Proof about real numbers

Question

Question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.

Page 28. Exercise 1. If $x$ and $y$ are arbitrary real numbers with $x<y$, prove that there is at least one real $z$ satisfying $x<z<y$.

Any hints on how to approach the problem would be appreciated, other exercises seem to be similar so if I solve this I should be able to solve others as well. Thank you in advance.

Anirban · Accepted Answer · 2014-10-01T04:59:29.123

3

Consider $z=\frac{x+y}{2}$. Then show that $x \lt z \lt y$.

edited Oct 01 '14 at 04:59

answered Sep 30 '14 at 08:36

Anirban

1,551

If i am not wrong, condition in the question is x<y. – Jasser Sep 30 '14 at 08:46
Yes. No problem with that. Since x<y; so according to the construction of z it is simple to show that x<z<y. – Anirban Sep 30 '14 at 08:51
@user291957 Your point being what? If $x<y$, then this answer constructs a $z$ such that $x<z<y$. – 5xum Sep 30 '14 at 08:52
In your answer you have written that $x \le z$ and $z\le y$ which implies that x can also be equal to y. – Jasser Sep 30 '14 at 08:57

score 2 · Answer 2 · answered Oct 01 '14 at 05:05

It's sufficient to construct a real number that satisfies the condition. The arithmetic, geometric and harmonic means of the two numbers $x$ and $y$ all satisfy the requirement, but the arithmetic mean is the easiest to compute (and has the least potential "complications" with negative numbers and division by zero, etc.) and does the job nicely.

score 1 · Answer 3 · answered Sep 30 '14 at 08:35

1

Hint:

First, try to find the number $z$ on some examples. For example, what is the number between $0$ and $2$? How about between $1$ and $2$? And between $1$ and $\frac{1}2$?

Can you find a pattern of how you find the number in between?

answered Sep 30 '14 at 08:35

5xum

123,496
6
128
204

score 1 · Answer 4 · answered Sep 30 '14 at 08:49

1

Hint : draw a real line and mark two different points such that $x \lt y$. Now how many numbers can you see in between x and y.

answered Sep 30 '14 at 08:49

Jasser

1,976

That would be $y-x$ amount of numbers. So that proves that at any $x,y\in\Bbb R$ where $x<y$, since $y-x>0$, there exists $z\in\Bbb R$ such that $x<z<y$. Right? or is it incomplete? – George Apriashvili Sep 30 '14 at 11:19
There are infinite numbers between x and y. I suspect why in the question it is emphasized to show that there exist atleast one such number. – Jasser Sep 30 '14 at 11:27

Proof about real numbers

4 Answers4