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I'm attempting to evaluate the truth of the following statement:

∃a∀b((a < b) → (a^2 < b^2)), where a and b are real numbers.

I have tested multiple values (whole numbers and fractions) and have come to the conclusion that the statement is true. However, I "feel" like the statement may not be true and my thinking about evaluating it is not robust enough. Could anyone provide some direction for me?

Huss
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    You are trying to show that there is an $a$ with a certain property. If you can show that (say) $10$ has the property, you will have completely finished. Is it true that if $10\lt b$, then $100\lt b^2$? Sure, on the positive reals, $f(x)=x^2$ is an increasing function. – André Nicolas Apr 01 '13 at 06:49
  • @AndréNicolas I may be misreading the statement, but does this not mean that for every b there will be an a<b and a^2<b^2. If the question was changed to ∃a∃b then your answer would be applicable? I'm not saying your wrong, but that is how I thought the question is to be read. – Huss Apr 01 '13 at 06:56
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    @Huss No. You're switching the quantifiers. The statement you're describing is $\forall b \exists a ( a < b \implies a^2 < b^2 )$. – A.S Apr 01 '13 at 06:59
  • Apart from needing to pay attention to parentheses, logic statements are read, like English, from left to right. The sentence says there is an $a$ such that $\dots$. – André Nicolas Apr 01 '13 at 07:03
  • @AndréNicolas Thanks, I'm still learning. If you wanted to put your comment into an answer I would accept it. – Huss Apr 01 '13 at 07:08
  • This is not a full answer, but have you tried applying induction? It smells like an induction homework/test-question. Just testing it for a few values is obviously not a full proof. –  Apr 01 '13 at 06:40
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    Induction is out of the question: it’s a statement about real numbers, not integers. – Brian M. Scott Apr 01 '13 at 06:40
  • I see. Well, perhaps and perhaps not. This paper seems to allow it in principle: http://math.uga.edu/~pete/realinduction.pdf. Anyway, my mistake. I didn't realize the proposition might be false! –  Apr 01 '13 at 06:55
  • The proposition is true. For the rest, I was assuming that by induction you meant one of the usual varieties, as the suggestion would otherwise be more than a bit misleading. – Brian M. Scott Apr 01 '13 at 07:01

2 Answers2

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HINT: What happens if take $a$ to be $0$?

Brian M. Scott
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We read the assertion for left to right. You are trying to show that there exists a number $a$ which has a certain property. What property? That whatever number $b$ is, if $a\lt b$ then $a^2\lt b^2$.

If you can show that (say) $10$ has the property, you will be completely finished. To show that there exists an $a$ with a certain property, all you need to do is toexhibit an $a$ that has the property. (There may also be indirect ways to show such an $a$ exists.)

So is it true that for any (all) $b$, if $10<b$, then $100<b^2$? Sure, on the positive reals, $f(x)=x^2$ is an increasing function.

Remark: Pick $a=-17$. Then $a$ does not have the required property. For $-17 \lt 2$, but $(-17)^2 \gt 2^2$. So there are "bad" $a$. But the statement just claimed that there is a good $a$.

You might note that we could have picked $a=0$, for $0$ has the desired property. Actually, nothing smaller than $0$ has the property.

André Nicolas
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