So basically because the number is irrational, then that number times itself any amount of times will also yield an irrational product.

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- Aug 29th 2013, 01:17 PMlamentofkingRe: How can a be irrational if it's exponent is an integer?
So basically because the number is irrational, then that number times itself any amount of times will also yield an irrational product.

- Aug 29th 2013, 01:32 PMemakarovRe: How can a be irrational if it's exponent is an integer?
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. - Aug 29th 2013, 02:43 PMHartlwRe: How can a be irrational if it's exponent is an integer?
- Aug 29th 2013, 03:18 PMHartlwRe: How can a be irrational if it's exponent is an integer?
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.

- Aug 30th 2013, 01:49 PMlamentofkingRe: How can a be irrational if it's exponent is an integer?
So in order to prove this statement I need to prove why a cannot be rational?

- Aug 30th 2013, 01:54 PMemakarovRe: How can a be irrational if it's exponent is an integer?
- Aug 30th 2013, 02:16 PMHartlwRe: How can a be irrational if it's exponent is an integer?
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. - Aug 30th 2013, 09:36 PMibduttRe: How can a be irrational if it's exponent is an integer?
Why? a variable might be real that means rational / irrational and also it may be an imaginary one. it depends on the context were we are using it.

- Aug 30th 2013, 09:38 PMibduttRe: How can a be irrational if it's exponent is an integer?
I agree this could be one approach.

let a be a rational number. that means a = c/d. now a^b = (c/d)^b = (c^b)/ (d^b) that means a^b is rational similarly proceed for when a is irrational. - Aug 30th 2013, 11:57 PMzzephodRe: How can a be irrational if it's exponent is an integer?
- Aug 31st 2013, 07:35 AMHartlwRe: How can a be irrational if it's exponent is an integer?
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) - Aug 31st 2013, 01:44 PMlamentofkingRe: How can a be irrational if it's exponent is an integer?
Thank you everyone. I do believe the problem has been made clear now. :-)