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Prove: $ 0 \le a \lt b$ implies $ 0 \le a^2 \lt b^2 $ and $0 \le \sqrt{a^3} \lt \sqrt{b^3}$.

Now show that the statement is false if the hypothesis $a \ge 0$ or $a \lt 0$ is removed.

EDIT: Someone mentioned I should give context. I know how to disprove the statement. I am stuck on how to prove it. I do not know how to prove directly, but after an attempt on using the contradiction, it seemed to be the way to go. I'm not sure what to do and am seeking verification.

Daniel W. Farlow
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user210592
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  • What thoughts do you have on this matter? We do not like to give answers without some work or thought shown. – Cameron L. Williams Jan 26 '15 at 02:42
  • Let $a=-3$, and $b=1$. – André Nicolas Jan 26 '15 at 02:43
  • I know how to prove it when it is false. I just use a counterexample like Andre mentioned, but originally used a= -1 and b = 1. – user210592 Jan 26 '15 at 03:14
  • However, when I attempt to prove it. I try to use a contradiction but it hasn't worked out so well for me. – user210592 Jan 26 '15 at 03:14
  • @user210592 So you know how to disprove it when either of the conditions $a\geq 0$ or $a<0$ are removed? All you need is a direct proof? – Daniel W. Farlow Jan 26 '15 at 03:30
  • @induktio Yes. I know how to disprove it. But how do I prove it? A direct proof didn't work when I tried, are you saying I should try that method again? The contradiction seemed to be a better option, but I am confused so.. Lol. – user210592 Jan 26 '15 at 03:35
  • @user210592 I just provided a direct proof of the implication(s) you made. If you do not understand my proof(s), then I am not sure how much else I or others can help. – Daniel W. Farlow Jan 26 '15 at 03:38
  • @induktio I thought it would have been more complicated? I was trying to use the theorems and definitions the book gave for the chapter. Thank you for the very simple break down. – user210592 Jan 26 '15 at 03:40
  • @user210592 Glad it makes sense now :) – Daniel W. Farlow Jan 26 '15 at 03:41

2 Answers2

1

Step 1: Multiply $a<b$ on both sides by the nonnegative value $a$.

Step 2: Multiply $a<b$ on boht sides by the nonnegative value $b$.

Step 3: Combine the results of the previous two steps.

vadim123
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1

You said you know how to disprove it when the mentioned conditions are removed; thus, I am going to show you how to do the direct proof(s) since you mentioned you do not know how or can't do these.

For $0\leq a < b\longrightarrow 0\leq a^2 < b^2$, we have the following: \begin{align} 0\leq a < b &\Longleftrightarrow 0\leq a < b\tag{given}\\[0.5em] &\Longleftrightarrow 0\leq a^2 < b^2\tag{square all quantities} \end{align} This is the first direct proof.

For $0\leq a < b\longrightarrow 0\leq \sqrt{a^3} < \sqrt{b^3}$, we have the following: \begin{align} 0\leq a <b &\Longleftrightarrow 0\leq a < b\tag{given}\\[0.5em] &\Longleftrightarrow 0\leq a^3 < b^3\tag{cube everything}\\[0.5em] &\Longleftrightarrow 0\leq \sqrt{a^3} <\sqrt{b^3}\tag{"root everything"} \end{align}

Together, these direct proofs fulfill the implication(s) you wanted to make when the conditions are true. Since you know how to do the other part, I think this is a full answer to your question I hope.

Daniel W. Farlow
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