Mixing
Proving irrationality of $sqrt[3]{3}+sqrt[3]{9}$ [duplicate]
$begingroup$
This question already has an answer here:
If $sqrt[3]{a} + sqrt[3]{b}$ is rational then prove $sqrt[3]{a}$ and $sqrt[3]{b}$ are rational
1 answer
I need to prove $$sqrt[3]{3}+sqrt[3]{9}$$
is irrational, I assumed
$$sqrt[3]{3}+sqrt[3]{9} = frac{m}{n}$$
I cubed both sides and got
$$sqrt[3]{3}+sqrt[3]{9} = frac{m^3-12n^2}{9n^3}$$
I tried setting $$frac{m^3-12n^2}{9n^3} = frac{m}{n}$$ but that led me nowhere. so what can I do?
number-theory radicals irrational-numbers
edited Jan 28 at 18:42
Asaf Karagila♦
307k33439771
asked Jan 28 at 8:49
Guysudai1Guysudai1
18011
$endgroup$
marked as duplicate by Watson, José Carlos Santos, Adrian Keister, max_zorn, hardmath Jan 29 at 17:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
If $sqrt[3]{a} + sqrt[3]{b}$ is rational then prove $sqrt[3]{a}$ and $sqrt[3]{b}$ are rational
1 answer
I need to prove $$sqrt[3]{3}+sqrt[3]{9}$$
is irrational, I assumed
$$sqrt[3]{3}+sqrt[3]{9} = frac{m}{n}$$
I cubed both sides and got
$$sqrt[3]{3}+sqrt[3]{9} = frac{m^3-12n^2}{9n^3}$$
I tried setting $$frac{m^3-12n^2}{9n^3} = frac{m}{n}$$ but that led me nowhere. so what can I do?
number-theory radicals irrational-numbers
edited Jan 28 at 18:42
Asaf Karagila♦
307k33439771
asked Jan 28 at 8:49
Guysudai1Guysudai1
18011
$endgroup$
marked as duplicate by Watson, José Carlos Santos, Adrian Keister, max_zorn, hardmath Jan 29 at 17:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
Please use descriptive titles. You have 150 characters to use, and you're more than welcomed to use $rmLaTeX$ as well. "Proving the irrationality of a number" is a meaningless title. Same goes to your other post from earlier.
$endgroup$
– Asaf Karagila♦
Jan 28 at 18:43
$begingroup$
Closely related: math.stackexchange.com/questions/1542708
$endgroup$
– Watson
Jan 29 at 9:06
$begingroup$
It may be that this exercise was intended to reinforce specific material covered in your textbook or other course materials, but your Readers will not be aware of that context unless you share it. It is certainly a problem that is amenable to various attacks, so knowing what you are studying will help Readers respond in a way you find useful.
$endgroup$
– hardmath
Jan 29 at 17:18
add a comment |
$begingroup$
This question already has an answer here:
If $sqrt[3]{a} + sqrt[3]{b}$ is rational then prove $sqrt[3]{a}$ and $sqrt[3]{b}$ are rational
1 answer
I need to prove $$sqrt[3]{3}+sqrt[3]{9}$$
is irrational, I assumed
$$sqrt[3]{3}+sqrt[3]{9} = frac{m}{n}$$
I cubed both sides and got
$$sqrt[3]{3}+sqrt[3]{9} = frac{m^3-12n^2}{9n^3}$$
I tried setting $$frac{m^3-12n^2}{9n^3} = frac{m}{n}$$ but that led me nowhere. so what can I do?
number-theory radicals irrational-numbers
edited Jan 28 at 18:42
Asaf Karagila♦
307k33439771
asked Jan 28 at 8:49
Guysudai1Guysudai1
18011
$endgroup$
This question already has an answer here:
If $sqrt[3]{a} + sqrt[3]{b}$ is rational then prove $sqrt[3]{a}$ and $sqrt[3]{b}$ are rational
1 answer
I need to prove $$sqrt[3]{3}+sqrt[3]{9}$$
is irrational, I assumed
$$sqrt[3]{3}+sqrt[3]{9} = frac{m}{n}$$
I cubed both sides and got
$$sqrt[3]{3}+sqrt[3]{9} = frac{m^3-12n^2}{9n^3}$$
I tried setting $$frac{m^3-12n^2}{9n^3} = frac{m}{n}$$ but that led me nowhere. so what can I do?
This question already has an answer here:
If $sqrt[3]{a} + sqrt[3]{b}$ is rational then prove $sqrt[3]{a}$ and $sqrt[3]{b}$ are rational
1 answer
number-theory radicals irrational-numbers
number-theory radicals irrational-numbers
edited Jan 28 at 18:42
Asaf Karagila♦
307k33439771
asked Jan 28 at 8:49
Guysudai1Guysudai1
18011
edited Jan 28 at 18:42
Asaf Karagila♦
307k33439771
asked Jan 28 at 8:49
Guysudai1Guysudai1
18011
edited Jan 28 at 18:42
Asaf Karagila♦
307k33439771
edited Jan 28 at 18:42
Asaf Karagila♦
307k33439771
edited Jan 28 at 18:42
Asaf Karagila♦
307k33439771
307k33439771
asked Jan 28 at 8:49
Guysudai1Guysudai1
18011
asked Jan 28 at 8:49
Guysudai1Guysudai1
18011
asked Jan 28 at 8:49
Guysudai1Guysudai1
18011
18011
marked as duplicate by Watson, José Carlos Santos, Adrian Keister, max_zorn, hardmath Jan 29 at 17:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Watson, José Carlos Santos, Adrian Keister, max_zorn, hardmath Jan 29 at 17:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
Please use descriptive titles. You have 150 characters to use, and you're more than welcomed to use $rmLaTeX$ as well. "Proving the irrationality of a number" is a meaningless title. Same goes to your other post from earlier.
$endgroup$
– Asaf Karagila♦
Jan 28 at 18:43
$begingroup$
Closely related: math.stackexchange.com/questions/1542708
$endgroup$
– Watson
Jan 29 at 9:06
$begingroup$
It may be that this exercise was intended to reinforce specific material covered in your textbook or other course materials, but your Readers will not be aware of that context unless you share it. It is certainly a problem that is amenable to various attacks, so knowing what you are studying will help Readers respond in a way you find useful.
$endgroup$
– hardmath
Jan 29 at 17:18
add a comment |
2
$begingroup$
Please use descriptive titles. You have 150 characters to use, and you're more than welcomed to use $rmLaTeX$ as well. "Proving the irrationality of a number" is a meaningless title. Same goes to your other post from earlier.
$endgroup$
– Asaf Karagila♦
Jan 28 at 18:43
$begingroup$
Closely related: math.stackexchange.com/questions/1542708
$endgroup$
– Watson
Jan 29 at 9:06
$begingroup$
It may be that this exercise was intended to reinforce specific material covered in your textbook or other course materials, but your Readers will not be aware of that context unless you share it. It is certainly a problem that is amenable to various attacks, so knowing what you are studying will help Readers respond in a way you find useful.
$endgroup$
– hardmath
Jan 29 at 17:18
2
2
$begingroup$
Please use descriptive titles. You have 150 characters to use, and you're more than welcomed to use $rmLaTeX$ as well. "Proving the irrationality of a number" is a meaningless title. Same goes to your other post from earlier.
$endgroup$
– Asaf Karagila♦
Jan 28 at 18:43
$begingroup$
Please use descriptive titles. You have 150 characters to use, and you're more than welcomed to use $rmLaTeX$ as well. "Proving the irrationality of a number" is a meaningless title. Same goes to your other post from earlier.
$endgroup$
– Asaf Karagila♦
Jan 28 at 18:43
$begingroup$
Closely related: math.stackexchange.com/questions/1542708
$endgroup$
– Watson
Jan 29 at 9:06
$begingroup$
Closely related: math.stackexchange.com/questions/1542708
$endgroup$
– Watson
Jan 29 at 9:06
$begingroup$
It may be that this exercise was intended to reinforce specific material covered in your textbook or other course materials, but your Readers will not be aware of that context unless you share it. It is certainly a problem that is amenable to various attacks, so knowing what you are studying will help Readers respond in a way you find useful.
$endgroup$
– hardmath
Jan 29 at 17:18
$begingroup$
It may be that this exercise was intended to reinforce specific material covered in your textbook or other course materials, but your Readers will not be aware of that context unless you share it. It is certainly a problem that is amenable to various attacks, so knowing what you are studying will help Readers respond in a way you find useful.
$endgroup$
– hardmath
Jan 29 at 17:18
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $sqrt[3]3+sqrt[3]9=r$.
Thus, since for all reals $a$, $b$ and $c$ we have:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we obtain:
$$3+9-r^3+9r=0.$$
Now, let $r=frac{m}{n},$ where $m$ and $n$ are naturals with $gcd=1.$
Thus, $$m^3-9n^2m-12n^3=0,$$ which says that $m$ is divisible by $3$.
Let $m=3m'$, where $m'$ is a natural number.
Thus, $$9m'^3-9n^2m'=4n^3,$$ which says that $n$ is divisible by $3$, which is a contradiction.
answered Jan 28 at 9:01
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
3
$begingroup$
What did you set as a, b, and c?
$endgroup$
– Guysudai1
Jan 28 at 9:07
1
$begingroup$
@Guysudai1 They are any real numbers.
$endgroup$
– Michael Rozenberg
Jan 28 at 9:07
4
$begingroup$
@Guysudai1 Michael seems to have chosen $a=sqrt[3]{3}$, $b=sqrt[3]{9}$, and $c=-r$. Then $a+b+c=0$ and one can study the left-hand side.
$endgroup$
– Joonas Ilmavirta
Jan 28 at 13:43
add a comment |
$begingroup$
Here are two other takes.
Take 1
Let $alpha = sqrt[3]{3}+sqrt[3]{9}$. Then $alpha^3 = 9 alpha + 12$.
By the rational root theorem, if $alpha$ is rational, then $alpha$ is an integer.
Now $1 < sqrt[3]{3} < 2 $ and $2 < sqrt[3]{9} < 3 $, and so $3 < alpha < 5$.
Since $x=4$ is not a root of $x^3 = 9 x + 12$, $alpha$ is not an integer and so $alpha$ is irrational.
Take 2
We have $alpha = beta+beta^2$, where $beta=sqrt[3]{3}$.
If $alpha$ were rational, then $beta$ would be a root of a quadratic polynomial with rational coefficients. However, the polynomial with rational coefficients with least degree having $beta$ as a root is $x^3-3$.
answered Jan 28 at 11:07
lhflhf
167k11172403
$endgroup$
$begingroup$
It is fine what $beta$ is a root of $x^3 - 3$ but how do you know that it is of least degree polynomial with rational coefficients. Are you using some standard result?
$endgroup$
– prashant sharma
Feb 4 at 7:26
$begingroup$
@prashantsharma, $x^3 - 3$ is a cubic with no rational roots and so is irreducible.
$endgroup$
– lhf
Feb 4 at 7:58
$begingroup$
@Ihf Means are you suggesting that any equation with no rational roots is irreducible?
$endgroup$
– prashant sharma
Feb 5 at 9:24
$begingroup$
@prashantsharma, not any equation, but cubics, yes.
$endgroup$
– lhf
Feb 5 at 10:12
$begingroup$
@Ihf Can you please supply a proof?
$endgroup$
– prashant sharma
Feb 5 at 10:34
|
show 2 more comments
$begingroup$
Let $a=sqrt[3]{3}$ and $b=sqrt[3]{9}$, and suppose that $a+b=r$ is rational. Then $$r^3 =(a+b)^3 = a^3+3ab(a+b)+b^3 = 12+9(a+b)=12+9r$$
So $r$ is a solution of equation $x^3-9x-12=0$ which has no rational root.
answered Jan 28 at 9:11
Maria MazurMaria Mazur
48.7k1260122
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $sqrt[3]3+sqrt[3]9=r$.
Thus, since for all reals $a$, $b$ and $c$ we have:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we obtain:
$$3+9-r^3+9r=0.$$
Now, let $r=frac{m}{n},$ where $m$ and $n$ are naturals with $gcd=1.$
Thus, $$m^3-9n^2m-12n^3=0,$$ which says that $m$ is divisible by $3$.
Let $m=3m'$, where $m'$ is a natural number.
Thus, $$9m'^3-9n^2m'=4n^3,$$ which says that $n$ is divisible by $3$, which is a contradiction.
answered Jan 28 at 9:01
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
3
$begingroup$
What did you set as a, b, and c?
$endgroup$
– Guysudai1
Jan 28 at 9:07
1
$begingroup$
@Guysudai1 They are any real numbers.
$endgroup$
– Michael Rozenberg
Jan 28 at 9:07
4
$begingroup$
@Guysudai1 Michael seems to have chosen $a=sqrt[3]{3}$, $b=sqrt[3]{9}$, and $c=-r$. Then $a+b+c=0$ and one can study the left-hand side.
$endgroup$
– Joonas Ilmavirta
Jan 28 at 13:43
add a comment |
$begingroup$
Let $sqrt[3]3+sqrt[3]9=r$.
Thus, since for all reals $a$, $b$ and $c$ we have:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we obtain:
$$3+9-r^3+9r=0.$$
Now, let $r=frac{m}{n},$ where $m$ and $n$ are naturals with $gcd=1.$
Thus, $$m^3-9n^2m-12n^3=0,$$ which says that $m$ is divisible by $3$.
Let $m=3m'$, where $m'$ is a natural number.
Thus, $$9m'^3-9n^2m'=4n^3,$$ which says that $n$ is divisible by $3$, which is a contradiction.
answered Jan 28 at 9:01
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
3
$begingroup$
What did you set as a, b, and c?
$endgroup$
– Guysudai1
Jan 28 at 9:07
1
$begingroup$
@Guysudai1 They are any real numbers.
$endgroup$
– Michael Rozenberg
Jan 28 at 9:07
4
$begingroup$
@Guysudai1 Michael seems to have chosen $a=sqrt[3]{3}$, $b=sqrt[3]{9}$, and $c=-r$. Then $a+b+c=0$ and one can study the left-hand side.
$endgroup$
– Joonas Ilmavirta
Jan 28 at 13:43
add a comment |
$begingroup$
Let $sqrt[3]3+sqrt[3]9=r$.
Thus, since for all reals $a$, $b$ and $c$ we have:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we obtain:
$$3+9-r^3+9r=0.$$
Now, let $r=frac{m}{n},$ where $m$ and $n$ are naturals with $gcd=1.$
Thus, $$m^3-9n^2m-12n^3=0,$$ which says that $m$ is divisible by $3$.
Let $m=3m'$, where $m'$ is a natural number.
Thus, $$9m'^3-9n^2m'=4n^3,$$ which says that $n$ is divisible by $3$, which is a contradiction.
answered Jan 28 at 9:01
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
Let $sqrt[3]3+sqrt[3]9=r$.
Thus, since for all reals $a$, $b$ and $c$ we have:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we obtain:
$$3+9-r^3+9r=0.$$
Now, let $r=frac{m}{n},$ where $m$ and $n$ are naturals with $gcd=1.$
Thus, $$m^3-9n^2m-12n^3=0,$$ which says that $m$ is divisible by $3$.
Let $m=3m'$, where $m'$ is a natural number.
Thus, $$9m'^3-9n^2m'=4n^3,$$ which says that $n$ is divisible by $3$, which is a contradiction.
answered Jan 28 at 9:01
Michael RozenbergMichael Rozenberg
109k1896201
edited Jan 28 at 9:07
answered Jan 28 at 9:01
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 28 at 9:01
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 28 at 9:01
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
3
$begingroup$
What did you set as a, b, and c?
$endgroup$
– Guysudai1
Jan 28 at 9:07
1
$begingroup$
@Guysudai1 They are any real numbers.
$endgroup$
– Michael Rozenberg
Jan 28 at 9:07
4
$begingroup$
@Guysudai1 Michael seems to have chosen $a=sqrt[3]{3}$, $b=sqrt[3]{9}$, and $c=-r$. Then $a+b+c=0$ and one can study the left-hand side.
$endgroup$
– Joonas Ilmavirta
Jan 28 at 13:43
add a comment |
3
$begingroup$
What did you set as a, b, and c?
$endgroup$
– Guysudai1
Jan 28 at 9:07
1
$begingroup$
@Guysudai1 They are any real numbers.
$endgroup$
– Michael Rozenberg
Jan 28 at 9:07
4
$begingroup$
@Guysudai1 Michael seems to have chosen $a=sqrt[3]{3}$, $b=sqrt[3]{9}$, and $c=-r$. Then $a+b+c=0$ and one can study the left-hand side.
$endgroup$
– Joonas Ilmavirta
Jan 28 at 13:43
3
3
$begingroup$
What did you set as a, b, and c?
$endgroup$
– Guysudai1
Jan 28 at 9:07
$begingroup$
What did you set as a, b, and c?
$endgroup$
– Guysudai1
Jan 28 at 9:07
1
1
$begingroup$
@Guysudai1 They are any real numbers.
$endgroup$
– Michael Rozenberg
Jan 28 at 9:07
$begingroup$
@Guysudai1 They are any real numbers.
$endgroup$
– Michael Rozenberg
Jan 28 at 9:07
4
4
$begingroup$
@Guysudai1 Michael seems to have chosen $a=sqrt[3]{3}$, $b=sqrt[3]{9}$, and $c=-r$. Then $a+b+c=0$ and one can study the left-hand side.
$endgroup$
– Joonas Ilmavirta
Jan 28 at 13:43
$begingroup$
@Guysudai1 Michael seems to have chosen $a=sqrt[3]{3}$, $b=sqrt[3]{9}$, and $c=-r$. Then $a+b+c=0$ and one can study the left-hand side.
$endgroup$
– Joonas Ilmavirta
Jan 28 at 13:43
add a comment |
$begingroup$
Here are two other takes.
Take 1
Let $alpha = sqrt[3]{3}+sqrt[3]{9}$. Then $alpha^3 = 9 alpha + 12$.
By the rational root theorem, if $alpha$ is rational, then $alpha$ is an integer.
Now $1 < sqrt[3]{3} < 2 $ and $2 < sqrt[3]{9} < 3 $, and so $3 < alpha < 5$.
Since $x=4$ is not a root of $x^3 = 9 x + 12$, $alpha$ is not an integer and so $alpha$ is irrational.
Take 2
We have $alpha = beta+beta^2$, where $beta=sqrt[3]{3}$.
If $alpha$ were rational, then $beta$ would be a root of a quadratic polynomial with rational coefficients. However, the polynomial with rational coefficients with least degree having $beta$ as a root is $x^3-3$.
answered Jan 28 at 11:07
lhflhf
167k11172403
$endgroup$
$begingroup$
It is fine what $beta$ is a root of $x^3 - 3$ but how do you know that it is of least degree polynomial with rational coefficients. Are you using some standard result?
$endgroup$
– prashant sharma
Feb 4 at 7:26
$begingroup$
@prashantsharma, $x^3 - 3$ is a cubic with no rational roots and so is irreducible.
$endgroup$
– lhf
Feb 4 at 7:58
$begingroup$
@Ihf Means are you suggesting that any equation with no rational roots is irreducible?
$endgroup$
– prashant sharma
Feb 5 at 9:24
$begingroup$
@prashantsharma, not any equation, but cubics, yes.
$endgroup$
– lhf
Feb 5 at 10:12
$begingroup$
@Ihf Can you please supply a proof?
$endgroup$
– prashant sharma
Feb 5 at 10:34
|
show 2 more comments
$begingroup$
Here are two other takes.
Take 1
Let $alpha = sqrt[3]{3}+sqrt[3]{9}$. Then $alpha^3 = 9 alpha + 12$.
By the rational root theorem, if $alpha$ is rational, then $alpha$ is an integer.
Now $1 < sqrt[3]{3} < 2 $ and $2 < sqrt[3]{9} < 3 $, and so $3 < alpha < 5$.
Since $x=4$ is not a root of $x^3 = 9 x + 12$, $alpha$ is not an integer and so $alpha$ is irrational.
Take 2
We have $alpha = beta+beta^2$, where $beta=sqrt[3]{3}$.
If $alpha$ were rational, then $beta$ would be a root of a quadratic polynomial with rational coefficients. However, the polynomial with rational coefficients with least degree having $beta$ as a root is $x^3-3$.
answered Jan 28 at 11:07
lhflhf
167k11172403
$endgroup$
$begingroup$
It is fine what $beta$ is a root of $x^3 - 3$ but how do you know that it is of least degree polynomial with rational coefficients. Are you using some standard result?
$endgroup$
– prashant sharma
Feb 4 at 7:26
$begingroup$
@prashantsharma, $x^3 - 3$ is a cubic with no rational roots and so is irreducible.
$endgroup$
– lhf
Feb 4 at 7:58
$begingroup$
@Ihf Means are you suggesting that any equation with no rational roots is irreducible?
$endgroup$
– prashant sharma
Feb 5 at 9:24
$begingroup$
@prashantsharma, not any equation, but cubics, yes.
$endgroup$
– lhf
Feb 5 at 10:12
$begingroup$
@Ihf Can you please supply a proof?
$endgroup$
– prashant sharma
Feb 5 at 10:34
|
show 2 more comments
$begingroup$
Here are two other takes.
Take 1
Let $alpha = sqrt[3]{3}+sqrt[3]{9}$. Then $alpha^3 = 9 alpha + 12$.
By the rational root theorem, if $alpha$ is rational, then $alpha$ is an integer.
Now $1 < sqrt[3]{3} < 2 $ and $2 < sqrt[3]{9} < 3 $, and so $3 < alpha < 5$.
Since $x=4$ is not a root of $x^3 = 9 x + 12$, $alpha$ is not an integer and so $alpha$ is irrational.
Take 2
We have $alpha = beta+beta^2$, where $beta=sqrt[3]{3}$.
If $alpha$ were rational, then $beta$ would be a root of a quadratic polynomial with rational coefficients. However, the polynomial with rational coefficients with least degree having $beta$ as a root is $x^3-3$.
answered Jan 28 at 11:07
lhflhf
167k11172403
$endgroup$
Here are two other takes.
Take 1
Let $alpha = sqrt[3]{3}+sqrt[3]{9}$. Then $alpha^3 = 9 alpha + 12$.
By the rational root theorem, if $alpha$ is rational, then $alpha$ is an integer.
Now $1 < sqrt[3]{3} < 2 $ and $2 < sqrt[3]{9} < 3 $, and so $3 < alpha < 5$.
Since $x=4$ is not a root of $x^3 = 9 x + 12$, $alpha$ is not an integer and so $alpha$ is irrational.
Take 2
We have $alpha = beta+beta^2$, where $beta=sqrt[3]{3}$.
If $alpha$ were rational, then $beta$ would be a root of a quadratic polynomial with rational coefficients. However, the polynomial with rational coefficients with least degree having $beta$ as a root is $x^3-3$.
answered Jan 28 at 11:07
lhflhf
167k11172403
edited Jan 28 at 11:13
answered Jan 28 at 11:07
lhflhf
167k11172403
answered Jan 28 at 11:07
lhflhf
167k11172403
answered Jan 28 at 11:07
lhflhf
167k11172403
167k11172403
$begingroup$
It is fine what $beta$ is a root of $x^3 - 3$ but how do you know that it is of least degree polynomial with rational coefficients. Are you using some standard result?
$endgroup$
– prashant sharma
Feb 4 at 7:26
$begingroup$
@prashantsharma, $x^3 - 3$ is a cubic with no rational roots and so is irreducible.
$endgroup$
– lhf
Feb 4 at 7:58
$begingroup$
@Ihf Means are you suggesting that any equation with no rational roots is irreducible?
$endgroup$
– prashant sharma
Feb 5 at 9:24
$begingroup$
@prashantsharma, not any equation, but cubics, yes.
$endgroup$
– lhf
Feb 5 at 10:12
$begingroup$
@Ihf Can you please supply a proof?
$endgroup$
– prashant sharma
Feb 5 at 10:34
|
show 2 more comments
$begingroup$
It is fine what $beta$ is a root of $x^3 - 3$ but how do you know that it is of least degree polynomial with rational coefficients. Are you using some standard result?
$endgroup$
– prashant sharma
Feb 4 at 7:26
$begingroup$
@prashantsharma, $x^3 - 3$ is a cubic with no rational roots and so is irreducible.
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– lhf
Feb 4 at 7:58
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@Ihf Means are you suggesting that any equation with no rational roots is irreducible?
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– prashant sharma
Feb 5 at 9:24
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@prashantsharma, not any equation, but cubics, yes.
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– lhf
Feb 5 at 10:12
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@Ihf Can you please supply a proof?
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– prashant sharma
Feb 5 at 10:34
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It is fine what $beta$ is a root of $x^3 - 3$ but how do you know that it is of least degree polynomial with rational coefficients. Are you using some standard result?
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– prashant sharma
Feb 4 at 7:26
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It is fine what $beta$ is a root of $x^3 - 3$ but how do you know that it is of least degree polynomial with rational coefficients. Are you using some standard result?
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– prashant sharma
Feb 4 at 7:26
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@prashantsharma, $x^3 - 3$ is a cubic with no rational roots and so is irreducible.
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– lhf
Feb 4 at 7:58
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@prashantsharma, $x^3 - 3$ is a cubic with no rational roots and so is irreducible.
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– lhf
Feb 4 at 7:58
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@Ihf Means are you suggesting that any equation with no rational roots is irreducible?
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– prashant sharma
Feb 5 at 9:24
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@Ihf Means are you suggesting that any equation with no rational roots is irreducible?
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– prashant sharma
Feb 5 at 9:24
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@prashantsharma, not any equation, but cubics, yes.
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– lhf
Feb 5 at 10:12
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@prashantsharma, not any equation, but cubics, yes.
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– lhf
Feb 5 at 10:12
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@Ihf Can you please supply a proof?
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– prashant sharma
Feb 5 at 10:34
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@Ihf Can you please supply a proof?
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– prashant sharma
Feb 5 at 10:34
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show 2 more comments
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Let $a=sqrt[3]{3}$ and $b=sqrt[3]{9}$, and suppose that $a+b=r$ is rational. Then $$r^3 =(a+b)^3 = a^3+3ab(a+b)+b^3 = 12+9(a+b)=12+9r$$
So $r$ is a solution of equation $x^3-9x-12=0$ which has no rational root.
answered Jan 28 at 9:11
Maria MazurMaria Mazur
48.7k1260122
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add a comment |
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Let $a=sqrt[3]{3}$ and $b=sqrt[3]{9}$, and suppose that $a+b=r$ is rational. Then $$r^3 =(a+b)^3 = a^3+3ab(a+b)+b^3 = 12+9(a+b)=12+9r$$
So $r$ is a solution of equation $x^3-9x-12=0$ which has no rational root.
answered Jan 28 at 9:11
Maria MazurMaria Mazur
48.7k1260122
$endgroup$
add a comment |
$begingroup$
Let $a=sqrt[3]{3}$ and $b=sqrt[3]{9}$, and suppose that $a+b=r$ is rational. Then $$r^3 =(a+b)^3 = a^3+3ab(a+b)+b^3 = 12+9(a+b)=12+9r$$
So $r$ is a solution of equation $x^3-9x-12=0$ which has no rational root.
answered Jan 28 at 9:11
Maria MazurMaria Mazur
48.7k1260122
$endgroup$
Let $a=sqrt[3]{3}$ and $b=sqrt[3]{9}$, and suppose that $a+b=r$ is rational. Then $$r^3 =(a+b)^3 = a^3+3ab(a+b)+b^3 = 12+9(a+b)=12+9r$$
So $r$ is a solution of equation $x^3-9x-12=0$ which has no rational root.
answered Jan 28 at 9:11
Maria MazurMaria Mazur
48.7k1260122
answered Jan 28 at 9:11
Maria MazurMaria Mazur
48.7k1260122
answered Jan 28 at 9:11
Maria MazurMaria Mazur
48.7k1260122
answered Jan 28 at 9:11
Maria MazurMaria Mazur
48.7k1260122
48.7k1260122
add a comment |
add a comment |
2
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Please use descriptive titles. You have 150 characters to use, and you're more than welcomed to use $rmLaTeX$ as well. "Proving the irrationality of a number" is a meaningless title. Same goes to your other post from earlier.
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– Asaf Karagila♦
Jan 28 at 18:43
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Closely related: math.stackexchange.com/questions/1542708
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– Watson
Jan 29 at 9:06
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It may be that this exercise was intended to reinforce specific material covered in your textbook or other course materials, but your Readers will not be aware of that context unless you share it. It is certainly a problem that is amenable to various attacks, so knowing what you are studying will help Readers respond in a way you find useful.
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– hardmath
Jan 29 at 17:18