Mixing
Proving $sqrt3 + sqrt[3]{2}$ to be irrational
$begingroup$
In a test I tried to solve recently I came across the following question:
Prove $$sqrt3 + sqrt[3]{2}$$ is irrational
I tried proving it by saying it is equal to some rational number $$sqrt3 + sqrt[3]{2} = frac{m}{n}$$ and reaching a contradiction.
I tried squaring / cubing both sides and that didnt work. So how do I go about solving this problem?
number-theory irrational-numbers
edited Jan 28 at 8:52
Asaf Karagila♦
307k33439770
asked Jan 28 at 7:38
Guysudai1Guysudai1
18011
$endgroup$
add a comment |
$begingroup$
In a test I tried to solve recently I came across the following question:
Prove $$sqrt3 + sqrt[3]{2}$$ is irrational
I tried proving it by saying it is equal to some rational number $$sqrt3 + sqrt[3]{2} = frac{m}{n}$$ and reaching a contradiction.
I tried squaring / cubing both sides and that didnt work. So how do I go about solving this problem?
number-theory irrational-numbers
edited Jan 28 at 8:52
Asaf Karagila♦
307k33439770
asked Jan 28 at 7:38
Guysudai1Guysudai1
18011
$endgroup$
1
$begingroup$
Possible duplicate of Prove that $5^{1/3}+7^{1/2}$ is irrational
$endgroup$
– Watson
Jan 29 at 9:08
add a comment |
$begingroup$
In a test I tried to solve recently I came across the following question:
Prove $$sqrt3 + sqrt[3]{2}$$ is irrational
I tried proving it by saying it is equal to some rational number $$sqrt3 + sqrt[3]{2} = frac{m}{n}$$ and reaching a contradiction.
I tried squaring / cubing both sides and that didnt work. So how do I go about solving this problem?
number-theory irrational-numbers
edited Jan 28 at 8:52
Asaf Karagila♦
307k33439770
asked Jan 28 at 7:38
Guysudai1Guysudai1
18011
$endgroup$
In a test I tried to solve recently I came across the following question:
Prove $$sqrt3 + sqrt[3]{2}$$ is irrational
I tried proving it by saying it is equal to some rational number $$sqrt3 + sqrt[3]{2} = frac{m}{n}$$ and reaching a contradiction.
I tried squaring / cubing both sides and that didnt work. So how do I go about solving this problem?
number-theory irrational-numbers
number-theory irrational-numbers
edited Jan 28 at 8:52
Asaf Karagila♦
307k33439770
asked Jan 28 at 7:38
Guysudai1Guysudai1
18011
edited Jan 28 at 8:52
Asaf Karagila♦
307k33439770
asked Jan 28 at 7:38
Guysudai1Guysudai1
18011
edited Jan 28 at 8:52
Asaf Karagila♦
307k33439770
edited Jan 28 at 8:52
Asaf Karagila♦
307k33439770
edited Jan 28 at 8:52
Asaf Karagila♦
307k33439770
307k33439770
asked Jan 28 at 7:38
Guysudai1Guysudai1
18011
asked Jan 28 at 7:38
Guysudai1Guysudai1
18011
asked Jan 28 at 7:38
Guysudai1Guysudai1
18011
18011
1
$begingroup$
Possible duplicate of Prove that $5^{1/3}+7^{1/2}$ is irrational
$endgroup$
– Watson
Jan 29 at 9:08
add a comment |
1
$begingroup$
Possible duplicate of Prove that $5^{1/3}+7^{1/2}$ is irrational
$endgroup$
– Watson
Jan 29 at 9:08
1
1
$begingroup$
Possible duplicate of Prove that $5^{1/3}+7^{1/2}$ is irrational
$endgroup$
– Watson
Jan 29 at 9:08
$begingroup$
Possible duplicate of Prove that $5^{1/3}+7^{1/2}$ is irrational
$endgroup$
– Watson
Jan 29 at 9:08
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
It might be useful to first prove the following two statements - the sum of a rational and irrational number is irrational, and the product of a rational and irrational number is irrational.
You can then move the $sqrt 3$ and get:
$sqrt[3] 2=frac{m}{n}-sqrt 3$
$2=(frac{m}{n}-sqrt 3)^3$
After expanding the term at the RHS, the two lemmas you proved might come in handy.
answered Jan 28 at 7:49
GSoferGSofer
781313
$endgroup$
add a comment |
$begingroup$
It's possible to find a polynomial that has $sqrt{3}+sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots.
$$left((x-sqrt{3})^3-2right)left((x+sqrt{3})^3-2right)=0$$
$$left(x^3-3sqrt{3}x^2+9x-3sqrt{3}-2right)left(x^3+3sqrt{3}x^2+9x+3sqrt{3}-2right)=0$$
$$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$
By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $sqrt{3}+sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $pm 23$ the $x^6$ term is much larger than everything else combined.
This is not the easy way, of course.
answered Jan 28 at 8:07
jmerryjmerry
16.5k11633
$endgroup$
add a comment |
$begingroup$
Suppose $sqrt 3 + sqrt[3]2$ were rational; that is,
$sqrt 3 + sqrt[3]2 = r in Bbb Q: tag 1$
then
$sqrt[3]2 = r - sqrt 3; tag 2$
we cube:
$2 = r^3 - 3r^2sqrt 3 + 3r(sqrt 3)^2 - (sqrt 3)^3, tag 3$
or
$2 = r^3 - 3r^2 sqrt 3 + 9r - 3sqrt 3, tag 4$
or
$2 = r^3 + 9r - (3r^2 + 3)sqrt 3, tag 5$
or
$sqrt 3 = dfrac{2 - r^3 -9r}{3r^2 + 3} in Bbb Q, tag 6$
which contradicts the fact that
$sqrt 3 notin Bbb Q; tag 7$
therefore, (1) is false. $OEDelta$.
answered Jan 28 at 8:03
Robert LewisRobert Lewis
48.4k23167
$endgroup$
add a comment |
$begingroup$
Let $sqrt3+sqrt[3]2=rinmathbb Q$.
Thus, $$2=(r-sqrt3)^3$$ or
$$2=r^3-3sqrt3r^2+9r-3sqrt3$$ or
$$sqrt3=frac{r^3+9r-2}{3(r^2+1)}inmathbb Q,$$
which is a contradiction.
answered Jan 28 at 8:06
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Squaring is simpler than cubing. ;-)
Let $u=sqrt[3]{2}+sqrt{3}$. Then $sqrt{3}=u-sqrt[3]{2}$ and therefore
$$
(u^2-3)-2usqrt[3]{2}+sqrt[3]{4}=0
$$
If $uinmathbb{Q}$, this contradicts ${1,sqrt[3]{2},sqrt[3]{4}}$ being a basis of $mathbb{Q}(sqrt[3]{2})$ over $mathbb{Q}$, which stems from $x^3-2$ being irreducible over $mathbb{Q}$ (Eisenstein) and standard facts on simple algebraic extensions.
Even simpler: if $u$ is rational, then $mathbb{Q}(sqrt{3})$ is a subfield of $mathbb{Q}(sqrt[3]{2})$, which is ruled out by the dimension theorem.
answered Jan 28 at 10:14
egregegreg
185k1486206
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It might be useful to first prove the following two statements - the sum of a rational and irrational number is irrational, and the product of a rational and irrational number is irrational.
You can then move the $sqrt 3$ and get:
$sqrt[3] 2=frac{m}{n}-sqrt 3$
$2=(frac{m}{n}-sqrt 3)^3$
After expanding the term at the RHS, the two lemmas you proved might come in handy.
answered Jan 28 at 7:49
GSoferGSofer
781313
$endgroup$
add a comment |
$begingroup$
It might be useful to first prove the following two statements - the sum of a rational and irrational number is irrational, and the product of a rational and irrational number is irrational.
You can then move the $sqrt 3$ and get:
$sqrt[3] 2=frac{m}{n}-sqrt 3$
$2=(frac{m}{n}-sqrt 3)^3$
After expanding the term at the RHS, the two lemmas you proved might come in handy.
answered Jan 28 at 7:49
GSoferGSofer
781313
$endgroup$
add a comment |
$begingroup$
It might be useful to first prove the following two statements - the sum of a rational and irrational number is irrational, and the product of a rational and irrational number is irrational.
You can then move the $sqrt 3$ and get:
$sqrt[3] 2=frac{m}{n}-sqrt 3$
$2=(frac{m}{n}-sqrt 3)^3$
After expanding the term at the RHS, the two lemmas you proved might come in handy.
answered Jan 28 at 7:49
GSoferGSofer
781313
$endgroup$
It might be useful to first prove the following two statements - the sum of a rational and irrational number is irrational, and the product of a rational and irrational number is irrational.
You can then move the $sqrt 3$ and get:
$sqrt[3] 2=frac{m}{n}-sqrt 3$
$2=(frac{m}{n}-sqrt 3)^3$
After expanding the term at the RHS, the two lemmas you proved might come in handy.
answered Jan 28 at 7:49
GSoferGSofer
781313
answered Jan 28 at 7:49
GSoferGSofer
781313
answered Jan 28 at 7:49
GSoferGSofer
781313
answered Jan 28 at 7:49
GSoferGSofer
781313
781313
add a comment |
add a comment |
$begingroup$
It's possible to find a polynomial that has $sqrt{3}+sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots.
$$left((x-sqrt{3})^3-2right)left((x+sqrt{3})^3-2right)=0$$
$$left(x^3-3sqrt{3}x^2+9x-3sqrt{3}-2right)left(x^3+3sqrt{3}x^2+9x+3sqrt{3}-2right)=0$$
$$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$
By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $sqrt{3}+sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $pm 23$ the $x^6$ term is much larger than everything else combined.
This is not the easy way, of course.
answered Jan 28 at 8:07
jmerryjmerry
16.5k11633
$endgroup$
add a comment |
$begingroup$
It's possible to find a polynomial that has $sqrt{3}+sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots.
$$left((x-sqrt{3})^3-2right)left((x+sqrt{3})^3-2right)=0$$
$$left(x^3-3sqrt{3}x^2+9x-3sqrt{3}-2right)left(x^3+3sqrt{3}x^2+9x+3sqrt{3}-2right)=0$$
$$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$
By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $sqrt{3}+sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $pm 23$ the $x^6$ term is much larger than everything else combined.
This is not the easy way, of course.
answered Jan 28 at 8:07
jmerryjmerry
16.5k11633
$endgroup$
add a comment |
$begingroup$
It's possible to find a polynomial that has $sqrt{3}+sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots.
$$left((x-sqrt{3})^3-2right)left((x+sqrt{3})^3-2right)=0$$
$$left(x^3-3sqrt{3}x^2+9x-3sqrt{3}-2right)left(x^3+3sqrt{3}x^2+9x+3sqrt{3}-2right)=0$$
$$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$
By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $sqrt{3}+sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $pm 23$ the $x^6$ term is much larger than everything else combined.
This is not the easy way, of course.
answered Jan 28 at 8:07
jmerryjmerry
16.5k11633
$endgroup$
It's possible to find a polynomial that has $sqrt{3}+sqrt[3]{2}$ as a root, along with several other variations from choosing different square and cube roots.
$$left((x-sqrt{3})^3-2right)left((x+sqrt{3})^3-2right)=0$$
$$left(x^3-3sqrt{3}x^2+9x-3sqrt{3}-2right)left(x^3+3sqrt{3}x^2+9x+3sqrt{3}-2right)=0$$
$$x^6-9x^4-4x^3+27x^2-36x-23 = 0$$
By the rational root theorem, the only possible rational roots of that polynomial are $-23,-1,1,$ and $23$. $sqrt{3}+sqrt[3]{2}$ is between $1+1=2$ and $2+2=4$, so it isn't any of them. Also, as is easily verified, none of those possible roots actually are roots. The values at $pm 1$ are $1-9-4+27-36-23=-44$ and $1-9+4+27+36-23=36$, while at $pm 23$ the $x^6$ term is much larger than everything else combined.
This is not the easy way, of course.
answered Jan 28 at 8:07
jmerryjmerry
16.5k11633
edited Jan 28 at 8:18
answered Jan 28 at 8:07
jmerryjmerry
16.5k11633
answered Jan 28 at 8:07
jmerryjmerry
16.5k11633
answered Jan 28 at 8:07
jmerryjmerry
16.5k11633
16.5k11633
add a comment |
add a comment |
$begingroup$
Suppose $sqrt 3 + sqrt[3]2$ were rational; that is,
$sqrt 3 + sqrt[3]2 = r in Bbb Q: tag 1$
then
$sqrt[3]2 = r - sqrt 3; tag 2$
we cube:
$2 = r^3 - 3r^2sqrt 3 + 3r(sqrt 3)^2 - (sqrt 3)^3, tag 3$
or
$2 = r^3 - 3r^2 sqrt 3 + 9r - 3sqrt 3, tag 4$
or
$2 = r^3 + 9r - (3r^2 + 3)sqrt 3, tag 5$
or
$sqrt 3 = dfrac{2 - r^3 -9r}{3r^2 + 3} in Bbb Q, tag 6$
which contradicts the fact that
$sqrt 3 notin Bbb Q; tag 7$
therefore, (1) is false. $OEDelta$.
answered Jan 28 at 8:03
Robert LewisRobert Lewis
48.4k23167
$endgroup$
add a comment |
$begingroup$
Suppose $sqrt 3 + sqrt[3]2$ were rational; that is,
$sqrt 3 + sqrt[3]2 = r in Bbb Q: tag 1$
then
$sqrt[3]2 = r - sqrt 3; tag 2$
we cube:
$2 = r^3 - 3r^2sqrt 3 + 3r(sqrt 3)^2 - (sqrt 3)^3, tag 3$
or
$2 = r^3 - 3r^2 sqrt 3 + 9r - 3sqrt 3, tag 4$
or
$2 = r^3 + 9r - (3r^2 + 3)sqrt 3, tag 5$
or
$sqrt 3 = dfrac{2 - r^3 -9r}{3r^2 + 3} in Bbb Q, tag 6$
which contradicts the fact that
$sqrt 3 notin Bbb Q; tag 7$
therefore, (1) is false. $OEDelta$.
answered Jan 28 at 8:03
Robert LewisRobert Lewis
48.4k23167
$endgroup$
add a comment |
$begingroup$
Suppose $sqrt 3 + sqrt[3]2$ were rational; that is,
$sqrt 3 + sqrt[3]2 = r in Bbb Q: tag 1$
then
$sqrt[3]2 = r - sqrt 3; tag 2$
we cube:
$2 = r^3 - 3r^2sqrt 3 + 3r(sqrt 3)^2 - (sqrt 3)^3, tag 3$
or
$2 = r^3 - 3r^2 sqrt 3 + 9r - 3sqrt 3, tag 4$
or
$2 = r^3 + 9r - (3r^2 + 3)sqrt 3, tag 5$
or
$sqrt 3 = dfrac{2 - r^3 -9r}{3r^2 + 3} in Bbb Q, tag 6$
which contradicts the fact that
$sqrt 3 notin Bbb Q; tag 7$
therefore, (1) is false. $OEDelta$.
answered Jan 28 at 8:03
Robert LewisRobert Lewis
48.4k23167
$endgroup$
Suppose $sqrt 3 + sqrt[3]2$ were rational; that is,
$sqrt 3 + sqrt[3]2 = r in Bbb Q: tag 1$
then
$sqrt[3]2 = r - sqrt 3; tag 2$
we cube:
$2 = r^3 - 3r^2sqrt 3 + 3r(sqrt 3)^2 - (sqrt 3)^3, tag 3$
or
$2 = r^3 - 3r^2 sqrt 3 + 9r - 3sqrt 3, tag 4$
or
$2 = r^3 + 9r - (3r^2 + 3)sqrt 3, tag 5$
or
$sqrt 3 = dfrac{2 - r^3 -9r}{3r^2 + 3} in Bbb Q, tag 6$
which contradicts the fact that
$sqrt 3 notin Bbb Q; tag 7$
therefore, (1) is false. $OEDelta$.
answered Jan 28 at 8:03
Robert LewisRobert Lewis
48.4k23167
answered Jan 28 at 8:03
Robert LewisRobert Lewis
48.4k23167
answered Jan 28 at 8:03
Robert LewisRobert Lewis
48.4k23167
answered Jan 28 at 8:03
Robert LewisRobert Lewis
48.4k23167
48.4k23167
add a comment |
add a comment |
$begingroup$
Let $sqrt3+sqrt[3]2=rinmathbb Q$.
Thus, $$2=(r-sqrt3)^3$$ or
$$2=r^3-3sqrt3r^2+9r-3sqrt3$$ or
$$sqrt3=frac{r^3+9r-2}{3(r^2+1)}inmathbb Q,$$
which is a contradiction.
answered Jan 28 at 8:06
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Let $sqrt3+sqrt[3]2=rinmathbb Q$.
Thus, $$2=(r-sqrt3)^3$$ or
$$2=r^3-3sqrt3r^2+9r-3sqrt3$$ or
$$sqrt3=frac{r^3+9r-2}{3(r^2+1)}inmathbb Q,$$
which is a contradiction.
answered Jan 28 at 8:06
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Let $sqrt3+sqrt[3]2=rinmathbb Q$.
Thus, $$2=(r-sqrt3)^3$$ or
$$2=r^3-3sqrt3r^2+9r-3sqrt3$$ or
$$sqrt3=frac{r^3+9r-2}{3(r^2+1)}inmathbb Q,$$
which is a contradiction.
answered Jan 28 at 8:06
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
Let $sqrt3+sqrt[3]2=rinmathbb Q$.
Thus, $$2=(r-sqrt3)^3$$ or
$$2=r^3-3sqrt3r^2+9r-3sqrt3$$ or
$$sqrt3=frac{r^3+9r-2}{3(r^2+1)}inmathbb Q,$$
which is a contradiction.
answered Jan 28 at 8:06
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 28 at 8:06
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 28 at 8:06
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 28 at 8:06
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
add a comment |
add a comment |
$begingroup$
Squaring is simpler than cubing. ;-)
Let $u=sqrt[3]{2}+sqrt{3}$. Then $sqrt{3}=u-sqrt[3]{2}$ and therefore
$$
(u^2-3)-2usqrt[3]{2}+sqrt[3]{4}=0
$$
If $uinmathbb{Q}$, this contradicts ${1,sqrt[3]{2},sqrt[3]{4}}$ being a basis of $mathbb{Q}(sqrt[3]{2})$ over $mathbb{Q}$, which stems from $x^3-2$ being irreducible over $mathbb{Q}$ (Eisenstein) and standard facts on simple algebraic extensions.
Even simpler: if $u$ is rational, then $mathbb{Q}(sqrt{3})$ is a subfield of $mathbb{Q}(sqrt[3]{2})$, which is ruled out by the dimension theorem.
answered Jan 28 at 10:14
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Squaring is simpler than cubing. ;-)
Let $u=sqrt[3]{2}+sqrt{3}$. Then $sqrt{3}=u-sqrt[3]{2}$ and therefore
$$
(u^2-3)-2usqrt[3]{2}+sqrt[3]{4}=0
$$
If $uinmathbb{Q}$, this contradicts ${1,sqrt[3]{2},sqrt[3]{4}}$ being a basis of $mathbb{Q}(sqrt[3]{2})$ over $mathbb{Q}$, which stems from $x^3-2$ being irreducible over $mathbb{Q}$ (Eisenstein) and standard facts on simple algebraic extensions.
Even simpler: if $u$ is rational, then $mathbb{Q}(sqrt{3})$ is a subfield of $mathbb{Q}(sqrt[3]{2})$, which is ruled out by the dimension theorem.
answered Jan 28 at 10:14
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Squaring is simpler than cubing. ;-)
Let $u=sqrt[3]{2}+sqrt{3}$. Then $sqrt{3}=u-sqrt[3]{2}$ and therefore
$$
(u^2-3)-2usqrt[3]{2}+sqrt[3]{4}=0
$$
If $uinmathbb{Q}$, this contradicts ${1,sqrt[3]{2},sqrt[3]{4}}$ being a basis of $mathbb{Q}(sqrt[3]{2})$ over $mathbb{Q}$, which stems from $x^3-2$ being irreducible over $mathbb{Q}$ (Eisenstein) and standard facts on simple algebraic extensions.
Even simpler: if $u$ is rational, then $mathbb{Q}(sqrt{3})$ is a subfield of $mathbb{Q}(sqrt[3]{2})$, which is ruled out by the dimension theorem.
answered Jan 28 at 10:14
egregegreg
185k1486206
$endgroup$
Squaring is simpler than cubing. ;-)
Let $u=sqrt[3]{2}+sqrt{3}$. Then $sqrt{3}=u-sqrt[3]{2}$ and therefore
$$
(u^2-3)-2usqrt[3]{2}+sqrt[3]{4}=0
$$
If $uinmathbb{Q}$, this contradicts ${1,sqrt[3]{2},sqrt[3]{4}}$ being a basis of $mathbb{Q}(sqrt[3]{2})$ over $mathbb{Q}$, which stems from $x^3-2$ being irreducible over $mathbb{Q}$ (Eisenstein) and standard facts on simple algebraic extensions.
Even simpler: if $u$ is rational, then $mathbb{Q}(sqrt{3})$ is a subfield of $mathbb{Q}(sqrt[3]{2})$, which is ruled out by the dimension theorem.
answered Jan 28 at 10:14
egregegreg
185k1486206
answered Jan 28 at 10:14
egregegreg
185k1486206
answered Jan 28 at 10:14
egregegreg
185k1486206
answered Jan 28 at 10:14
egregegreg
185k1486206
185k1486206
add a comment |
add a comment |
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$begingroup$
Possible duplicate of Prove that $5^{1/3}+7^{1/2}$ is irrational
$endgroup$
– Watson
Jan 29 at 9:08