Arithmetic Mean Proof

In this post, we show that if a and b are real numbers and a is less than b, then their arithmetic mean is greater than a. That is, a < b implies

a < \displaystyle \frac{a + b}{2}.

Proof 1: Direct Proof

We know that a < b.

Dividing both sides by 2, we have

\displaystyle \frac{a}{2} < \displaystyle \frac{b}{2}.

Adding \frac{a}{2} to both sides results to 

\frac{a}{2} + \frac{a}{2} < \frac{b}{2} + \frac{a}{2}

which equals

a < \displaystyle \frac{a + b}{2}.

Proof 2: Indirect Proof

Let us assume that for a < b, a \geq \frac{a + b}{2}.

Multiplying both sides by 2 give us 2a \geq a + b.

Subtracting a from both sides results to a \geq b.

This is in contradiction with the assumption above.

This means that the contradiction is false.

Therefore, a < \displaystyle \frac{a + b}{2}.

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