Thread: 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

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 and be arbitrary real numbers such that . then the distance between and is given by . Take half this distance. thus we have the quantity which is greater than zero

Clearly we have . thus we can let

So we have found one z that satisfies the conditions.