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Determine the truth value of each statement, assuming that x and y are real numbers, and justify your answer.

  1. $\forall$x, $\exists$y such that xy=1
  2. $\exists$y such that $\forall$x , xy=1

I understand the first problem to be true. Since the equation can be "solved" in terms of y, I found that y was a reciprocal of x. So now xy=1 is true. The second problem confuses me. The books says its truth value is false. The only reasoning I have for that answer is the question is the reverse of the first. Can anyone help me understand why the second question is false?

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  • $\begingroup$ The first statement is false if $x=0$. $\endgroup$ – quasi Aug 27 '17 at 15:45
  • $\begingroup$ The first one is "given a real, pick it's inverse" the second one is "pick an inverse for all reals" which is impossible. The ordering of the quantifiers is the difficulty here. $\endgroup$ – Daniel Gratzer Aug 27 '17 at 15:45
  • $\begingroup$ Ok! Now I understand. Thank you. $\endgroup$ – ErinA Aug 27 '17 at 15:48
  • $\begingroup$ The second is problematic, when x = 0, also. There is no y, such that, when x = 0, $xy= 1$. $\endgroup$ – amWhy Aug 27 '17 at 15:51
  • $\begingroup$ Even if we exclude $0$ from the domain of x, y, There is no specific y, such that $\forall x$, $xy = 1$. $\endgroup$ – amWhy Aug 27 '17 at 15:57

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