Question

Prove or disprove: There exists a bijective function
\[f:\mathbb{Q} \rightarrow \mathbb{R}\]

I believe this is not true simply because Q is countable and R is not. I think I need more than that for my proof though. Would it be enough to give an example, i.e. there is no \[x \in \mathbb{Q}\ that\ is\ equal\ \to\ \pi?\]

Answer 1

Your first argument is sufficient, what I think you're saying for your second argument isn't true.

Answer 2

I think the first argument is sufficient as well personally. I guess it just seems to easy lol. But I think my second argument is true as well since Q is the set of rationals, and pi is irrational.

Answer 3

Simply saying an element in the first set doesn't occur in the second set doesn't prove anything about the function. Hence I assumed you meant there was no x in Q such that x is mapped to pi, which again isn't necessarily true. So yeah, pi doesnt occur in Q, you are right.

Answer 4

right, I see what you are saying in the sense that I could arbitrarily map something to pi, but I couldn't create a function that would get me to pi using only elements of Q.

I guess I will go with my first argument since I know that it's true. I'll just try to find that explicitly stated in my textbook so I can quote it.

Answer 5

If you want a nicer proof. Define any bijective f. Then w.l.o.g. swap the elements so for x in Q, f(x) = x, then all of Q is mapped to an element, but not all of R is mapped to, hence it cant be a bijection.
Hopefully that makes sense.

Answer 6

And the w.l.o.g. occurs because you're not changing the size of either set.

Answer 7

have you already proven that Q and R do not have the same cardinality?

Answer 8

Ok, not directly myself, but I have a Theorem in the book that Q is countable, and it was well established in the previous chapter that R is uncountable, so I have my proof.

Answer 9

Thank you both for working through this with me.

Answer 10

No problem, glad to have good questions!