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the_morbidus
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Homework Statement
prove that if 0<a<b, then
a < [itex]\sqrt{ab}[/itex] < (a+b)/2 < b
Notice that the inequality [itex]\sqrt{ab}[/itex] [itex]\leq[/itex] (a+b)/2 holds for all a,b [itex]\geq[/itex] 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 [itex]\in[/itex] R via
c(x,y) = (cx, 0)
Using these two modified operations, is V a vector space? Justify your answer.
Homework Equations
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.
[itex]\sqrt{ab}[/itex] = [itex]\sqrt{k(k+1)}[/itex] = [itex]\sqrt{k^2 +k}[/itex] = k + [itex]\sqrt{k}[/itex]
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 [itex]\sqrt{k}[/itex] = 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??
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