I need help setting up the following proof:
Prove that $0 ≤ a < b$ implies $0 ≤ a^2 < b^2$ and $0 ≤ a^{(1/2)} < b^{(1/2)}$.
I am thinking proof by contradiction is the right method, I just dont know how to start it.
Asked
Active
Viewed 1,571 times
0
-
Not necessarily; consider the non trivial part: $a < b$. But $a \ge 0$ and thus we can multiply both terms of the inequality without changing its "direction", i.e. $aa < ab$. And the same with $b \ge 0$ : $ab < bb$. Thus, "joining them": $0 \le aa < ab < bb$. – Mauro ALLEGRANZA Sep 11 '17 at 16:14
2 Answers
0
assuming $$a^2\geq b^2$$ this is equivalent to $$a^2-b^2\geq 0$$ or $$(a-b)(a+b)\geq 0$$ this means $$a\geq b$$ which is a contradiction to $a<b$. in the case of $$a^{1/2}\geq b^{1/2}$$ we get by squaring (all positive) $$a\geq b$$ which contrdicts $$0\le a<b$$
Dr. Sonnhard Graubner
- 95,283
0
You can multiply inequalities memberwise when the terms are non-negative and you can write
$$0\le a<b\implies 0\le a^2<b^2$$ directly.
Then by definition
$$0\le\sqrt a$$
and $$\sqrt a\ge\sqrt b\implies a\ge b,$$ which would be contradictory with $a<b$.