Mixing
Prove that for all $x in R$ at least one of $sqrt{3}-x$ or $sqrt{3}+x$ is irrational
1
$begingroup$
Prove that for all $x in R$ at least one of $sqrt{3}-x$ or $sqrt{3}+x$ is irrational.
Could you not just prove this using contradiction by plugging in $1$ for
$x$?
I'm having difficulty even starting this proof.
proof-verification proof-writing proof-explanation
asked Jan 28 at 2:38
JohnJohn
446
$endgroup$
add a comment |
1
$begingroup$
Prove that for all $x in R$ at least one of $sqrt{3}-x$ or $sqrt{3}+x$ is irrational.
Could you not just prove this using contradiction by plugging in $1$ for
$x$?
I'm having difficulty even starting this proof.
proof-verification proof-writing proof-explanation
asked Jan 28 at 2:38
JohnJohn
446
$endgroup$
1
$begingroup$
You are asked to prove it for all $x$, so plugging in one value will not work. If you wanted to disprove the claim you could find one value where it failed, but here the claim is true.
$endgroup$
– Ross Millikan
Jan 28 at 2:47
add a comment |
1
1
1
2
$begingroup$
Prove that for all $x in R$ at least one of $sqrt{3}-x$ or $sqrt{3}+x$ is irrational.
Could you not just prove this using contradiction by plugging in $1$ for
$x$?
I'm having difficulty even starting this proof.
proof-verification proof-writing proof-explanation
asked Jan 28 at 2:38
JohnJohn
446
$endgroup$
Prove that for all $x in R$ at least one of $sqrt{3}-x$ or $sqrt{3}+x$ is irrational.
Could you not just prove this using contradiction by plugging in $1$ for
$x$?
I'm having difficulty even starting this proof.
proof-verification proof-writing proof-explanation
proof-verification proof-writing proof-explanation
asked Jan 28 at 2:38
JohnJohn
446
asked Jan 28 at 2:38
JohnJohn
446
asked Jan 28 at 2:38
JohnJohn
446
asked Jan 28 at 2:38
JohnJohn
446
asked Jan 28 at 2:38
JohnJohn
446
446
1
$begingroup$
You are asked to prove it for all $x$, so plugging in one value will not work. If you wanted to disprove the claim you could find one value where it failed, but here the claim is true.
$endgroup$
– Ross Millikan
Jan 28 at 2:47
add a comment |
1
$begingroup$
You are asked to prove it for all $x$, so plugging in one value will not work. If you wanted to disprove the claim you could find one value where it failed, but here the claim is true.
$endgroup$
– Ross Millikan
Jan 28 at 2:47
1
1
$begingroup$
You are asked to prove it for all $x$, so plugging in one value will not work. If you wanted to disprove the claim you could find one value where it failed, but here the claim is true.
$endgroup$
– Ross Millikan
Jan 28 at 2:47
$begingroup$
You are asked to prove it for all $x$, so plugging in one value will not work. If you wanted to disprove the claim you could find one value where it failed, but here the claim is true.
$endgroup$
– Ross Millikan
Jan 28 at 2:47
add a comment |
1 Answer
1
active
oldest
votes
3
$begingroup$
If for any $x in Bbb R$,
$sqrt 3 + x, ; sqrt 3 - x in Bbb Q, tag 1$
then
$2sqrt 3 = (sqrt 3 + x) + (sqrt 3 - x) in Bbb Q, tag 2$
whence
$sqrt 3 in Bbb Q, tag 3$
contradicting the irrationality of $sqrt 3$; thus, (1) is false.
answered Jan 28 at 2:46
Robert LewisRobert Lewis
48.4k23167
$endgroup$
1
$begingroup$
So the clue to an approach is "at least one is". Assume "both aren't", and get the contradiction. :-)
$endgroup$
– timtfj
Jan 28 at 3:48
$begingroup$
@timtfj: yes. Simple logic. $not exists x p(x) equiv forall x neg p(x)$.
$endgroup$
– Robert Lewis
Jan 28 at 3:59
1
$begingroup$
I'm afraid I think of it as $A+B=overline {overline A cdot overline B}$ for this one, but that's my background showing itself ;-) (And having just typed that, I can see why nobody here is using overbars in boolean expressions! It was a nightmare.)
$endgroup$
– timtfj
Jan 28 at 4:30
1
$begingroup$
Thank you for your answer. Is there any other approaches to this proof?
$endgroup$
– John
Jan 28 at 23:25
$begingroup$
@John: probably, but I'd have to think about it to find an example. Thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 29 at 0:19
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
});
}
});
draft saved
draft discarded
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%2f3090383%2fprove-that-for-all-x-in-r-at-least-one-of-sqrt3-x-or-sqrt3x-is-irr%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
3
$begingroup$
If for any $x in Bbb R$,
$sqrt 3 + x, ; sqrt 3 - x in Bbb Q, tag 1$
then
$2sqrt 3 = (sqrt 3 + x) + (sqrt 3 - x) in Bbb Q, tag 2$
whence
$sqrt 3 in Bbb Q, tag 3$
contradicting the irrationality of $sqrt 3$; thus, (1) is false.
answered Jan 28 at 2:46
Robert LewisRobert Lewis
48.4k23167
$endgroup$
1
$begingroup$
So the clue to an approach is "at least one is". Assume "both aren't", and get the contradiction. :-)
$endgroup$
– timtfj
Jan 28 at 3:48
$begingroup$
@timtfj: yes. Simple logic. $not exists x p(x) equiv forall x neg p(x)$.
$endgroup$
– Robert Lewis
Jan 28 at 3:59
1
$begingroup$
I'm afraid I think of it as $A+B=overline {overline A cdot overline B}$ for this one, but that's my background showing itself ;-) (And having just typed that, I can see why nobody here is using overbars in boolean expressions! It was a nightmare.)
$endgroup$
– timtfj
Jan 28 at 4:30
1
$begingroup$
Thank you for your answer. Is there any other approaches to this proof?
$endgroup$
– John
Jan 28 at 23:25
$begingroup$
@John: probably, but I'd have to think about it to find an example. Thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 29 at 0:19
add a comment |
3
$begingroup$
If for any $x in Bbb R$,
$sqrt 3 + x, ; sqrt 3 - x in Bbb Q, tag 1$
then
$2sqrt 3 = (sqrt 3 + x) + (sqrt 3 - x) in Bbb Q, tag 2$
whence
$sqrt 3 in Bbb Q, tag 3$
contradicting the irrationality of $sqrt 3$; thus, (1) is false.
answered Jan 28 at 2:46
Robert LewisRobert Lewis
48.4k23167
$endgroup$
1
$begingroup$
So the clue to an approach is "at least one is". Assume "both aren't", and get the contradiction. :-)
$endgroup$
– timtfj
Jan 28 at 3:48
$begingroup$
@timtfj: yes. Simple logic. $not exists x p(x) equiv forall x neg p(x)$.
$endgroup$
– Robert Lewis
Jan 28 at 3:59
1
$begingroup$
I'm afraid I think of it as $A+B=overline {overline A cdot overline B}$ for this one, but that's my background showing itself ;-) (And having just typed that, I can see why nobody here is using overbars in boolean expressions! It was a nightmare.)
$endgroup$
– timtfj
Jan 28 at 4:30
1
$begingroup$
Thank you for your answer. Is there any other approaches to this proof?
$endgroup$
– John
Jan 28 at 23:25
$begingroup$
@John: probably, but I'd have to think about it to find an example. Thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 29 at 0:19
add a comment |
3
3
3
$begingroup$
If for any $x in Bbb R$,
$sqrt 3 + x, ; sqrt 3 - x in Bbb Q, tag 1$
then
$2sqrt 3 = (sqrt 3 + x) + (sqrt 3 - x) in Bbb Q, tag 2$
whence
$sqrt 3 in Bbb Q, tag 3$
contradicting the irrationality of $sqrt 3$; thus, (1) is false.
answered Jan 28 at 2:46
Robert LewisRobert Lewis
48.4k23167
$endgroup$
If for any $x in Bbb R$,
$sqrt 3 + x, ; sqrt 3 - x in Bbb Q, tag 1$
then
$2sqrt 3 = (sqrt 3 + x) + (sqrt 3 - x) in Bbb Q, tag 2$
whence
$sqrt 3 in Bbb Q, tag 3$
contradicting the irrationality of $sqrt 3$; thus, (1) is false.
answered Jan 28 at 2:46
Robert LewisRobert Lewis
48.4k23167
answered Jan 28 at 2:46
Robert LewisRobert Lewis
48.4k23167
answered Jan 28 at 2:46
Robert LewisRobert Lewis
48.4k23167
answered Jan 28 at 2:46
Robert LewisRobert Lewis
48.4k23167
48.4k23167
1
$begingroup$
So the clue to an approach is "at least one is". Assume "both aren't", and get the contradiction. :-)
$endgroup$
– timtfj
Jan 28 at 3:48
$begingroup$
@timtfj: yes. Simple logic. $not exists x p(x) equiv forall x neg p(x)$.
$endgroup$
– Robert Lewis
Jan 28 at 3:59
1
$begingroup$
I'm afraid I think of it as $A+B=overline {overline A cdot overline B}$ for this one, but that's my background showing itself ;-) (And having just typed that, I can see why nobody here is using overbars in boolean expressions! It was a nightmare.)
$endgroup$
– timtfj
Jan 28 at 4:30
1
$begingroup$
Thank you for your answer. Is there any other approaches to this proof?
$endgroup$
– John
Jan 28 at 23:25
$begingroup$
@John: probably, but I'd have to think about it to find an example. Thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 29 at 0:19
add a comment |
1
$begingroup$
So the clue to an approach is "at least one is". Assume "both aren't", and get the contradiction. :-)
$endgroup$
– timtfj
Jan 28 at 3:48
$begingroup$
@timtfj: yes. Simple logic. $not exists x p(x) equiv forall x neg p(x)$.
$endgroup$
– Robert Lewis
Jan 28 at 3:59
1
$begingroup$
I'm afraid I think of it as $A+B=overline {overline A cdot overline B}$ for this one, but that's my background showing itself ;-) (And having just typed that, I can see why nobody here is using overbars in boolean expressions! It was a nightmare.)
$endgroup$
– timtfj
Jan 28 at 4:30
1
$begingroup$
Thank you for your answer. Is there any other approaches to this proof?
$endgroup$
– John
Jan 28 at 23:25
$begingroup$
@John: probably, but I'd have to think about it to find an example. Thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 29 at 0:19
1
1
$begingroup$
So the clue to an approach is "at least one is". Assume "both aren't", and get the contradiction. :-)
$endgroup$
– timtfj
Jan 28 at 3:48
$begingroup$
So the clue to an approach is "at least one is". Assume "both aren't", and get the contradiction. :-)
$endgroup$
– timtfj
Jan 28 at 3:48
$begingroup$
@timtfj: yes. Simple logic. $not exists x p(x) equiv forall x neg p(x)$.
$endgroup$
– Robert Lewis
Jan 28 at 3:59
$begingroup$
@timtfj: yes. Simple logic. $not exists x p(x) equiv forall x neg p(x)$.
$endgroup$
– Robert Lewis
Jan 28 at 3:59
1
1
$begingroup$
I'm afraid I think of it as $A+B=overline {overline A cdot overline B}$ for this one, but that's my background showing itself ;-) (And having just typed that, I can see why nobody here is using overbars in boolean expressions! It was a nightmare.)
$endgroup$
– timtfj
Jan 28 at 4:30
$begingroup$
I'm afraid I think of it as $A+B=overline {overline A cdot overline B}$ for this one, but that's my background showing itself ;-) (And having just typed that, I can see why nobody here is using overbars in boolean expressions! It was a nightmare.)
$endgroup$
– timtfj
Jan 28 at 4:30
1
1
$begingroup$
Thank you for your answer. Is there any other approaches to this proof?
$endgroup$
– John
Jan 28 at 23:25
$begingroup$
Thank you for your answer. Is there any other approaches to this proof?
$endgroup$
– John
Jan 28 at 23:25
$begingroup$
@John: probably, but I'd have to think about it to find an example. Thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 29 at 0:19
$begingroup$
@John: probably, but I'd have to think about it to find an example. Thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 29 at 0:19
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3090383%2fprove-that-for-all-x-in-r-at-least-one-of-sqrt3-x-or-sqrt3x-is-irr%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
1
$begingroup$
You are asked to prove it for all $x$, so plugging in one value will not work. If you wanted to disprove the claim you could find one value where it failed, but here the claim is true.
$endgroup$
– Ross Millikan
Jan 28 at 2:47