Prove that a+b is irrational?

• Sep 17th 2009, 03:42 PM
GreenDay14
Prove that a+b is irrational?
okay so the question is: Suppose that a is a rational number and that b is an irrational number. Prove that a+b is irrational.

So i went to prove A: ab is irrational by contradiction, assuming A is false.

Therefore ab is rational, so ab = m/n for integers m and , n is nonzero. So if a is given to be rational then a = k/l for integers k and l, l and k are both nonzero.

So b = m/n + a .... then i completely got lost...

Could someone help me from there, or let me know if I am completely wrong? Thanks.
• Sep 17th 2009, 04:42 PM
Mush
Quote:

Originally Posted by GreenDay14
okay so the question is: Suppose that a is a rational number and that b is an irrational number. Prove that a+b is irrational.

So i went to prove A: ab is irrational by contradiction, assuming A is false.

Therefore ab is rational, so ab = m/n for integers m and , n is nonzero. So if a is given to be rational then a = k/l for integers k and l, l and k are both nonzero.

So b = m/n + a .... then i completely got lost...

Could someone help me from there, or let me know if I am completely wrong? Thanks.

$\displaystyle a$ is rational and $\displaystyle b$ is irrational.

Therefore $\displaystyle a = \frac{p}{q}$.

Now, let's assume that $\displaystyle a+b = \frac{l}{m}$, i.e., assume that a+b is rational.

Now, since $\displaystyle a = \frac{p}{q}$, we can write $\displaystyle a+b = \frac{p}{q} + b$

Therefore $\displaystyle \frac{p}{q} + b = \frac{l}{m}$

Therefore $\displaystyle b = \frac{l}{m} - \frac{p}{q} = \frac{ql - pm}{mq}$

This is a contradiction since $\displaystyle \frac{ql - pm}{mq}$ is rational, and b was defined as irrational. Therefore, the assumption that a+b is rational must be false, and a+b must be irrational.