# Math Help - real number system

1. ## real number system

I would appreciated if someone knows how to do this:

If x and y are arbitrary real numbers with x<y, prove that there is at least one z satisfying x < z < y.

Thanks

2. Originally Posted by RB06
I would appreciated if someone knows how to do this:

If x and y are arbitrary real numbers with x<y, prove that there is at least one z satisfying x < z < y.

Thanks
well, i don't know how rigorous you want to be here, but it seems to me that all we have to do is find a number that satisfies the requirements to be z.

So, let $x$ and $y$ be arbitrary real numbers such that $x < y$. then the distance between $x$ and $y$ is given by $|y - x|$. Take half this distance. thus we have the quantity $\frac {|y - x|}2$ which is greater than zero

Clearly we have $x < x + \frac {|y - x|}2 < y$. thus we can let $z = x + \frac {|y - x|}2$

So we have found one z that satisfies the conditions.

3. $x< \frac{x+y}{2} < y$