# Irrational Numbers

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#1
Could someone check my work please? The Problem is: Prove that the cube root of any irrational number is an irrational number.

I let be some irrational number. Then I tried to use proof by contradiction. I assumed that , where and are integers in lowest terms. Therefore and since p and q are in lowest terms this equation implies that , however is an irrational number and so it's impossible for it to be divisible by another integer. This is a contradiction. Our initial assumption that was rational must therefore be false and so the cube root of must be irrational.
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2 years ago
#2
(Original post by FXLander)
Could someone check my work please? The Problem is: Prove that the cube root of any irrational number is an irrational number.

I let be some irrational number. Then I tried to use proof by contradiction. I assumed that , where and are integers in lowest terms. Therefore and since p and q are in lowest terms this equation implies that , however is an irrational number and so it's impossible for it to be divisible by another integer. This is a contradiction. Our initial assumption that was rational must therefore be false and so the cube root of must be irrational.
No, take x^1/3 to be 4^1/3, which is irrational. So Then x = 4 which is rational.

You can consider a prime factorisation of x though and use the fact that if a is prime a|b^n if and only if a|b.

So in your working you should have x is prime and x|p or some prime factor of x divides p.
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2 years ago
#3
This is not true. The equation implies x and q^3 divides p^3. Work from there.

You can just observe that p^3 and q^3 are both rational, and that irrational mult by (non-zero) rational must be irrational.
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2 years ago
#4
(Original post by NotNotBatman)

You can consider a prime factorisation of x though and use the fact that if a is prime a|b^n if and only if a|b.

So in your working you should have x is prime and x|p or some prime factor of x divides p.
x is irrational, not sure why he needs to assume its prime to deduce that the cube root is irrational. It doesn’t have prime factors.
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#5
(Original post by RDKGames)
This is not true. The equation implies x and q^3 divides p^3. Work from there.

You can just observe that p^3 and q^3 are both rational, and that irrational mult by (non-zero) rational must be irrational.
So from xq^3 = p^3, we can say that p^3 is irrational and similarly q^3 is irrational as well. So is it fine to say that since p and q were assumed to be integers a product of integers can't result in an irrational number, which is the contradiction we need?
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2 years ago
#6
(Original post by FXLander)
So from xq^3 = p^3, we can say that p^3 is irrational and similarly q^3 is irrational as well. So is it fine to say that since p and q were assumed to be integers a product of integers can't result in an irrational number, which is the contradiction we need?
No.

You assumed p,q are integers with q being non-zero. Hence, p^3 and q^3 are rational. But we have xq^3=p^3 where x is irrational.

If you can’t see the contradiction here, it might be good to prove a simple lemma of rational*irrational=irrational

Tbh you should see the problem just from x=p^3/q^3
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#7
(Original post by RDKGames)
No.

You assumed p,q are integers with q being non-zero. Hence, p^3 and q^3 are rational. But we have xq^3=p^3 where x is irrational.

If you can’t see the contradiction here, it might be good to prove a simple lemma of rational*irrational=irrational

Tbh you should see the problem just from x=p^3/q^3
So p^3 and q^3 are both rational. But xq^3 = p^3 is irrational *rational, so that implies p^3 is irrational, which is a contradiction. (Why don't we say that p^3 and q^3 are integers as p and q are both integers, does it matter? Or is that just because every integer is a rational number and it's better to refer to rational numbers in this case?)

With regards to seeing it from x = p^3/q^3, do you mean that the LHS is irrational, whereas the RHS is a rational number divided by another rational number i.e. a rational number, so that is also a contradiction?
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2 years ago
#8
(Original post by FXLander)
So p^3 and q^3 are both rational. But xq^3 = p^3 is irrational *rational, so that implies p^3 is irrational, which is a contradiction. (Why don't we say that p^3 and q^3 are integers as p and q are both integers, does it matter? Or is that just because every integer is a rational number and it's better to refer to rational numbers in this case?)

With regards to seeing it from x = p^3/q^3, do you mean that the LHS is irrational, whereas the RHS is a rational number divided by another rational number i.e. a rational number, so that is also a contradiction?
First part: yes, you can say they are integers but it’s better to use rational/irrational since that’s what we’re looking at and comparing.

Second part: exactly.
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2 years ago
#9
(Original post by RDKGames)
x is irrational, not sure why he needs to assume its prime to deduce that the cube root is irrational. It doesn’t have prime factors.
Oh okay. Didn't see the irrational bit,

Then Op the answer is pretty much there.
Non zero Rational times irrational is irrational; which can be shown indirectly in one line, by showing what x is.
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2 years ago
#10
(Original post by RDKGames)
First part: yes, you can say they are integers but it’s better to use rational/irrational since that’s what we’re looking at and comparing.
It’s not better. It makes no difference.
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