In the previous post, we proved that

$latex \displaystyle a < \frac{a + b}{2}$

both by direct and indirect proof. The proof that we have used in the previous post was proof by contradiction. In proof by contradiction, we show that by assuming the proposition to false would imply a contradiction. One of the most famous of this proof is the proof that there are infinitely many prime numbers.

In this post, we will prove that $latex 0.999 \cdots 1$. Recall that we already proved this theorem using direct proof. In proving by contradiction, first, we assume that

$latex 0.999 \cdots < 1$

then find a contradiction somewhere. We then conclude that our assumption cannot be be false. Continue reading