# Rational and irrational numbers. (semi- )

1. Sep 21, 2009

### Sven

Rational and irrational numbers. (semi-urgent)

I need to figure this out by tomorrow =/

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

none.

3. The attempt at a solution

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

2. Sep 21, 2009

### VietDao29

Re: Rational and irrational numbers. (semi-urgent)

So far so good. :)

When finding a proof seems hopeless, one should try to search for a counter-example.

According to part a,

$$1 + \sqrt{2}$$, and $$-\sqrt{2}$$ are both irrational, what if you take the sum of them?

First case, okay. :)

Second case:

Well, just do as you did in part a. Like this:

$$a \in \mathbb{Q} \backslash \{ 0 \} 0, b \notin \mathbb{Q}$$

Assume that
$$ab = r$$, where r is a rational number.
$$\Rightarrow b = ra ^ {-1}$$

What can you say about b in the above expression?

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If a, and b are both irrational numbers, is ab also irrational?

Think about 4-th root. :)