Question

Please help me prove:
If x does not equal 0, then x^2 > 0

Answer 1

let x=1 which is greater then 0.
\[x^{2} = 1^{2} = 1\]

Now let x = -1 which is less then 0.
\[x^{2} = -1^{2} = 1\]
here in both case i.e greater then and less then 0 the result is greater then 0.

Answer 2

x does not equal 0 so x>0 (i) or x<0 (ii)

(i) x>0 multiply both sides by x

x.x>0.x=0 ---> x^2>0

(ii) x<0 multiply both sides by x (note that x is a negative so < will change to >)

x<0
x.x>0.x=0 ---> x^2>0
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now suppose x^2>0 we must show x\(\neq\)0

x^2>0 take square root\[\sqrt{x^2}>0\]\[|x|>0\]so x>0 or x<0 and x\(\neq\)0