i would think u a genuis if u can zsolve this proof or disprove..?

Question

If there are two integers a and b such that (b/2) < a < b, then a³ > b².

can u explain it also?? i don't understand it.

Answer 1

This can be disproved by counter example.

Let
a = 2
b = 3
b/2 = 3/2

So b/2 < a < b

a³ = 2³ = 8
b² = 3² = 9

a³ < b²

So the statement if false.

Answer 2

b is always bigger than a

but the thing is that when A is cubed, or A * A * A, it is more than B squared, or B * B. That is why it is showing that B divided by 2 is smaller than A, and A is smaller than B just normally. So technically A and B are close answers to any coordinates on the coordinate plane

Answer 3

(b/2) < a < b, then a³ > b². False

We already know that
b > a from the first statement and a, b are integers.
if we take cubes on both sides
b^3 > a^3 for positive integers.

So a^3 > b^3 is false