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From my understanding what the question asks for is for us to prove them by

$$p_1 \to p_2,\quad p_2 \to p_3,\quad p_3 \to p_1$$

But how do we actually do this? The question is as follows:

The following statements are equivalent for all non-negative integers $a$ and $b$:

  • $a < b$
  • $(a + b)^2 < 4b^2$
  • $4a^2 < (a + b)^2$
Kamil Jarosz
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user315656
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  • Equivalent ask for iff condition between statements, what is different of if alone (implication). – Masacroso Feb 18 '16 at 18:24
  • @Masacroso The three implications $p_1 \implies p_2, p_2 \implies p_3, p_3 \implies p_1$ give you equivalence; by transitivity, you can construct the other three implications $p_2 \implies p_1, p_3 \implies p_2, p_1 \implies p_3$. – DylanSp Feb 18 '16 at 18:36
  • This is not what Im talking about @DylanSp. The implications are created by the OP but the real text is about equivalence relations, so is not correct the way the OP represent the text. – Masacroso Feb 18 '16 at 18:38
  • @Masacroso I'm not seeing how equivalence relations are relevant, unless there's something in the source textbook that's not mentioned in the question. – DylanSp Feb 18 '16 at 18:41
  • @DylanSp then ask to the author of the book, not to me. For me the interpretation is clear. I must be more specific: logic equivalence relation. – Masacroso Feb 18 '16 at 18:46
  • @Masacroso I'm still not sure what you're saying. Saying $p_1$, $p_2$, and $p_3$ means $p_1 \iff p_2$, $p_1 \iff p_3$, and $p_2 \iff p_3$. Proving $p_1 \implies p_2$, $p_1 \implies p_3$, and $p_2 \implies p_3$ suffices to prove this, as my earlier comment mentioned. As it happens, we don't need to use this method for this problem (see Jimmy's answer), but the general method is still sound. – DylanSp Feb 18 '16 at 18:52

1 Answers1

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Actually you can show that $p_1\iff p_2$ and $p_1\iff p_3$

  1. $p_1\implies p_2$ $$a<b \implies a+b<b+b \overset{a,b\ge 0}\implies (a+b)^2<(2b)^2 \ $$
  2. $p_2\implies p_1$ $$(a+b)^2<4b^2 \implies \sqrt{(a+b)^2}<\sqrt{4b^2}\overset{a,b\ge0}\implies a+b<2b \implies a<b \ $$

Similarly you can show that $p_1\iff p_3$ which combined (i.e. by transitivity) gives the required equivalence.

Jimmy R.
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  • Proving $p_1\implies p_2 \implies p_3$ is commonly used to prove the equivalence of three statements (so your understanding is not wrong) but here I found that this other method is simpler (worked faster for me at least). – Jimmy R. Feb 18 '16 at 18:56
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    Thank you very much, it is a great and simple breakdown of what I needed. – user315656 Feb 18 '16 at 19:35