#There exists a rational number a and an irrational number b so that a^b is irrational.
raw rapidsBOT
umbral badger
you can do this without necessarily formalizing all of what it means to raise a number to an irrational number, you just need to know some properties about raising numbers to a power. Namely that x^y is not equal to x^z if y is not equal to z and x is not equal to 1.. 1. ||If you know that the rational numbers are countable,|| 2.||and that the irrational numbers are uncountable,|| 3. ||and you also know that 2^x is strictly increasing,|| then you are already on the right track to prove the statement. 4. (do not look at this) ||go by contradiction and (seriously don't look at the next part)|| || use the pigeonhole principle||
oh directly
well you can do it this way too, and if you're not satisfied with this then we can talk about the formalisms
can you post the original question?
stuck frigate
This was the orginial question. We are only allowed to use things that were presented in class, and we must formalize the rest ourselves, which makes it hard to get help with. We essentially only learned things like the division algorithm, the definition of a rational number, greatest common divisor, topics like that. We are using hammock's book of proof if you are familiar and we are in chapter 8ish. I can follow what you are saying but the issue with it and what is frustrating me about this problem so much is that i cannot use a lot of things.
last leaf
To prove that there exists a rational number a and an irrational number b such that a^b is irrational, you can start by considering the case where b is an irrational number such as √2. Then, let a be some rational number like 2.
Now, consider a^b = 2^√2. If we can show that 2^√2 is irrational, then we have proved the statement.
Next, assume for the sake of contradiction that 2^√2 is rational. This would mean that 2^√2 can be expressed as a ratio of two integers, say p/q where p and q are integers and q≠0.
Therefore, we have: 2^√2 = p/q
Taking the square of both sides, we get: 2^2√2 = (p/q)^2
This simplifies to: 2√2 = p^2 / q^2
Now, we can isolate √2 on one side of the equation to get: √2 = p^2 / (q^2 * 2)
This shows that √2 is rational, which is a contradiction since we know that √2 is irrational.
Therefore, our initial assumption that 2^√2 is rational must be false, and hence 2^√2 is irrational.
Thus, we have found a rational number a = 2 and an irrational number b = √2 such that a^b = 2^√2 is irrational, as required.
stuck frigate
Thanks for your help!
However, I am confused at these steps
Taking the square of both sides, we get: 2^2√2 = (p/q)^2
This simplifies to: 2√2 = p^2 / q^2
How is 2^(2sqrt(2)) = 2sqrt(2)?