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The domain for x, y, z is real numbers.

i) $\forall x \exists y (y^2<x)$ FALSE counterexample: $x=0$

ii) $\forall x \exists y (y^3<x)$ TRUE

iii) $\forall x\exists y \forall z ((y>0) \land ((z^2<y) \rightarrow (z^2+1<x^4)))$ (Not sure how exactly x,y,z relate to each other but from my current understanding it is False) Update: It's false because the second part of the if-then statement, $(z^2+1<x^4)$ does not always hold true for all $x$ and for all $z$ in the reals.

iv) $\exists y \forall x (y^2<x)$ FALSE counterexample: all $x \lt 0$

please let me know if my answers are right or wrong, if wrong, please provide an explanation. For iii) please provide an explanation in either case.

Thanks!

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    $\begingroup$ For number 1, let x =0. Then what y of the reals is less than zero when it's squared? x= 0 is thus a counter example. $\endgroup$ – James Bender Oct 14 '15 at 1:20
  • $\begingroup$ @JamesBender Thanks! I have edited the question to reflect the updated answer. Please let me know whether my other answers are correct. $\endgroup$ – shoestringfries Oct 14 '15 at 1:27
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    $\begingroup$ Two is true. If x=0 take y=-1. $\endgroup$ – Jean-François Gagnon Oct 14 '15 at 1:30
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    $\begingroup$ @shoestringfries Why did you change your answer to number 2? Given x = 0, let y = -1. Certainly -1 cubed is less than zero, so 0 is not a counter example in this case. See the logic? Try this: Whenever you see a for all quantifier, pretend someone has given you some number from the reals and whenever you see exists you need to find some number that relates to the given. And re-think your answers to ii and iv. $\endgroup$ – James Bender Oct 14 '15 at 1:31
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    $\begingroup$ Surely (iv) is false. Let $x=-17$. $\endgroup$ – André Nicolas Oct 14 '15 at 1:33
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i) false — correct; ii) true – correct.

iii) says: for every $x$, there's $y > 0$ such that $$\forall z (z^2 < y \to z^2 + 1 < x^4) \tag{*} $$ Let $x = 1/2$, the statement claims there is a $yy > 0$ such that (*) is true. But then for some/any $z$ with $z^2 < y$ (and there are such $z$), we would have $1 \le z^2 + 1 < \frac 1 {16}$. So iii) is false.

iv) is false, and your counterexample works.

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