If a and b are integers with $a < b$, then the average of a and b is greater than a and less than b

Question

I need a direct proof for "If a and b are integers with $a < b$, then the average of a and b is greater than a and less than b." I have so far:

Hypothesis: A and B are integers and $A < B$

Conclusion: $ A < \frac{A + B}{2} < B $

But I'm not really sure where to go from there. Thank you!

@RobArthan I tried $2A<A+B<2B$ but I didn't see the next step working backwards from that. — GiantDuck, Mar 28 '17 at 22:23
Do you know that if $X<Y$ and $Z>0$ then $XZ<YZ$? Choose $Z=\frac{1}{2}$. — Thomas Andrews, Mar 28 '17 at 22:25
"I didn't see the next step working backwards from that" ...Dividing by two — JMoravitz, Mar 28 '17 at 22:25

score 2 · Accepted Answer · answered Mar 28 '17 at 22:25

2

$b>a \implies b+a>a+a \implies \frac{b+a}{2} > \frac{a+a}{2} = a $

Also we have $b>a \implies b+b > a+b \implies b>\frac{a+b}{2}$

Then we have $a<\frac{a+b}{2}<b$ as required

answered Mar 28 '17 at 22:25

mrnovice

score 1 · Answer 2 · answered Mar 28 '17 at 22:25

1

$A < B \implies A + A < A + B < B + B \implies 2A < A+B < 2B \implies A < \frac {A+B}2 < B$.

answered Mar 28 '17 at 22:25

fleablood

2 Answers2