Let a>0 and b>0. Then prove that a<b↔a^2<b^2↔√a<√b.

Answer 1

GIVEN: A< B

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if a < b then a^2 = a * a < a * b < b * b = b^2, so a^2 < b^2

sqrt(b) - sqrt(a) > sqrt(a) - sqrt(a) = 0

so sqrt(b) > sqrt(a)

GIVEN a^2 < b^2

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By contradiction, suppose a>b.

0 < b^2 - a^2 = (b+a)(b-a)

So either (b+a) and (b-a) are both positive OR

(b+a) and (b-a) are both negative

In the former case, b+a>0 ---> b >0> -a and b-a > 0 ---> b >a which is a contradiction..

In the latter case , b+a<0 ---> b < -a which is a contradiction, since they are both positive

Therefore a< b

As previously shown, sqrt(b) - sqrt(a) > sqrt(a) - sqrt(a) = 0 , so sqrt(b) > sqrt(a).

GIVEN: sqrt(a) < sqrt(b)

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Again by contradiction, suppose a>b.

then sqrt(a) - sqrt(b) > sqrt(b) - sqrt(b) = 0, so sqrt(a) > sqrt(b) which is a contradiction.

Therefore a<b.

Again, as previously shown. a^2 = a *a < a*b < b*b = b^2, so a^2 < b^2

;[end of proof]