# Proof by Contradiction: 0.999… = 1

In the previous post, we proved that $\displaystyle a < \frac{a + b}{2}$

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

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

Theorem $0.999 \cdots = 1$.

Proof

We assume that $0.999 \cdots < 1$, then by the previous theorem $0.999 \cdots < \displaystyle \frac{0.999 + 1}{2}$

which means that $0.999 \cdots < \displaystyle \frac{1.999 \cdots}{2}$.

By long division, the right hand side of the inequality is also equal to $0.999 \cdots$ which results to $0.999 \cdots < 0.999 \cdots$

which is impossible.

This means that our assumption is false. Therefore, $0.999 \cdots \geq 1$.  Clearly, $0.999 \cdots > 1$ is false, so $0.999 \cdots = 1$.