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?

  • $\begingroup$ 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. $\endgroup$ – user70390 Apr 1 '13 at 6:40
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    $\begingroup$ Induction is out of the question: it’s a statement about real numbers, not integers. $\endgroup$ – Brian M. Scott Apr 1 '13 at 6:40
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    $\begingroup$ 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. $\endgroup$ – André Nicolas Apr 1 '13 at 6:49
  • $\begingroup$ I see. Well, perhaps and perhaps not. This paper seems to allow it in principle: math.uga.edu/~pete/realinduction.pdf. Anyway, my mistake. I didn't realize the proposition might be false! $\endgroup$ – user70390 Apr 1 '13 at 6:55
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    $\begingroup$ @Huss No. You're switching the quantifiers. The statement you're describing is $\forall b \exists a ( a < b \implies a^2 < b^2 )$. $\endgroup$ – A.S Apr 1 '13 at 6:59

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.


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


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