There exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2

a) use similar idea to prove that there exists a rational number a and an irrational number b so that a^b is irrational

b) prove that if 2^√2is irrational then √2^√2is irrational.

a) To prove that there exist a rational number a and an irrational number b such that a^b is irrational, we can use a similar idea to the given example.

Consider the number √2, which is known to be irrational. Now, let's assume that (√2)^√2 is rational. In this case, we can take a = (√2)^√2 and b = √2.

Since b = √2 is irrational and a = (√2)^√2 is rational (according to our assumption), we have found a pair of a rational number and an irrational number such that a^b is rational.

However, if we consider (√2)^√2^√2, we notice that (√2^√2)^√2 = (√2)^(√2*√2) = (√2)^2 = 2. This means that (√2^√2)^√2 is irrational, as 2 is rational. This contradiction shows that our assumption (√2)^√2 being rational is false.

Therefore, our proof is complete, and we have proven that there exist a rational number a and an irrational number b such that a^b is irrational.

b) To prove that if 2^√2 is irrational, then √2^√2 is irrational, we can use a proof by contradiction.

Assume that 2^√2 is irrational, but √2^√2 is rational. Now, let's consider (√2^√2)^√2. According to the laws of exponentiation, (√2^√2)^√2 = (√2)^(√2*√2) = (√2)^2 = 2.

Since 2 is rational, this implies that (√2^√2)^√2 is rational. However, this contradicts our assumption that √2^√2 is rational. Therefore, our assumption must be false.

Hence, we can conclude that if 2^√2 is irrational, then √2^√2 is also irrational.