Understanding proof that between any two unequal rationals there is a third rational.

Question

I'm taking "Introduction to Mathematical Thinking" on Coursera platform and following proof is given :

Prove that between any two unequal rationals there is a third rational.

Step 1 $$ let \enspace x,y\in Q, x<y $$

then

Step 2 $$ x=\frac{p}{q},y=\frac{r}{s}, where \enspace p,q,r,s \in Z\ $$

then

Step 3 $$ \frac{x+y}{2}=\frac{p/q+r/s}{2}=\frac{\frac{ps+qr}{qs}}{2}=\frac{ps+qr}{2qs}\in Q $$

But

Step 4 $$ x < \frac{x+y}{2}<y $$

Done

Why does $\frac{x+y}{2}$ 'appear' ? What is the logic that causes this $\frac{x+y}{2}$ to be introduced ?

Is it introduced as all $\frac{x+y}{2}$ are rational where x , y is rational ?

I think I'm missing some understanding between step 3 and step 4 as '$x < \frac{x+y}{2}<y$' appears to be introduced without explanation ?

Answer 1

The appearance of $(x+y)/2$ is not something that follows from the previous work; it is not something that was solved for or deduced.

Instead, it's the result of a fit of creativity — the result of understanding some of the basic nature of rational numbers and their computation.

For example, one might arrive at it by the line of thought:

• the midpoint of two rational numbers lies between them
• I can compute the midpoint through elementary arithmetic
• (calculate)
• the midpoint should be $(x+y)/2$

And now that we have a guess as to an explicit number between $x$ and $y$, one produces a proof by verifying it really is a rational number between $x$ and $y$.

I think that, in step 4, the author presumes that it's obvious enough why the inequalities hold that they didn't bother adding any detail. However, given the level of the rest of the proof, I think that's a very bad presumption, and detail should have been given.

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Now, there are lots of fits of inspiration one might have; for example, my previous experience tells me that, if the denominators are positive, then

$$ \frac{p}{r} < \frac{p+q}{r+s} < \frac{q}{s} $$

so I might divine a proof around this approach.

Alternatively, maybe I'm a fan of decimal numbers, and I might devise a construction by finding the first digit where the decimals of the two rational numbers differ, and use that to write down a decimal expansion denoting a rational number between them.

Or, maybe I would think about starting from zero and incrementally taking little tiny steps, expecting that at least one step must lie between the two rational numbers. A proof would then work to figure out how small the steps need to be, and then argue that what is expected really happens.

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The point is, I wanted to convey that there isn't a single right answer here; don't think that your attempts to prove something like this must produce exactly the proof they came up with, or even something resembling it.

The $(x+y)/2$ approach, however, does have the advantage of being one of the simplest proofs to actually carry out. Some of the ones I described above would actually be fairly complicated to actually write out in a rigorous fashion.

Answer 2

If I take any two real numbers $x<y$, then it's always true that $x<\frac{x+y}2<y$.

Now, in the case that $x$ and $y$ are rational numbers, we don't automatically know that $\frac{x+y}2$ is a rational number as well. This is why the author writes the line above to explicitly show that $\frac{x+y}2\in\Bbb Q$. Then on the next line, they return to say that we have indeed found a rational number $z$ (i.e. $z=\frac{x+y}2$) such that $x<z<y$, as we wanted from the beginning.

Answer 3

The proof provides a number, shows that it is a rational number (step 3) and that it lies between $x$ and $y$ (step 4). One could extend step 4: $$ \frac{x+y}{2} = \frac{2x + (y-x)}{2} = x + \underbrace{\frac{y-x}{2}}_{>0} > x \\ \frac{x+y}{2} = \frac{-(y-x) + 2y}{2} = y - \underbrace{\frac{y-x}{2}}_{>0} < y $$

Answer 4

You want to prove that between any rational numbers , say , x and y , there is a nother rational number .

Now : for any two numbers x and y
with x less than y

The number (x+y)/2 lies in

middle point between x and y

So we can take it and show that it is a rational
That way (x+y)/2 is introduced