let x>2, now write proof for (x^3)>8, i was supposed to derive the answer from answering a former question, let x>2, now write proof for (x^2)>4. The answer was, since x>2, x is real and x >0. by multiply by x, (x^2)>2x, since 2>0, by multiplying 2, 2x>4. Through transitivity ->, (x^2)>4. but i cant derive answer. help tks
2<x so of course x>0 2^3<x^3 --> 8<x^3
i need to write proper proof.
this is a proper proof assume x>2 then by transitivity x>0 since x>0 x>2 iff x^2>4 iff x^3>8 qed.
since x>0, x>2 iff x^2>4 iff x^3>8? then how do i proof x^2>4 and x^3>8? if just proven, isn't it only vacuously true?
you have proven that x^2>4 that is x^3>4x, you agree? then, since x>2, 4x>8, thus, x^3>8. You see, nice and easy!
omg, how can i not think of this? thank you!
could you please take a look at the question I just posted? thx!
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