Original post by boiali5
Show that if 0 < a < b then a^2 < b^2. If a^2 < b^2, is it necessarily true that a < b? Give an
example to illustrate your answer.
example to illustrate your answer.
What have you done so far?
To expand a bit, we want to know what you can or can't use.
For instance, I doubt we can use the fact that f(x)=x^2 is strictly increasing on the positive reals. But if we were allowed to use it, there is nothing to show.
For instance, I doubt we can use the fact that f(x)=x^2 is strictly increasing on the positive reals. But if we were allowed to use it, there is nothing to show.
a=n, b=n+1 ; n E R+
a2 = b2=
hence a2<b2
I prefer a sketch of the x2 graph though.
a2 = b2=
hence a2<b2
I prefer a sketch of the x2 graph though.
first part is just that, strictly increasing, therefore true. second part, since we are no longer restricted to the R+, think of ways that the inequality fails.
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