Mixing
Can $sqrt{n} + sqrt{m}$ be rational if neither $n,m$ are perfect squares?
$begingroup$
Can the expression $sqrt{n} + sqrt{m}$ be rational if neither $n,m in mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $sqrt{m} = a-sqrt{n}, a in mathbb{Q}$, which I don't think is possible, but how to show it?
number-theory discrete-mathematics irrational-numbers rationality-testing faq
edited Nov 23 '15 at 20:04
Martin Sleziak
44.7k9117272
asked Aug 1 '13 at 16:42
Spine FeastSpine Feast
1,86812349
$endgroup$
add a comment |
$begingroup$
Can the expression $sqrt{n} + sqrt{m}$ be rational if neither $n,m in mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $sqrt{m} = a-sqrt{n}, a in mathbb{Q}$, which I don't think is possible, but how to show it?
number-theory discrete-mathematics irrational-numbers rationality-testing faq
edited Nov 23 '15 at 20:04
Martin Sleziak
44.7k9117272
asked Aug 1 '13 at 16:42
Spine FeastSpine Feast
1,86812349
$endgroup$
1
$begingroup$
For a general case see this question and an earlier version
$endgroup$
– Jyrki Lahtonen
May 31 '14 at 11:49
$begingroup$
Related: math.stackexchange.com/questions/890821
$endgroup$
– Watson
Nov 25 '18 at 17:46
add a comment |
$begingroup$
Can the expression $sqrt{n} + sqrt{m}$ be rational if neither $n,m in mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $sqrt{m} = a-sqrt{n}, a in mathbb{Q}$, which I don't think is possible, but how to show it?
number-theory discrete-mathematics irrational-numbers rationality-testing faq
edited Nov 23 '15 at 20:04
Martin Sleziak
44.7k9117272
asked Aug 1 '13 at 16:42
Spine FeastSpine Feast
1,86812349
$endgroup$
Can the expression $sqrt{n} + sqrt{m}$ be rational if neither $n,m in mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $sqrt{m} = a-sqrt{n}, a in mathbb{Q}$, which I don't think is possible, but how to show it?
number-theory discrete-mathematics irrational-numbers rationality-testing faq
number-theory discrete-mathematics irrational-numbers rationality-testing faq
edited Nov 23 '15 at 20:04
Martin Sleziak
44.7k9117272
asked Aug 1 '13 at 16:42
Spine FeastSpine Feast
1,86812349
edited Nov 23 '15 at 20:04
Martin Sleziak
44.7k9117272
asked Aug 1 '13 at 16:42
Spine FeastSpine Feast
1,86812349
edited Nov 23 '15 at 20:04
Martin Sleziak
44.7k9117272
edited Nov 23 '15 at 20:04
Martin Sleziak
44.7k9117272
edited Nov 23 '15 at 20:04
Martin Sleziak
44.7k9117272
44.7k9117272
asked Aug 1 '13 at 16:42
Spine FeastSpine Feast
1,86812349
asked Aug 1 '13 at 16:42
Spine FeastSpine Feast
1,86812349
asked Aug 1 '13 at 16:42
Spine FeastSpine Feast
1,86812349
1,86812349
1
$begingroup$
For a general case see this question and an earlier version
$endgroup$
– Jyrki Lahtonen
May 31 '14 at 11:49
$begingroup$
Related: math.stackexchange.com/questions/890821
$endgroup$
– Watson
Nov 25 '18 at 17:46
add a comment |
1
$begingroup$
For a general case see this question and an earlier version
$endgroup$
– Jyrki Lahtonen
May 31 '14 at 11:49
$begingroup$
Related: math.stackexchange.com/questions/890821
$endgroup$
– Watson
Nov 25 '18 at 17:46
1
1
$begingroup$
For a general case see this question and an earlier version
$endgroup$
– Jyrki Lahtonen
May 31 '14 at 11:49
$begingroup$
For a general case see this question and an earlier version
$endgroup$
– Jyrki Lahtonen
May 31 '14 at 11:49
$begingroup$
Related: math.stackexchange.com/questions/890821
$endgroup$
– Watson
Nov 25 '18 at 17:46
$begingroup$
Related: math.stackexchange.com/questions/890821
$endgroup$
– Watson
Nov 25 '18 at 17:46
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Squaring we get, $m=a^2+n-2asqrt nimplies sqrt n=frac{a^2+n-m}{2a}$ which is rational
answered Aug 1 '13 at 16:44
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
@SujaanKunalan, thanks for your feedback
$endgroup$
– lab bhattacharjee
Aug 1 '13 at 17:52
add a comment |
$begingroup$
Assume $m$ is a non-square integer. Then $sqrt{m}$ is irrational, and if $x=sqrt{m}+sqrt{n}$, then
$$(x-sqrt{m})^2=x^2-2xsqrt{m}+m=n$$
Or
$$frac{x^2+m-n}{2x}=sqrt{m}$$
If $x$ is rational, then the LHS is also rational. However the RHS is irrational, contradiction, so $x$ is irrational.
Same argument as here and here. It should be put in the FAQ :-)
edited Apr 13 '17 at 12:21
Community♦
1
answered Sep 22 '14 at 13:54
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
$endgroup$
add a comment |
$begingroup$
If $sqrt{n} + sqrt{m}$ is rational, then since
($sqrt{n} + sqrt{m})(sqrt{n} - sqrt{m}) = n - m,$
$sqrt{n} - sqrt{m}$ is rational. Thus
$sqrt{n}, sqrt{m}$ are rational, n,m are squares.
answered Aug 12 '18 at 11:15
William ElliotWilliam Elliot
7,7322720
$endgroup$
$begingroup$
Just wanted to say that this is a surprisingly slick solution, and deserves more recognition.
$endgroup$
– platty
Nov 28 '18 at 6:44
add a comment |
$begingroup$
Nice way to see thinks
Assume that, $$(sqrt{n}+sqrt{m})=frac{p}{q}$$
Then we have
$$(sqrt{n}+sqrt{m})=frac{p}{q}inBbb Q implies
n+m+2sqrt{nm} =(sqrt{n}+sqrt{m})^2 =frac{p^2}{q^2}inBbb Q\implies sqrt{nm} =frac{n+m}{2}+frac{p^2}{2q^2}inBbb Q $$
But if $ nm $ is not a perfect square then $sqrt{nm}not inBbb Q ,$ (This can be easily prove using the fundamental theorem of number theory: Decomposition into prime numbers)
Hence in this case we have $$sqrt{n}+sqrt{m}not inBbb Q$$
Remark $~~~~~$1. $mn$ can be a perfect square even though neither $n$ nor $m$ is a perfect square. (see the example below)
- We can still have $sqrt{n}+sqrt{m}notin Bbb Q$ even if $mn$ is perfect square.(see the example below)
Example: $n= 3$ and $ m = 12$ are not perfect square and $ nm = 36 =6^2.$
Moreover,
$$sqrt{n}+sqrt{m} = sqrt{3}+sqrt{12} =3sqrt 3 not inBbb Q$$
answered Oct 16 '17 at 19:55
Guy FsoneGuy Fsone
17.1k42873
$endgroup$
add a comment |
$begingroup$
It's easy to show that if the result is rational, it has to be natural.
Hint:
$$m=(lfloorsqrt{m}rfloor+epsilon)^{2}=lfloorsqrt m rfloor ^2+2epsilon lfloorsqrt m rfloor + epsilon^2$$
$$n=(lceil sqrt n rceil-epsilon)^2=lceil sqrt n rceil^2-2epsilonlceil sqrt n rceil+epsilon^2$$
Subtract these two from each other and show that $epsilon$ has to be rational.
answered Sep 22 '14 at 13:43
AliAli
1,144924
$endgroup$
$begingroup$
m+n+2*sqrt(m*n) has to be rational
$endgroup$
– Mahdi
Sep 22 '14 at 13:53
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%2f457382%2fcan-sqrtn-sqrtm-be-rational-if-neither-n-m-are-perfect-squares%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Squaring we get, $m=a^2+n-2asqrt nimplies sqrt n=frac{a^2+n-m}{2a}$ which is rational
answered Aug 1 '13 at 16:44
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
@SujaanKunalan, thanks for your feedback
$endgroup$
– lab bhattacharjee
Aug 1 '13 at 17:52
add a comment |
$begingroup$
Squaring we get, $m=a^2+n-2asqrt nimplies sqrt n=frac{a^2+n-m}{2a}$ which is rational
answered Aug 1 '13 at 16:44
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
@SujaanKunalan, thanks for your feedback
$endgroup$
– lab bhattacharjee
Aug 1 '13 at 17:52
add a comment |
$begingroup$
Squaring we get, $m=a^2+n-2asqrt nimplies sqrt n=frac{a^2+n-m}{2a}$ which is rational
answered Aug 1 '13 at 16:44
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
Squaring we get, $m=a^2+n-2asqrt nimplies sqrt n=frac{a^2+n-m}{2a}$ which is rational
answered Aug 1 '13 at 16:44
lab bhattacharjeelab bhattacharjee
224k15156274
answered Aug 1 '13 at 16:44
lab bhattacharjeelab bhattacharjee
224k15156274
answered Aug 1 '13 at 16:44
lab bhattacharjeelab bhattacharjee
224k15156274
answered Aug 1 '13 at 16:44
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
$begingroup$
@SujaanKunalan, thanks for your feedback
$endgroup$
– lab bhattacharjee
Aug 1 '13 at 17:52
add a comment |
$begingroup$
@SujaanKunalan, thanks for your feedback
$endgroup$
– lab bhattacharjee
Aug 1 '13 at 17:52
$begingroup$
@SujaanKunalan, thanks for your feedback
$endgroup$
– lab bhattacharjee
Aug 1 '13 at 17:52
$begingroup$
@SujaanKunalan, thanks for your feedback
$endgroup$
– lab bhattacharjee
Aug 1 '13 at 17:52
add a comment |
$begingroup$
Assume $m$ is a non-square integer. Then $sqrt{m}$ is irrational, and if $x=sqrt{m}+sqrt{n}$, then
$$(x-sqrt{m})^2=x^2-2xsqrt{m}+m=n$$
Or
$$frac{x^2+m-n}{2x}=sqrt{m}$$
If $x$ is rational, then the LHS is also rational. However the RHS is irrational, contradiction, so $x$ is irrational.
Same argument as here and here. It should be put in the FAQ :-)
edited Apr 13 '17 at 12:21
Community♦
1
answered Sep 22 '14 at 13:54
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
$endgroup$
add a comment |
$begingroup$
Assume $m$ is a non-square integer. Then $sqrt{m}$ is irrational, and if $x=sqrt{m}+sqrt{n}$, then
$$(x-sqrt{m})^2=x^2-2xsqrt{m}+m=n$$
Or
$$frac{x^2+m-n}{2x}=sqrt{m}$$
If $x$ is rational, then the LHS is also rational. However the RHS is irrational, contradiction, so $x$ is irrational.
Same argument as here and here. It should be put in the FAQ :-)
edited Apr 13 '17 at 12:21
Community♦
1
answered Sep 22 '14 at 13:54
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
$endgroup$
add a comment |
$begingroup$
Assume $m$ is a non-square integer. Then $sqrt{m}$ is irrational, and if $x=sqrt{m}+sqrt{n}$, then
$$(x-sqrt{m})^2=x^2-2xsqrt{m}+m=n$$
Or
$$frac{x^2+m-n}{2x}=sqrt{m}$$
If $x$ is rational, then the LHS is also rational. However the RHS is irrational, contradiction, so $x$ is irrational.
Same argument as here and here. It should be put in the FAQ :-)
edited Apr 13 '17 at 12:21
Community♦
1
answered Sep 22 '14 at 13:54
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
$endgroup$
Assume $m$ is a non-square integer. Then $sqrt{m}$ is irrational, and if $x=sqrt{m}+sqrt{n}$, then
$$(x-sqrt{m})^2=x^2-2xsqrt{m}+m=n$$
Or
$$frac{x^2+m-n}{2x}=sqrt{m}$$
If $x$ is rational, then the LHS is also rational. However the RHS is irrational, contradiction, so $x$ is irrational.
Same argument as here and here. It should be put in the FAQ :-)
edited Apr 13 '17 at 12:21
Community♦
1
answered Sep 22 '14 at 13:54
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Sep 22 '14 at 13:54
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
answered Sep 22 '14 at 13:54
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
answered Sep 22 '14 at 13:54
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
14.7k63464
add a comment |
add a comment |
$begingroup$
If $sqrt{n} + sqrt{m}$ is rational, then since
($sqrt{n} + sqrt{m})(sqrt{n} - sqrt{m}) = n - m,$
$sqrt{n} - sqrt{m}$ is rational. Thus
$sqrt{n}, sqrt{m}$ are rational, n,m are squares.
answered Aug 12 '18 at 11:15
William ElliotWilliam Elliot
7,7322720
$endgroup$
$begingroup$
Just wanted to say that this is a surprisingly slick solution, and deserves more recognition.
$endgroup$
– platty
Nov 28 '18 at 6:44
add a comment |
$begingroup$
If $sqrt{n} + sqrt{m}$ is rational, then since
($sqrt{n} + sqrt{m})(sqrt{n} - sqrt{m}) = n - m,$
$sqrt{n} - sqrt{m}$ is rational. Thus
$sqrt{n}, sqrt{m}$ are rational, n,m are squares.
answered Aug 12 '18 at 11:15
William ElliotWilliam Elliot
7,7322720
$endgroup$
$begingroup$
Just wanted to say that this is a surprisingly slick solution, and deserves more recognition.
$endgroup$
– platty
Nov 28 '18 at 6:44
add a comment |
$begingroup$
If $sqrt{n} + sqrt{m}$ is rational, then since
($sqrt{n} + sqrt{m})(sqrt{n} - sqrt{m}) = n - m,$
$sqrt{n} - sqrt{m}$ is rational. Thus
$sqrt{n}, sqrt{m}$ are rational, n,m are squares.
answered Aug 12 '18 at 11:15
William ElliotWilliam Elliot
7,7322720
$endgroup$
If $sqrt{n} + sqrt{m}$ is rational, then since
($sqrt{n} + sqrt{m})(sqrt{n} - sqrt{m}) = n - m,$
$sqrt{n} - sqrt{m}$ is rational. Thus
$sqrt{n}, sqrt{m}$ are rational, n,m are squares.
answered Aug 12 '18 at 11:15
William ElliotWilliam Elliot
7,7322720
edited Jan 6 at 0:31
answered Aug 12 '18 at 11:15
William ElliotWilliam Elliot
7,7322720
answered Aug 12 '18 at 11:15
William ElliotWilliam Elliot
7,7322720
answered Aug 12 '18 at 11:15
William ElliotWilliam Elliot
7,7322720
7,7322720
$begingroup$
Just wanted to say that this is a surprisingly slick solution, and deserves more recognition.
$endgroup$
– platty
Nov 28 '18 at 6:44
add a comment |
$begingroup$
Just wanted to say that this is a surprisingly slick solution, and deserves more recognition.
$endgroup$
– platty
Nov 28 '18 at 6:44
$begingroup$
Just wanted to say that this is a surprisingly slick solution, and deserves more recognition.
$endgroup$
– platty
Nov 28 '18 at 6:44
$begingroup$
Just wanted to say that this is a surprisingly slick solution, and deserves more recognition.
$endgroup$
– platty
Nov 28 '18 at 6:44
add a comment |
$begingroup$
Nice way to see thinks
Assume that, $$(sqrt{n}+sqrt{m})=frac{p}{q}$$
Then we have
$$(sqrt{n}+sqrt{m})=frac{p}{q}inBbb Q implies
n+m+2sqrt{nm} =(sqrt{n}+sqrt{m})^2 =frac{p^2}{q^2}inBbb Q\implies sqrt{nm} =frac{n+m}{2}+frac{p^2}{2q^2}inBbb Q $$
But if $ nm $ is not a perfect square then $sqrt{nm}not inBbb Q ,$ (This can be easily prove using the fundamental theorem of number theory: Decomposition into prime numbers)
Hence in this case we have $$sqrt{n}+sqrt{m}not inBbb Q$$
Remark $~~~~~$1. $mn$ can be a perfect square even though neither $n$ nor $m$ is a perfect square. (see the example below)
- We can still have $sqrt{n}+sqrt{m}notin Bbb Q$ even if $mn$ is perfect square.(see the example below)
Example: $n= 3$ and $ m = 12$ are not perfect square and $ nm = 36 =6^2.$
Moreover,
$$sqrt{n}+sqrt{m} = sqrt{3}+sqrt{12} =3sqrt 3 not inBbb Q$$
answered Oct 16 '17 at 19:55
Guy FsoneGuy Fsone
17.1k42873
$endgroup$
add a comment |
$begingroup$
Nice way to see thinks
Assume that, $$(sqrt{n}+sqrt{m})=frac{p}{q}$$
Then we have
$$(sqrt{n}+sqrt{m})=frac{p}{q}inBbb Q implies
n+m+2sqrt{nm} =(sqrt{n}+sqrt{m})^2 =frac{p^2}{q^2}inBbb Q\implies sqrt{nm} =frac{n+m}{2}+frac{p^2}{2q^2}inBbb Q $$
But if $ nm $ is not a perfect square then $sqrt{nm}not inBbb Q ,$ (This can be easily prove using the fundamental theorem of number theory: Decomposition into prime numbers)
Hence in this case we have $$sqrt{n}+sqrt{m}not inBbb Q$$
Remark $~~~~~$1. $mn$ can be a perfect square even though neither $n$ nor $m$ is a perfect square. (see the example below)
- We can still have $sqrt{n}+sqrt{m}notin Bbb Q$ even if $mn$ is perfect square.(see the example below)
Example: $n= 3$ and $ m = 12$ are not perfect square and $ nm = 36 =6^2.$
Moreover,
$$sqrt{n}+sqrt{m} = sqrt{3}+sqrt{12} =3sqrt 3 not inBbb Q$$
answered Oct 16 '17 at 19:55
Guy FsoneGuy Fsone
17.1k42873
$endgroup$
add a comment |
$begingroup$
Nice way to see thinks
Assume that, $$(sqrt{n}+sqrt{m})=frac{p}{q}$$
Then we have
$$(sqrt{n}+sqrt{m})=frac{p}{q}inBbb Q implies
n+m+2sqrt{nm} =(sqrt{n}+sqrt{m})^2 =frac{p^2}{q^2}inBbb Q\implies sqrt{nm} =frac{n+m}{2}+frac{p^2}{2q^2}inBbb Q $$
But if $ nm $ is not a perfect square then $sqrt{nm}not inBbb Q ,$ (This can be easily prove using the fundamental theorem of number theory: Decomposition into prime numbers)
Hence in this case we have $$sqrt{n}+sqrt{m}not inBbb Q$$
Remark $~~~~~$1. $mn$ can be a perfect square even though neither $n$ nor $m$ is a perfect square. (see the example below)
- We can still have $sqrt{n}+sqrt{m}notin Bbb Q$ even if $mn$ is perfect square.(see the example below)
Example: $n= 3$ and $ m = 12$ are not perfect square and $ nm = 36 =6^2.$
Moreover,
$$sqrt{n}+sqrt{m} = sqrt{3}+sqrt{12} =3sqrt 3 not inBbb Q$$
answered Oct 16 '17 at 19:55
Guy FsoneGuy Fsone
17.1k42873
$endgroup$
Nice way to see thinks
Assume that, $$(sqrt{n}+sqrt{m})=frac{p}{q}$$
Then we have
$$(sqrt{n}+sqrt{m})=frac{p}{q}inBbb Q implies
n+m+2sqrt{nm} =(sqrt{n}+sqrt{m})^2 =frac{p^2}{q^2}inBbb Q\implies sqrt{nm} =frac{n+m}{2}+frac{p^2}{2q^2}inBbb Q $$
But if $ nm $ is not a perfect square then $sqrt{nm}not inBbb Q ,$ (This can be easily prove using the fundamental theorem of number theory: Decomposition into prime numbers)
Hence in this case we have $$sqrt{n}+sqrt{m}not inBbb Q$$
Remark $~~~~~$1. $mn$ can be a perfect square even though neither $n$ nor $m$ is a perfect square. (see the example below)
- We can still have $sqrt{n}+sqrt{m}notin Bbb Q$ even if $mn$ is perfect square.(see the example below)
Example: $n= 3$ and $ m = 12$ are not perfect square and $ nm = 36 =6^2.$
Moreover,
$$sqrt{n}+sqrt{m} = sqrt{3}+sqrt{12} =3sqrt 3 not inBbb Q$$
answered Oct 16 '17 at 19:55
Guy FsoneGuy Fsone
17.1k42873
edited Feb 8 '18 at 9:46
answered Oct 16 '17 at 19:55
Guy FsoneGuy Fsone
17.1k42873
answered Oct 16 '17 at 19:55
Guy FsoneGuy Fsone
17.1k42873
answered Oct 16 '17 at 19:55
Guy FsoneGuy Fsone
17.1k42873
17.1k42873
add a comment |
add a comment |
$begingroup$
It's easy to show that if the result is rational, it has to be natural.
Hint:
$$m=(lfloorsqrt{m}rfloor+epsilon)^{2}=lfloorsqrt m rfloor ^2+2epsilon lfloorsqrt m rfloor + epsilon^2$$
$$n=(lceil sqrt n rceil-epsilon)^2=lceil sqrt n rceil^2-2epsilonlceil sqrt n rceil+epsilon^2$$
Subtract these two from each other and show that $epsilon$ has to be rational.
answered Sep 22 '14 at 13:43
AliAli
1,144924
$endgroup$
$begingroup$
m+n+2*sqrt(m*n) has to be rational
$endgroup$
– Mahdi
Sep 22 '14 at 13:53
add a comment |
$begingroup$
It's easy to show that if the result is rational, it has to be natural.
Hint:
$$m=(lfloorsqrt{m}rfloor+epsilon)^{2}=lfloorsqrt m rfloor ^2+2epsilon lfloorsqrt m rfloor + epsilon^2$$
$$n=(lceil sqrt n rceil-epsilon)^2=lceil sqrt n rceil^2-2epsilonlceil sqrt n rceil+epsilon^2$$
Subtract these two from each other and show that $epsilon$ has to be rational.
answered Sep 22 '14 at 13:43
AliAli
1,144924
$endgroup$
$begingroup$
m+n+2*sqrt(m*n) has to be rational
$endgroup$
– Mahdi
Sep 22 '14 at 13:53
add a comment |
$begingroup$
It's easy to show that if the result is rational, it has to be natural.
Hint:
$$m=(lfloorsqrt{m}rfloor+epsilon)^{2}=lfloorsqrt m rfloor ^2+2epsilon lfloorsqrt m rfloor + epsilon^2$$
$$n=(lceil sqrt n rceil-epsilon)^2=lceil sqrt n rceil^2-2epsilonlceil sqrt n rceil+epsilon^2$$
Subtract these two from each other and show that $epsilon$ has to be rational.
answered Sep 22 '14 at 13:43
AliAli
1,144924
$endgroup$
It's easy to show that if the result is rational, it has to be natural.
Hint:
$$m=(lfloorsqrt{m}rfloor+epsilon)^{2}=lfloorsqrt m rfloor ^2+2epsilon lfloorsqrt m rfloor + epsilon^2$$
$$n=(lceil sqrt n rceil-epsilon)^2=lceil sqrt n rceil^2-2epsilonlceil sqrt n rceil+epsilon^2$$
Subtract these two from each other and show that $epsilon$ has to be rational.
answered Sep 22 '14 at 13:43
AliAli
1,144924
answered Sep 22 '14 at 13:43
AliAli
1,144924
answered Sep 22 '14 at 13:43
AliAli
1,144924
answered Sep 22 '14 at 13:43
AliAli
1,144924
1,144924
$begingroup$
m+n+2*sqrt(m*n) has to be rational
$endgroup$
– Mahdi
Sep 22 '14 at 13:53
add a comment |
$begingroup$
m+n+2*sqrt(m*n) has to be rational
$endgroup$
– Mahdi
Sep 22 '14 at 13:53
$begingroup$
m+n+2*sqrt(m*n) has to be rational
$endgroup$
– Mahdi
Sep 22 '14 at 13:53
$begingroup$
m+n+2*sqrt(m*n) has to be rational
$endgroup$
– Mahdi
Sep 22 '14 at 13:53
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%2f457382%2fcan-sqrtn-sqrtm-be-rational-if-neither-n-m-are-perfect-squares%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
$begingroup$
For a general case see this question and an earlier version
$endgroup$
– Jyrki Lahtonen
May 31 '14 at 11:49
$begingroup$
Related: math.stackexchange.com/questions/890821
$endgroup$
– Watson
Nov 25 '18 at 17:46