1a) Prove that there exist irrational numbers a and b so that a^b is rational. look at the numbers √2^√2 and (√2^√2)^√2
1b) use similar idea to prove that there exists a rational number a and an irrational number b so that a^b is irrational
1c) prove that if 2^√2is irrational then √2^√2is irrational.
Thanks for your help.
(√2^√2)^√2 = √2^(√2*√2) = √2^2 = 2
√2^√2 = (2^1/2)^√2 = 2^(1/√2)
That should get you started
To prove the statements, we will use the concept of rational and irrational numbers, as well as the laws of exponents.
1a) To prove that there exist irrational numbers a and b such that a^b is rational, we can consider the numbers √2^√2 and (√2^√2)^√2.
Let's start by assuming √2^√2 is rational. In that case, our proof is complete since we have found a pair of irrational numbers where a = √2 and b = √2.
Now, let's consider the case where √2^√2 is irrational. In that scenario, we raise (√2^√2) to the power of √2. So, we have (√2^√2)^√2.
By applying the law of exponents, the expression becomes (√2)^(√2 * √2) = (√2)^(√2 * 2) = (√2)^2 = 2.
Now, we have 2 raised to the power of √2, which is a rational number. Therefore, we have found a pair of irrational numbers where a = √2^√2 and b = √2, and a^b = 2, which is rational.
Hence, we have proved that there exist irrational numbers a and b such that a^b is rational.
1b) 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 as in the previous proof.
Consider the numbers √2 and (√2^√2).
Let's assume that (√2^√2) is irrational. In that case, we have our desired result, where a = (√2^√2) and b = √2 are irrational, and a^b = (√2^√2)^√2 is also irrational.
Now, suppose (√2^√2) is rational. Then, let's take a = 2, which is a rational number. Let b = log(2, (√2^√2)). Since log(2, (√2^√2)) is irrational, we have a rational number (a = 2) raised to an irrational number (b = log(2, (√2^√2))), resulting in a^b being irrational.
Therefore, we have proved that there exists a rational number a and an irrational number b such that a^b is irrational.
1c) To prove that if 2^√2 is irrational, then √2^√2 is irrational, we can use a proof by contradiction.
Suppose 2^√2 is irrational, and let's assume that √2^√2 is rational.
Since √2^√2 is rational, then (√2^√2)^√2 is also rational. By applying the law of exponents, we have (√2)^(√2 * √2) = (√2)^(√2 * 2) = (√2)^2 = 2.
However, we know that 2^√2 is irrational, which contradicts our assumption that √2^√2 is rational.
Therefore, our assumption was incorrect, and √2^√2 must be irrational if 2^√2 is irrational.
Hence, we have proved that if 2^√2 is irrational, then √2^√2 is also irrational.