# Thread: Prove the following statement: If r3 is irrational, then r is irrational. disprove

1. ## Prove the following statement: If r3 is irrational, then r is irrational. disprove

9. a. Prove the following statement: If r3 is irrational, then r is irrational. (by contrapositive)

b. Disprove the converse of the statement in part (a).

a) let r be rational number

$\displaystyle r=\frac{a}{b}$ for $\displaystyle a,b \in \mathbb{Z};b \neq 0$

$\displaystyle r^3=\frac{a^3}{b^3}$

$\displaystyle a^3r^3=b^3 \rightarrow a*a*a*r*r*r=b*b*b$

by definition integers are closed under multiplication, all integers are rational

therefore r^3 is rational

b) converse is if r is irrational then r^3 is irrational

counterexample??

2. ## Re: Prove the following statement: If r3 is irrational, then r is irrational. disprov

If you are told to use the contrapositive, you are NOT trying to reach a contradiction.

These two statements are equivalent: p ->q and ~q -> ~p. So if you can prove the second statement, then you prove the first one.

So since you are trying to show that if r^3 is irrational then r is irrational, one way to do this is to prove its contrapositive, that if r is rational, then r^3 is rational.

There is no disproof involved!

3. ## Re: Prove the following statement: If r3 is irrational, then r is irrational. disprov

To disprove part (b), yes you need a counter-example. So we need to find an irrational number, whose cube is rational.

The obvious candidate is $\sqrt[3]{2}$. Its cube is clearly rational. Can you prove it is irrational?

4. ## Re: Prove the following statement: If r3 is irrational, then r is irrational. disprov

Originally Posted by Prove It
If you are told to use the contrapositive, you are NOT trying to reach a contradiction.

These two statements are equivalent: p ->q and ~q -> ~p. So if you can prove the second statement, then you prove the first one.

So since you are trying to show that if r^3 is irrational then r is irrational, one way to do this is to prove its contrapositive, that if r is rational, then r^3 is rational.

There is no disproof involved!
didn't I prove it by contrapositive? I took p --> q and made it ~q-->~p, I let r be rational then showed r^3 is rational

Originally Posted by Deveno
To disprove part (b), yes you need a counter-example. So we need to find an irrational number, whose cube is rational.

The obvious candidate is $\sqrt[3]{2}$. Its cube is clearly rational. Can you prove it is irrational?
assume the cuberoot of 2 is rational
$\displaystyle \sqrt[3]{2} = \frac{a}{b}; a,b \in \mathbb{Z}, b \neq 0$ and a/b is lowest terms

$\displaystyle 2=\frac{a^3}{b^3}$

$\displaystyle 2a^3=b^3$ the left hand side is even, thus the right side must also be even

$\displaystyle 2a^3=(2m)^3; (2m)=b$

$\displaystyle a^3=\frac{8}{2}m^3$

both have a factor of 2, both are even therefore they were not in there lowest terms, it is not rational, therefore is irrational

5. ## Re: Prove the following statement: If r3 is irrational, then r is irrational. disprov

Originally Posted by Jonroberts74
9. a. Prove the following statement: If r3 is irrational, then r is irrational. (by contrapositive)

b. Disprove the converse of the statement in part (a).

a) let r be rational number

$\displaystyle r=\frac{a}{b}$ for $\displaystyle a,b \in \mathbb{Z};b \neq 0$

$\displaystyle r^3=\frac{a^3}{b^3}$

$\displaystyle a^3r^3=b^3 \rightarrow a*a*a*r*r*r=b*b*b$

by definition integers are closed under multiplication, all integers are rational

therefore r^3 is rational

b) converse is if r is irrational then r^3 is irrational

counterexample??
$\displaystyle r^3=\frac{a^3}{b^3}$

$\displaystyle a^3r^3=b^3 \rightarrow a*a*a*r*r*r=b*b*b$

this was wrong meant

$\displaystyle b^3r^3=a^3 \Rightarrow b*b*b*r*r*r=a*a*a$

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### prove that 3 √x is irrational number

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