Suppose that f is a function from A to B, where A and B are finite sets |A| = |B|. Show that f is one-to-one if and only if it is onto.
Lets say we were to translate this statement using variables. Where:
a: f is a function from A to B, where A and B are finite sets |A| = |B|
b: f is one-to-one
c: f is onto
Now, I believe this statement translates to "a -> (b <-> c)" symbolically. Is this correct?
This would mean that we need to prove the two sub-statements "a -> (b -> c)" and "a -> (c -> b)" in order to prove the overall statement "a -> (b <-> c)".
To prove "a -> (b -> c)", I'm thinking of first showing that "b -> c" is true using a proof by contradiction. We would then get "a -> T", which is always true (trivial proof). Is this correct?
To prove "a -> (c -> b)", we would use the same procedure.
Is this what the top answerer here does?
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