# Irrational Numbers

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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.

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

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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.

**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.

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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#3

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Therefore and since p and q are in lowest terms this equation implies that

**FXLander**)Therefore and since p and q are in lowest terms this equation implies that

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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#4

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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.

**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.

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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.

**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.

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#6

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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?

**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?

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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(Original post by

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

**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

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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#8

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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?

**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?

Second part: exactly.

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#9

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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.

**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.

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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#10

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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.

**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.

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