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 exists 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. Let's consider the number √2, which is irrational. Now, consider the number (√2)^(√2).
Assume for contradiction that (√2)^(√2) is rational. In that case, we have found a pair of irrational numbers a = √2 and b = √2 such that a^b is rational, which contradicts the assumption that it is irrational. Therefore, (√2)^(√2) must be irrational.
b) To prove that if 2^√2 is irrational, then (√2)^(√2) is irrational, we will use a proof by contradiction.
Assume for contradiction that 2^√2 is irrational and (√2)^(√2) is rational.
Let's consider the number (√2)^(√2)^√2. Simplifying this expression, we get (√2)^(√2*√2) = (√2)^2 = 2.
Now, if (√2)^(√2)^√2 is rational, then we have found a pair of irrational numbers (√2)^(√2) and √2 such that their power is rational, which contradicts the assumption that it is irrational.
Therefore, if 2^√2 is irrational, then (√2)^(√2) must also be irrational.