So basically because the number is irrational, then that number times itself any amount of times will also yield an irrational product.
No. For example, is an irrational number, but is rational. However, any integer power of a rational number is rational.
You need to learn how to prove universal statement, implications and negations. In this case, you fix an arbitrary and assume that there exists a such that is irrational. Then you need to show that is not rational. You prove this by contradiction, i.e., you assume that is rational. Then is rational as well, contrary to the assumption.
Can't edit, so: You can't extrapolate from my post 15 that irrationalxirrational is irrational (which I believe you meant) because multiplication is not defined for the set of irrational numbers. Within the set of real numbers, emakorov is correct, you can have irrational times irrational is rational.
I assume that's the statement you are referring to, in which case, as I stated much earlier:
If a were rational a^b would be rational. So a can't be rational, ie, it must be irrational since there are no other options in real numbers.
In my personal opinion, the issue has been confused by a lot of irrelevant discussion.
If a is rational, a^b can't be irrational. N/MxN/M=N^2/M^2
The context of the problem is assumed to be the set of real numbers.
rational number: a number which can be expressed as N/M where N and M are integers.
irrational number: a number which can't be expressed as N/M where N and M are integers, for ex: pi, sqrt2, e, ...
rational numbers and irrational numbers belong to the set of real numbers.
If a is rational, a^b is rational. If a is irrational, it does not follow that a^b is irrational. As emakorov pointed out sqrt2xsqrt2=2, or irrationalxirrational = rational. However, as we have pointed out countless times now, if a^b is irrational, a must be irrational (again, for if a were rational, a^b wouild be rational)