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DaniyalAlwani
chetan2u

x < y, then we can square both sides
x^2 < y^2

We can then rewrite it as x^2 - y^2 < 0 which is a third statement.

Can you please explain why this is not "must true" statement?

What if x is a very small negative integer?
Say x=-100 and y=1....x<y but x^2>y^2 as 10000>1
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you can square (AKA multiplying both sides by something) both sides of a inequality IF and ONLY IF you know that you are multiplying both sides by a positive number (or at least you must know the sign of what you are multiplying for).


Instead, notice how x^2 - y^2 < 0
can be transformed into the number I

x^2 - y^2 < 0
x^2 < y^2
radq(x^2) < radq(y^2)
abs(x) < abs(y) --> abs(x) is SMALLER than abs(y)

it is the same as the first one, and the first one is not valid because if x and y are both negative and x<y --> than we know by definition that in that case abs(x) > abs(y). example, x = -4, y = -3
x < y --> and abs(x) is GREATER than abs(y)
DaniyalAlwani
chetan2u

x < y, then we can square both sides
x^2 < y^2

We can then rewrite it as x^2 - y^2 < 0 which is a third statement.

Can you please explain why this is not "must true" statement?
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