Actually what i know is that i must assume that this statement is false and then try to come up with non sense statement.
Prove by the smallest counterexample technique the statement $$\binom {2n}n\leqslant4^n.$$
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3Suppose $k$ were the smallest counterexample. Check it's not $k = 0$. Then derive a contradiction from the assumption that $\binom{2(k-1)}{k-1} \leqslant 4^{k-1}$. It's induction in disguise. – Daniel Fischer Oct 30 '14 at 20:11
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Assume that $$\binom{2n}{n}>4^n$$ for some $n\in \mathbb{N}.$ Then
$$\binom{2n}{n}=\frac{2(2n-1)}{n}\binom{2(n-1)}{n-1}>4^{n-1}\implies \binom{2(n-1)}{n-1}>\frac{n4^n}{2(2n-1)}>4^{n-1}.$$
That is, if the inequality fails to hold for some $n\in \mathbb{N}$ then it fails to hold for any smaller natural. In particular, we would have
$$2=\binom{2}{1}>4^1=4,$$
which gives the desired contradiction.
mfl
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Even stronger we could use n=0 for the contradiction. But well done! – nathan.j.mcdougall Oct 30 '14 at 20:14