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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.

Firstly, is my proof correct?

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

FD_bfa
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Csoriburi
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    You proved the wrong implication. You proved that if either $\frac{a+b}{2}>a$ or $\frac{a+b}{2}>b$ then $a\ne b$. You're asked to prove the converse implication. – jjagmath May 10 '23 at 18:22
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    It seems to me that OP put forth effort, so I don't know why there are so many down/close votes – J. W. Tanner May 10 '23 at 18:46
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    You don't need any case distinction. $\frac{a+b}{2}>a$ is equivalent to $b>a$ and $\frac{a+b}{2}>b$ to $a>b$. So your title claim to prove is: "If $a,b$ are two distinct real numbers, then either $b>a$ or $a>b$". You should be able to prove this. – Dietrich Burde May 10 '23 at 19:17
  • [tag:solution-verification] should not be the only tag for a question – J. W. Tanner May 10 '23 at 19:26

3 Answers3

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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 $a$.

shark
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    Yes the problem asked here is probably trivial for those of you with 0.01% rankings postdocs, but this student has given some efforts to the problem, by cases which is not the fastest solution. Here I have suggested 3 approaches to the problem 1) geometric 2) contradiction 3) direct proof which would help the student see how to approach this and similar problems in the future. Whoever downvoted my solution just wants the student left stuck and don't see how one might approach a problem in different ways. – shark May 10 '23 at 18:51
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    @AnotherUser Why is this a low quality question? – jjagmath May 10 '23 at 19:10
  • @jjagmath I have changed my mind (and I have retracted my closing vote). But the question is very poorly written. – Another User May 10 '23 at 19:16
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    @AnotherUser Why is poorly written? The OP used MathJax (which is more than many users do), wrote his/her proof, which by the way is a correct proof (of the wrong implication, but that's not the point) and asked feedback about it. The only issue I see is that the OP wrote the question in the title but not in the body of the question. – jjagmath May 10 '23 at 19:20
  • @jjagmath That's a big issue, in my opinion. And it contains two question, not just one. – Another User May 10 '23 at 19:41
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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.")

J. W. Tanner
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Stephen Donovan
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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.

ConMan
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