# Hit a wall with Analysis, also with a problem in Linear Algebra

1. Sep 22, 2011

### the_morbidus

1. The problem statement, all variables and given/known data
prove that if 0<a<b, then

a < $\sqrt{ab}$ < (a+b)/2 < b

Notice that the inequality $\sqrt{ab}$ $\leq$ (a+b)/2 holds for all a,b $\geq$ 0.
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the linear algebra problem is

Let V be the set of pairs (x,Y) of real numbers and define a modified addition operation

(x,y) + (u,v) = (x+u, 0)

as well as a modified scalar multiplication by c $\in$ R via

c(x,y) = (cx, 0)

Using these two modified operations, is V a vector space? Justify your answer.

2. Relevant equations

3. The attempt at a solution

i tried to solve it by making a = k and b = k+1 so i would know that b's value is definitely greater than a.

$\sqrt{ab}$ = $\sqrt{k(k+1)}$ = $\sqrt{k^2 +k}$ = k + $\sqrt{k}$

now if i use numbers to demonstrate that this is consistent with what i'm suppose to prove, i'll have k = 1 and then a=1 but if i plug it above then k $\sqrt{k}$ = 2
now for the 3rd part (a+b)/2 then (k+k+1)/2 = (2k+1)/2 = k+ 1/2 and if k=1 then this is smaller than the 2nd step and contradicts what i'm trying to prove.

what should I do?

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ok so for the linear algebra problem I was going to introduce another vector U which u and v are part of, and v has a value that is opposite of y to cancel it out, but then that would not explain the (cx,0) part, well y can't be 0 because then y+v = v, unless y=v=0 could work for both, what do you guys think??

Last edited: Sep 22, 2011
2. Sep 22, 2011

### lanedance

this is not true, the last step is invalid (try squaring it and see what you get)

also, please post different problem as different threads otherwise the solutions will get too confusing

3. Sep 22, 2011

### dynamicsolo

Proving that the ends of the first inequality are true isn't too hard... The tougher part is the middle: $\sqrt{ab} \leq (a+b)/2$. Try squaring both sides (Why is this OK? Keep the conditions in mind.) and manipulate the inequality in a form that has to be true...