# Thread: semi-urgent: rational and irrational numbers!

1. ## semi-urgent: rational and irrational numbers!

Okay I don't get this =( I think I've made a decent attempt but I need some help getting there all the way. This is due tomorrow. can someone help me? =(

a. If a is rational and b is irrational, is a+b necessarily irrational? What if a and b are both irrational?
b. If a is rational and b is irrational, is ab necessarily irrational?
c. Is there a number of a such that a^2 is irrational, but a^4 is rational?
a. I think I have this first part. You can prove it by contradiction.

R= some rational number

a+b = R
b = R-a
A rational number minus a rational number is a rational number. This would mean b = rational, which is not true. therefore a+b is irrational.

This second part, if both are irrational? I was thinking:

a+b = R

a = R-b, or b = R-a. I'm not sure how this helps me x(
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b. If a is rational and b is irrational, is ab necessarily irrational?

No idea, but here's my attempt:

Two cases. If a = 0, ab = 0, and 0 is rational so ab is rational.

if a =/= 0...proof by contradiction maybe?

a*b = rational
a*b = a*b
b = a*b*a^-1?

But then b=b? And that doesn't help me.
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c Is there a number a such that a^2 is irrational, but a^4 is rational?

Well again I have no idea but here's my attempt:

a=b
a^2 = ab
ab = x
b = x/a
b = x * a^-1

can someone help me finish these? =( thanks in advance.

2. Originally Posted by Sven
a. If a is rational and b is irrational, is a+b necessarily irrational? What if a and b are both irrational?
b. If a is rational and b is irrational, is ab necessarily irrational?
c. Is there a number of a such that a^2 is irrational, but a^4 is rational?
a. No. All you need is a specific counterexample. eg. $\displaystyle a = -1$ and $\displaystyle b = 1 + \sqrt{2}$.

b. Assume $\displaystyle ab = c$ where c is rational and $\displaystyle a \neq 0$. Then $\displaystyle b = \frac{c}{a}$. What does this mean about b ....?

c. Yes. eg. $\displaystyle a = 2^{1/4}$.