If $a,b$ are two distinct real numbers, then either $\frac{a+b}{2}>a$ or $\frac{a+b}{2}>b$

Question

Prove: If $a,b$ are two distinct real numbers, then either $\frac{a+b}{2}>a$ or $\frac{a+b}{2}>b$

Here is my attempt:

Case 1: Multiplying both sides by $2$ and subtracting $b$ gives us: $a>b$.

Case 2: Repeating the same argument but now subtracting $a$ we get: $a<b$. Since $a,b$ are two distinct numbers one of them is greater than the other one.

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Firstly, is my proof correct?

Secondly, could the proof be done in a way that doesn't involve cases?

Answer 1

Let's try to visualize this $a+b\over2$ is simply the midpoint between the points $a,b$ on the real line. Then is it true that this midpoint must be greater than the smaller of $a$ and $b$?

If you insist on an algebraic proof, then suppose both inequality signs reverses direction, then try to derive a contraction.

Alternatively you can assume $a<b$ and prove directly that $a+b\over 2 $ is greater than $a$.

Answer 2

Your proof is definitely along the right lines; if we're being precise though I think that it should be made clear that your manipulations of the two inequalities are equivalences, rather than implications.

So $\frac{a + b}{2} > a$ if and only if $b > a,$ and $\frac{a + b}{2} > b$ if and only if $a > b.$ By constructive dilemma, this tells us that $\frac{a + b}{2} > a$ or $\frac{a + b}{2} > b$ if and only if $a > b$ or $b > a,$ and because we know that the second part has to be true when $a$ and $b$ are distinct, the first part does as well. That last part, going backwards from $a > b$ or $b > a$ to our conclusion, is why the fact that the equivalences go both ways is important.

As far as other ways to proceed, I agree with the other answer that a proof by contradiction is the way to go: for some distinct real numbers $a$ and $b$ to be a counterexample, we must have both $\frac{a + b}{2} < a$ and $\frac{a + b}{2} < b,$ right? So if we simply add inequalities (I'll leave the formal details of this to you) we'll get $a + b < a + b,$ which surely can't be true, so our counterexample cannot exist, meaning our original statement must be true. (I like to say this is because it "can't not be true.")

Answer 3

Your proof is kind of the wrong way around - you need to prove that the two possibilities given cover all possible cases, so to do it more properly you would do something like this:

Since $a$ and $b$ are distinct, either $a > b$ or $b > a$.

If $a > b$, then $a + b > b + b = 2b$ and so $\frac{a + b}{2} > b$. If $b > a$, then $a + b > a + a = 2b$ and so $\frac{a + b}{2} > a$.

The way you wrote it is fine for your own scratch work, which is a useful step in working towards a proper proof but is generally not great for writing proofs properly, because the proof needs to follow a clear stream of logic.

As for alternatives to a case-by-case proof, one option is to take advantage of a trick of logic - the statement "$A$ or $B$" is equivalent to "not $A$ implies $B$" because you can break "$A$ or $B$" into two cases - either $A$ is true (in which case $B$ doesn't matter), or $A$ is false and $B$ is true. So if we assume $A$ to be false and prove that it requires $B$ to be true, we've proven the original statement.

In this case, it would look something like this:

Assume $\frac{a + b}{2} \ngtr a$, so $\frac{a + b}{2} \leq a$. Then $a + b \leq 2a$, and so $b \leq a$. But $a$ and $b$ are distinct, so $b < a$. Then $a + b > b + b = 2b$ and hence $\frac{a + b}{2} > b$.

Is that a neater proof than the one by cases? Personally I think the first proof is tidier, but this can be a useful technique for other proofs.