Show $(3 + sqrt{2})^{2/3}$ is irrational using RZT
$begingroup$
I am asked to prove that $(3+sqrt{2})^{2/3}$ is irrational via the rational zeroes theorem.
This is what I have so far:
$ x = (3+sqrt{2})^{2/3} $
$ x^3 = (3+sqrt{2})^{2} $
$ x^3 - 11 - 6sqrt{2} = 0 $
From here, I do not know how to get it into the form admissible by the RZT. The only way I know how to proceed from here is to say that any rational solution of the form $r= frac{c}{d}$ $c,d, in mathbb{Z} $ would need to have $d= pm 1$ and $c$ divide $-11 -6sqrt{2}$.
So we have
$ -11 -6sqrt{2} = z c$, $z in mathbb{Z}$
$-6sqrt{2} = 11 +zc$
Which clearly no $c$ will satisfy. So it is proved, but I feel like I did not utilize the RZT the way I was supposed to. At least, it is different than the other example problems I have been doing, for example, proving $sqrt{3}$ is irrational.
irrational-numbers rational-numbers
asked Nov 30 '18 at 23:03
pmacpmac
8116
$endgroup$
add a comment |
$begingroup$
I am asked to prove that $(3+sqrt{2})^{2/3}$ is irrational via the rational zeroes theorem.
This is what I have so far:
$ x = (3+sqrt{2})^{2/3} $
$ x^3 = (3+sqrt{2})^{2} $
$ x^3 - 11 - 6sqrt{2} = 0 $
From here, I do not know how to get it into the form admissible by the RZT. The only way I know how to proceed from here is to say that any rational solution of the form $r= frac{c}{d}$ $c,d, in mathbb{Z} $ would need to have $d= pm 1$ and $c$ divide $-11 -6sqrt{2}$.
So we have
$ -11 -6sqrt{2} = z c$, $z in mathbb{Z}$
$-6sqrt{2} = 11 +zc$
Which clearly no $c$ will satisfy. So it is proved, but I feel like I did not utilize the RZT the way I was supposed to. At least, it is different than the other example problems I have been doing, for example, proving $sqrt{3}$ is irrational.
irrational-numbers rational-numbers
asked Nov 30 '18 at 23:03
pmacpmac
8116
$endgroup$
1
$begingroup$
$(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done.
$endgroup$
– Will Jagy
Nov 30 '18 at 23:08
$begingroup$
Would it count to use the argument: if it were rational, then its cube $(3+sqrt{2})^2 = 11 + 6sqrt{2}$ would be rational, implying $sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there?
$endgroup$
– Daniel Schepler
Nov 30 '18 at 23:09
add a comment |
$begingroup$
I am asked to prove that $(3+sqrt{2})^{2/3}$ is irrational via the rational zeroes theorem.
This is what I have so far:
$ x = (3+sqrt{2})^{2/3} $
$ x^3 = (3+sqrt{2})^{2} $
$ x^3 - 11 - 6sqrt{2} = 0 $
From here, I do not know how to get it into the form admissible by the RZT. The only way I know how to proceed from here is to say that any rational solution of the form $r= frac{c}{d}$ $c,d, in mathbb{Z} $ would need to have $d= pm 1$ and $c$ divide $-11 -6sqrt{2}$.
So we have
$ -11 -6sqrt{2} = z c$, $z in mathbb{Z}$
$-6sqrt{2} = 11 +zc$
Which clearly no $c$ will satisfy. So it is proved, but I feel like I did not utilize the RZT the way I was supposed to. At least, it is different than the other example problems I have been doing, for example, proving $sqrt{3}$ is irrational.
irrational-numbers rational-numbers
asked Nov 30 '18 at 23:03
pmacpmac
8116
$endgroup$
I am asked to prove that $(3+sqrt{2})^{2/3}$ is irrational via the rational zeroes theorem.
This is what I have so far:
$ x = (3+sqrt{2})^{2/3} $
$ x^3 = (3+sqrt{2})^{2} $
$ x^3 - 11 - 6sqrt{2} = 0 $
From here, I do not know how to get it into the form admissible by the RZT. The only way I know how to proceed from here is to say that any rational solution of the form $r= frac{c}{d}$ $c,d, in mathbb{Z} $ would need to have $d= pm 1$ and $c$ divide $-11 -6sqrt{2}$.
So we have
$ -11 -6sqrt{2} = z c$, $z in mathbb{Z}$
$-6sqrt{2} = 11 +zc$
Which clearly no $c$ will satisfy. So it is proved, but I feel like I did not utilize the RZT the way I was supposed to. At least, it is different than the other example problems I have been doing, for example, proving $sqrt{3}$ is irrational.
irrational-numbers rational-numbers
irrational-numbers rational-numbers
asked Nov 30 '18 at 23:03
pmacpmac
8116
asked Nov 30 '18 at 23:03
pmacpmac
8116
edited Nov 30 '18 at 23:21
pmac
asked Nov 30 '18 at 23:03
pmacpmac
8116
asked Nov 30 '18 at 23:03
pmacpmac
8116
asked Nov 30 '18 at 23:03
pmacpmac
8116
8116
1
$begingroup$
$(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done.
$endgroup$
– Will Jagy
Nov 30 '18 at 23:08
$begingroup$
Would it count to use the argument: if it were rational, then its cube $(3+sqrt{2})^2 = 11 + 6sqrt{2}$ would be rational, implying $sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there?
$endgroup$
– Daniel Schepler
Nov 30 '18 at 23:09
add a comment |
1
$begingroup$
$(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done.
$endgroup$
– Will Jagy
Nov 30 '18 at 23:08
$begingroup$
Would it count to use the argument: if it were rational, then its cube $(3+sqrt{2})^2 = 11 + 6sqrt{2}$ would be rational, implying $sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there?
$endgroup$
– Daniel Schepler
Nov 30 '18 at 23:09
1
1
$begingroup$
$(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done.
$endgroup$
– Will Jagy
Nov 30 '18 at 23:08
$begingroup$
$(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done.
$endgroup$
– Will Jagy
Nov 30 '18 at 23:08
$begingroup$
Would it count to use the argument: if it were rational, then its cube $(3+sqrt{2})^2 = 11 + 6sqrt{2}$ would be rational, implying $sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there?
$endgroup$
– Daniel Schepler
Nov 30 '18 at 23:09
$begingroup$
Would it count to use the argument: if it were rational, then its cube $(3+sqrt{2})^2 = 11 + 6sqrt{2}$ would be rational, implying $sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there?
$endgroup$
– Daniel Schepler
Nov 30 '18 at 23:09
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
From $x^3-11-6sqrt2=0$, you get that $(x^3-11)^2=72$; in other words, $x^6-22x^3+49=0$. But the only possyble rational roots of this polynomial are $pm1$, $pm7$, and $pm49$. However, none of them is.
answered Nov 30 '18 at 23:07
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
add a comment |
$begingroup$
You don't need the RZT and you already did all the hard work: from
$$x^3-11-6sqrt2=0$$
we can deduce as follows: if $;xinBbb Q;$ then also $;x^3inBbb Q;$ and thus also $;x^3-11=6sqrt2inBbb Q;$ , from where we get the straightforward contradiction that $;sqrt2inBbb Q;$ ...
answered Nov 30 '18 at 23:15
DonAntonioDonAntonio
178k1494230
$endgroup$
1
$begingroup$
Yes, but the questions asks to use RZT.
$endgroup$
– pmac
Nov 30 '18 at 23:16
$begingroup$
@pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
$endgroup$
– DonAntonio
Nov 30 '18 at 23:16
$begingroup$
But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
$endgroup$
– Travis
Dec 1 '18 at 11:01
$begingroup$
@Travis Yes and yes...so what? I don't understand what you're trying to tell...
$endgroup$
– DonAntonio
Dec 1 '18 at 11:21
add a comment |
$begingroup$
Note $(3 + sqrt 2)^{frac 23}$ doesn't have to be the only root.
You got that it is a solution to
$x^3 - (11 + 6sqrt 2) = 0$.
So it is a solution to $(x^3 - (11 + 6sqrt 2))(x^3 - (11 - 6sqrt 2)) =$
$=x^6 - 22x^3 - (11^2 - 72) = x^6 -22x^3 - 49=0$.
Now by the rational root test the only possible rational roots are $pm 1, pm 7, pm 49$ and not of them are roots.
====
D'oh. Just read Jose Carlos Santos answer. Yeah $x^3 - 11 -6sqrt2=0 iff (x^3 -11) = 6sqrt 2 implies (x^2 - 11)^2 = 72$ is a lot more obvious an insightful and easier than my idea of looking for conjugates.
They both work and have the same result but in terms of ease in seeing and teaching... His is better.
(Although theoretically his isolating a square root and squaring it and my multiplying by conjugates is basically the same thing.)
answered Nov 30 '18 at 23:25
fleabloodfleablood
70.5k22685
$endgroup$
$begingroup$
Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
$endgroup$
– pmac
Nov 30 '18 at 23:31
add a comment |
$begingroup$
As an alternative, we have that
$$x=(3 + sqrt{2})^2=13+6sqrt 2not in mathbb{Q}$$
now suppose by contradiction that
$$sqrt[3] x =yin mathbb{Q} iff y^3=x$$
which is impossible.
answered Nov 30 '18 at 23:08
gimusigimusi
92.8k84494
$endgroup$
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%2f3020769%2fshow-3-sqrt22-3-is-irrational-using-rzt%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From $x^3-11-6sqrt2=0$, you get that $(x^3-11)^2=72$; in other words, $x^6-22x^3+49=0$. But the only possyble rational roots of this polynomial are $pm1$, $pm7$, and $pm49$. However, none of them is.
answered Nov 30 '18 at 23:07
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
add a comment |
$begingroup$
From $x^3-11-6sqrt2=0$, you get that $(x^3-11)^2=72$; in other words, $x^6-22x^3+49=0$. But the only possyble rational roots of this polynomial are $pm1$, $pm7$, and $pm49$. However, none of them is.
answered Nov 30 '18 at 23:07
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
add a comment |
$begingroup$
From $x^3-11-6sqrt2=0$, you get that $(x^3-11)^2=72$; in other words, $x^6-22x^3+49=0$. But the only possyble rational roots of this polynomial are $pm1$, $pm7$, and $pm49$. However, none of them is.
answered Nov 30 '18 at 23:07
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
From $x^3-11-6sqrt2=0$, you get that $(x^3-11)^2=72$; in other words, $x^6-22x^3+49=0$. But the only possyble rational roots of this polynomial are $pm1$, $pm7$, and $pm49$. However, none of them is.
answered Nov 30 '18 at 23:07
José Carlos SantosJosé Carlos Santos
160k22127232
edited Nov 30 '18 at 23:09
answered Nov 30 '18 at 23:07
José Carlos SantosJosé Carlos Santos
160k22127232
answered Nov 30 '18 at 23:07
José Carlos SantosJosé Carlos Santos
160k22127232
answered Nov 30 '18 at 23:07
José Carlos SantosJosé Carlos Santos
160k22127232
160k22127232
add a comment |
add a comment |
$begingroup$
You don't need the RZT and you already did all the hard work: from
$$x^3-11-6sqrt2=0$$
we can deduce as follows: if $;xinBbb Q;$ then also $;x^3inBbb Q;$ and thus also $;x^3-11=6sqrt2inBbb Q;$ , from where we get the straightforward contradiction that $;sqrt2inBbb Q;$ ...
answered Nov 30 '18 at 23:15
DonAntonioDonAntonio
178k1494230
$endgroup$
1
$begingroup$
Yes, but the questions asks to use RZT.
$endgroup$
– pmac
Nov 30 '18 at 23:16
$begingroup$
@pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
$endgroup$
– DonAntonio
Nov 30 '18 at 23:16
$begingroup$
But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
$endgroup$
– Travis
Dec 1 '18 at 11:01
$begingroup$
@Travis Yes and yes...so what? I don't understand what you're trying to tell...
$endgroup$
– DonAntonio
Dec 1 '18 at 11:21
add a comment |
$begingroup$
You don't need the RZT and you already did all the hard work: from
$$x^3-11-6sqrt2=0$$
we can deduce as follows: if $;xinBbb Q;$ then also $;x^3inBbb Q;$ and thus also $;x^3-11=6sqrt2inBbb Q;$ , from where we get the straightforward contradiction that $;sqrt2inBbb Q;$ ...
answered Nov 30 '18 at 23:15
DonAntonioDonAntonio
178k1494230
$endgroup$
1
$begingroup$
Yes, but the questions asks to use RZT.
$endgroup$
– pmac
Nov 30 '18 at 23:16
$begingroup$
@pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
$endgroup$
– DonAntonio
Nov 30 '18 at 23:16
$begingroup$
But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
$endgroup$
– Travis
Dec 1 '18 at 11:01
$begingroup$
@Travis Yes and yes...so what? I don't understand what you're trying to tell...
$endgroup$
– DonAntonio
Dec 1 '18 at 11:21
add a comment |
$begingroup$
You don't need the RZT and you already did all the hard work: from
$$x^3-11-6sqrt2=0$$
we can deduce as follows: if $;xinBbb Q;$ then also $;x^3inBbb Q;$ and thus also $;x^3-11=6sqrt2inBbb Q;$ , from where we get the straightforward contradiction that $;sqrt2inBbb Q;$ ...
answered Nov 30 '18 at 23:15
DonAntonioDonAntonio
178k1494230
$endgroup$
You don't need the RZT and you already did all the hard work: from
$$x^3-11-6sqrt2=0$$
we can deduce as follows: if $;xinBbb Q;$ then also $;x^3inBbb Q;$ and thus also $;x^3-11=6sqrt2inBbb Q;$ , from where we get the straightforward contradiction that $;sqrt2inBbb Q;$ ...
answered Nov 30 '18 at 23:15
DonAntonioDonAntonio
178k1494230
answered Nov 30 '18 at 23:15
DonAntonioDonAntonio
178k1494230
answered Nov 30 '18 at 23:15
DonAntonioDonAntonio
178k1494230
answered Nov 30 '18 at 23:15
DonAntonioDonAntonio
178k1494230
178k1494230
1
$begingroup$
Yes, but the questions asks to use RZT.
$endgroup$
– pmac
Nov 30 '18 at 23:16
$begingroup$
@pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
$endgroup$
– DonAntonio
Nov 30 '18 at 23:16
$begingroup$
But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
$endgroup$
– Travis
Dec 1 '18 at 11:01
$begingroup$
@Travis Yes and yes...so what? I don't understand what you're trying to tell...
$endgroup$
– DonAntonio
Dec 1 '18 at 11:21
add a comment |
1
$begingroup$
Yes, but the questions asks to use RZT.
$endgroup$
– pmac
Nov 30 '18 at 23:16
$begingroup$
@pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
$endgroup$
– DonAntonio
Nov 30 '18 at 23:16
$begingroup$
But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
$endgroup$
– Travis
Dec 1 '18 at 11:01
$begingroup$
@Travis Yes and yes...so what? I don't understand what you're trying to tell...
$endgroup$
– DonAntonio
Dec 1 '18 at 11:21
1
1
$begingroup$
Yes, but the questions asks to use RZT.
$endgroup$
– pmac
Nov 30 '18 at 23:16
$begingroup$
Yes, but the questions asks to use RZT.
$endgroup$
– pmac
Nov 30 '18 at 23:16
$begingroup$
@pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
$endgroup$
– DonAntonio
Nov 30 '18 at 23:16
$begingroup$
@pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT.
$endgroup$
– DonAntonio
Nov 30 '18 at 23:16
$begingroup$
But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
$endgroup$
– Travis
Dec 1 '18 at 11:01
$begingroup$
But $sqrt{2}$ is a root of $t^2 - 2 in Bbb Z[t]$, and RZT shows that polynomial has no rational solutions.
$endgroup$
– Travis
Dec 1 '18 at 11:01
$begingroup$
@Travis Yes and yes...so what? I don't understand what you're trying to tell...
$endgroup$
– DonAntonio
Dec 1 '18 at 11:21
$begingroup$
@Travis Yes and yes...so what? I don't understand what you're trying to tell...
$endgroup$
– DonAntonio
Dec 1 '18 at 11:21
add a comment |
$begingroup$
Note $(3 + sqrt 2)^{frac 23}$ doesn't have to be the only root.
You got that it is a solution to
$x^3 - (11 + 6sqrt 2) = 0$.
So it is a solution to $(x^3 - (11 + 6sqrt 2))(x^3 - (11 - 6sqrt 2)) =$
$=x^6 - 22x^3 - (11^2 - 72) = x^6 -22x^3 - 49=0$.
Now by the rational root test the only possible rational roots are $pm 1, pm 7, pm 49$ and not of them are roots.
====
D'oh. Just read Jose Carlos Santos answer. Yeah $x^3 - 11 -6sqrt2=0 iff (x^3 -11) = 6sqrt 2 implies (x^2 - 11)^2 = 72$ is a lot more obvious an insightful and easier than my idea of looking for conjugates.
They both work and have the same result but in terms of ease in seeing and teaching... His is better.
(Although theoretically his isolating a square root and squaring it and my multiplying by conjugates is basically the same thing.)
answered Nov 30 '18 at 23:25
fleabloodfleablood
70.5k22685
$endgroup$
$begingroup$
Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
$endgroup$
– pmac
Nov 30 '18 at 23:31
add a comment |
$begingroup$
Note $(3 + sqrt 2)^{frac 23}$ doesn't have to be the only root.
You got that it is a solution to
$x^3 - (11 + 6sqrt 2) = 0$.
So it is a solution to $(x^3 - (11 + 6sqrt 2))(x^3 - (11 - 6sqrt 2)) =$
$=x^6 - 22x^3 - (11^2 - 72) = x^6 -22x^3 - 49=0$.
Now by the rational root test the only possible rational roots are $pm 1, pm 7, pm 49$ and not of them are roots.
====
D'oh. Just read Jose Carlos Santos answer. Yeah $x^3 - 11 -6sqrt2=0 iff (x^3 -11) = 6sqrt 2 implies (x^2 - 11)^2 = 72$ is a lot more obvious an insightful and easier than my idea of looking for conjugates.
They both work and have the same result but in terms of ease in seeing and teaching... His is better.
(Although theoretically his isolating a square root and squaring it and my multiplying by conjugates is basically the same thing.)
answered Nov 30 '18 at 23:25
fleabloodfleablood
70.5k22685
$endgroup$
$begingroup$
Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
$endgroup$
– pmac
Nov 30 '18 at 23:31
add a comment |
$begingroup$
Note $(3 + sqrt 2)^{frac 23}$ doesn't have to be the only root.
You got that it is a solution to
$x^3 - (11 + 6sqrt 2) = 0$.
So it is a solution to $(x^3 - (11 + 6sqrt 2))(x^3 - (11 - 6sqrt 2)) =$
$=x^6 - 22x^3 - (11^2 - 72) = x^6 -22x^3 - 49=0$.
Now by the rational root test the only possible rational roots are $pm 1, pm 7, pm 49$ and not of them are roots.
====
D'oh. Just read Jose Carlos Santos answer. Yeah $x^3 - 11 -6sqrt2=0 iff (x^3 -11) = 6sqrt 2 implies (x^2 - 11)^2 = 72$ is a lot more obvious an insightful and easier than my idea of looking for conjugates.
They both work and have the same result but in terms of ease in seeing and teaching... His is better.
(Although theoretically his isolating a square root and squaring it and my multiplying by conjugates is basically the same thing.)
answered Nov 30 '18 at 23:25
fleabloodfleablood
70.5k22685
$endgroup$
Note $(3 + sqrt 2)^{frac 23}$ doesn't have to be the only root.
You got that it is a solution to
$x^3 - (11 + 6sqrt 2) = 0$.
So it is a solution to $(x^3 - (11 + 6sqrt 2))(x^3 - (11 - 6sqrt 2)) =$
$=x^6 - 22x^3 - (11^2 - 72) = x^6 -22x^3 - 49=0$.
Now by the rational root test the only possible rational roots are $pm 1, pm 7, pm 49$ and not of them are roots.
====
D'oh. Just read Jose Carlos Santos answer. Yeah $x^3 - 11 -6sqrt2=0 iff (x^3 -11) = 6sqrt 2 implies (x^2 - 11)^2 = 72$ is a lot more obvious an insightful and easier than my idea of looking for conjugates.
They both work and have the same result but in terms of ease in seeing and teaching... His is better.
(Although theoretically his isolating a square root and squaring it and my multiplying by conjugates is basically the same thing.)
answered Nov 30 '18 at 23:25
fleabloodfleablood
70.5k22685
edited Nov 30 '18 at 23:31
answered Nov 30 '18 at 23:25
fleabloodfleablood
70.5k22685
answered Nov 30 '18 at 23:25
fleabloodfleablood
70.5k22685
answered Nov 30 '18 at 23:25
fleabloodfleablood
70.5k22685
70.5k22685
$begingroup$
Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
$endgroup$
– pmac
Nov 30 '18 at 23:31
add a comment |
$begingroup$
Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
$endgroup$
– pmac
Nov 30 '18 at 23:31
$begingroup$
Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
$endgroup$
– pmac
Nov 30 '18 at 23:31
$begingroup$
Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks.
$endgroup$
– pmac
Nov 30 '18 at 23:31
add a comment |
$begingroup$
As an alternative, we have that
$$x=(3 + sqrt{2})^2=13+6sqrt 2not in mathbb{Q}$$
now suppose by contradiction that
$$sqrt[3] x =yin mathbb{Q} iff y^3=x$$
which is impossible.
answered Nov 30 '18 at 23:08
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
As an alternative, we have that
$$x=(3 + sqrt{2})^2=13+6sqrt 2not in mathbb{Q}$$
now suppose by contradiction that
$$sqrt[3] x =yin mathbb{Q} iff y^3=x$$
which is impossible.
answered Nov 30 '18 at 23:08
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
As an alternative, we have that
$$x=(3 + sqrt{2})^2=13+6sqrt 2not in mathbb{Q}$$
now suppose by contradiction that
$$sqrt[3] x =yin mathbb{Q} iff y^3=x$$
which is impossible.
answered Nov 30 '18 at 23:08
gimusigimusi
92.8k84494
$endgroup$
As an alternative, we have that
$$x=(3 + sqrt{2})^2=13+6sqrt 2not in mathbb{Q}$$
now suppose by contradiction that
$$sqrt[3] x =yin mathbb{Q} iff y^3=x$$
which is impossible.
answered Nov 30 '18 at 23:08
gimusigimusi
92.8k84494
edited Nov 30 '18 at 23:31
answered Nov 30 '18 at 23:08
gimusigimusi
92.8k84494
answered Nov 30 '18 at 23:08
gimusigimusi
92.8k84494
answered Nov 30 '18 at 23:08
gimusigimusi
92.8k84494
92.8k84494
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%2f3020769%2fshow-3-sqrt22-3-is-irrational-using-rzt%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
$begingroup$
$(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done.
$endgroup$
– Will Jagy
Nov 30 '18 at 23:08
$begingroup$
Would it count to use the argument: if it were rational, then its cube $(3+sqrt{2})^2 = 11 + 6sqrt{2}$ would be rational, implying $sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there?
$endgroup$
– Daniel Schepler
Nov 30 '18 at 23:09