Is there a rational number between any two irrationals?

Question in the title.

I feel intuitively the answer should be yes, but can't really seem to prove this... Any help?

Oh ok thanks guys

Yes

Edit : as Noble said

Edit : even more what Noble said

Yes, all you have to do is look at the digits of both irrational numbers until there is a difference between two digits, truncate the irrationals after this digit (so you have a rational number) add the two rational numbers together, divide by two and you've got a rational number between the two irrationals.

So for example, suppose you've got the two following irrational numbers $\alpha,\beta$ and wlog write them as

$\alpha = 0.a_{1}a_{2}a_{3}...$

$\beta = 0.b_{1}b_{2}b_{3}...$

Where the $a_i, b_i$ are the digits.

The reason we can say they're between 0 and 1 is because it doesn't matter what the digits are to the left of the decimal point, it's just extra notation if we add them in.

Let's say that $a_i = b_i$ for $1 \leq i \leq n-1$, so what we're saying is there is a difference between the two irrationals at the digit in place $a_n$

So denote the rational numbers

$\alpha' = 0.a_{1}a_{2}a_{3}...a_{n}$

$\beta' = 0.b_{1}b_{2}b_{3}...b_n$

Then the number $\dfrac{\alpha' + \beta'}{2}$ is also rational, but note that

$\dfrac{\alpha' + \beta'}{2} = \dfrac{0.a_{1}a_{2}a_{3}...a_{n} + 0.b_{1}b_{2}b_{3}...b_n}{2}$

This number will have the first $n-2$ digits identical to the digits of $\alpha',\beta'$ but the last two digits will be such that $a < \dfrac{\alpha' + \beta'}{2} < b$

Why does $i$ have to be bounded? If $i$ was infinite could an argument be made that $\alpha$ and $\beta$ are still possibly different numbers?

I think you accidentally negged me, with that god awful -8 reputation power.

Either that or you're getting your own back for me finding your facebook picture the other day

Balls

Let me see if I can redress

xx

Haha don't worry about it, I've got plenty of rep.

You call a +6 "plenty"

Hah



















[obviously I have more cos i am soooooooo much older]

Haha don't worry about it, I've got plenty of rep.