- #1
jhson114
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Im trying to either prove or disprove that if a and b are rational numbers, then a^b is also rational. I tried doing it with a contradiction, but i can't seem to correctly arrive at a solution. this is how i started the problem
defn of rational number: a,b = {m/n: m,n are all nonzero integers}
1. a^b is irrational (hypothesis/assumption)
2. b^b is irrational (from 1)
3. (m/n)^(m/n) (from defn. of rational number)
4. [m^(m/n)]/[(n^(m/n)] (algebra)
i'm stuck right here. i need to prove that an integer raised to a rational number is either rational or irrational. any inputs will be really helpful. thank you
defn of rational number: a,b = {m/n: m,n are all nonzero integers}
1. a^b is irrational (hypothesis/assumption)
2. b^b is irrational (from 1)
3. (m/n)^(m/n) (from defn. of rational number)
4. [m^(m/n)]/[(n^(m/n)] (algebra)
i'm stuck right here. i need to prove that an integer raised to a rational number is either rational or irrational. any inputs will be really helpful. thank you