '18 at 23:53
egregegreg
181k1485203
answered Dec 6 '18 at 23:53
egregegreg
181k1485203
answered Dec 6 '18 at 23:53
egregegreg
181k1485203
answered Dec 6 '18 at 23:53
egregegreg
181k1485203
181k1485203
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%2f3029211%2fdetermining-whether-or-not-a-relation-involving-absolute-value-is-transitive%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
