I am working on the Solow book about how to do proofs.

Consider the problem of proving that, “If x and y are real numbers that $x^2 +6y^2 = 25$ and $y^2 +x = 3$, then $y = 2$.” In working forward the hypothesis, which of the following is not valid? Explain.

a) $y^2 = 3−x$

=> a) is considered valid. But why?

In my opinion a) is not valid because the hypothesis doesn't imply a)
Hypothesis: $x^2 +6y^2 = 25$ and $y^2 +x = 3$
Counter example:
$x^2 +6y^2 = 25 \implies x = \sqrt{(25-6y^2)}$
$y^2 +x = 3 \implies x = 3-y^2$
$\sqrt{(25-6y^2)} \neq 3-y^2$

  • $\begingroup$ How do you they're different for all values of $y$? $\endgroup$
    – Bernard
    Apr 19, 2018 at 20:07
  • $\begingroup$ I thought I only need to show one counter example (y=0). But then I would only show that a) doesn't imply the hypothesis. But that was not asked. I messed up the direction. $\endgroup$
    – siva
    Apr 19, 2018 at 20:13

1 Answer 1


The hypotesis is

  • $x^2 +6y^2 = 25$ and $y^2 +x = 3$

then $y^2 +x = 3$ is true.

From the hypotesis we obtain

  • $x^2 +6(3-x) = 25 \iff x^2-6x-7=0 \implies x=-1,7 \implies y=\pm 2,y=\pm 2i$

and then the statement “If x and y are real numbers that $x^2 +6y^2 = 25$ and $y^2 +x = 3$, then $y = 2$” is not true, indeed

  • $P\implies Q$

is equivalent to

  • $\neg Q \implies \neg P$

which is false.

  • $\begingroup$ In your second part, you have proofed the proposition to be wrong but I think that was part of the question. Anyway I think the main gist is that you can always imply something more narrow from a given statement, right? $\endgroup$
    – siva
    Apr 19, 2018 at 20:18
  • $\begingroup$ @siva the point is that if for $y\neq 2$ (Q false) we obtain that P is true then the implication is false. $\endgroup$
    – user
    Apr 19, 2018 at 20:29
  • $\begingroup$ I get that point. But I think the question doesn't care about $y \neg 2$. "In working forward the hypothesis which of the following is not valid". I should only verify if the hypothesis implies answer a. $\endgroup$
    – siva
    Apr 19, 2018 at 20:41
  • $\begingroup$ @siva ah ok! then it is true of course as already noted. $\endgroup$
    – user
    Apr 19, 2018 at 20:43

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