Proof: For all real numbers $a$ and $b$, if $a < b$ then $a < \frac{a+b}{2} < b$.

Question

Suppose $a$ and $b$ are real numbers in which $a < b$. By the definition of midpoint, $a <m < b$ for some real number $m$ where $m$ is the midpoint. Then:

Case 1:
$a<b$
$a+a<b+a$
$2a<b+a$
$\frac{2a}{2}<\frac{b+a}{2}$
$a<\frac{b+a}{2} $

Case 2:
$a<b$
$b+a<b+b$
$b+a<2b$
$\frac{b+a}{2}<\frac{2b}{2}$
$\frac{b+a}{2}<b$

Note that $a<\frac{a+b}{2}$ and $\frac{a+b}{2} <b$, therefore $a <\frac{a+b}{2}< b$ is valid.

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The problem that I have is whether or not the definition of midpoint is necessary.

By the definition of midpoint, a <m < b for some real number m where m is the midpoint.

And if there are missing definition i Should add to the proof.

Usually, when writing a direct proof, I want to know the definition of the concept I am working with. This is easy to do when writing the definition of odd, even, rational numbers, etc. The problem I always face is not knowing what definition to use when writing about concepts that don't have one specific definition.

Answer 1

I do think it is necessary to define the midpoint of $a$ and $b$. All your proof shows is that the number $\frac{a+b}{2}$ is between $a$ and $b$. But there are other numbers between $a$ and $b$. For example, $\frac{2a+b}{3}$ is between $a$ and $b$, as I'm sure you can prove.

So there does arise a question: Were you only asked to prove that the number $\frac{a+b}{2}$ lies between $a$ and $b$? Or were you instead asked to prove that the midpoint of $a$ and $b$ lies between $a$ and $b$?

The key point here is that if your proof does use the concept of the midpoint of the interval $[a,b]$, then midpoint should be defined: it is a number $m$ having the property that the distance between $a$ and $m$ is equal to the distance between $m$ and $b$. Here I am using the ordinary distance formula along the real number line, where the distance between $x$ and $y$ is $|x-y|$.

If we adopt that definition, and if your desire is to prove that the midpoint is between $a$ and $b$, then your proof is not yet complete: having proved that $\frac{a+b}{2}$ is between $a$ and $b$, you must still prove that the distance between $a$ and $\frac{a+b}{2}$ is equal to the distance between $\frac{a+b}{2}$ and $b$. In other words, you must prove the equation $\left| \frac{a+b}{2} - a \right| = \left| b - \frac{a+b}{2} \right|$

Answer 2

Yeah you don't need that definition; you never even used it.

Also don't separate your working into cases like that, that means you're dealing with mutually exclusive scenarios, like:

• $a>0$ and $b>0$
• $a>0$ and $b<0$
• $a<0$ and $b>0$
• $a<0$ and $b<0$

Answer 3

$$a\lt b\implies a+a\lt a+b\lt b+b\implies\frac{a+a}{2}\lt\frac{a+b}{2}\lt\frac{b+b}{2}\implies a\lt\frac{a+b}{2}\lt b$$