Proving X + Y / 2 is Between X & Y in R

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Problem:

"Suppose X, Y in R (Real Numbers), X < Y prove there exists Z in R such that X < Z < R."

I'm currently trying to prove that X + Y / 2 satisfies this but I'm getting stuck. I first show that X + Y / 2 cannot be = to either X or Y. I then try to show that X + Y / 2 is > X since X < Y but I cannot seem to tie this in. Any help would be appreciated.
 
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Nvm I think I got it.

Since X < Y
Y - X must be in Positive Reals
2 is in Positive Reals thus 1/2 is in Positive Reals
Y - X / 2 is thus in Positive Reals
You can then rearrange such that you can the equation

X + Y / 2 - X in positive reals
Thus X < X + Y / 2.

Right?
 
Dunno about that (you ought to bracket things up), but if x<y why don't you just add something to both sides?
 
matt grime said:
(you ought to bracket things up)
In case you missed it, matt's point is: x+ y/2 is NOT between x and y:
(x+ y)/2 IS!
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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