1
$\begingroup$

I'm learning how to determine the truth value of statements and I want to make sure that i'm understanding and answering the questions correctly. I'm struggling with determining if i'm reading the statements correctly. I'm reading $\forall$x $\exists$y as "for all x there exists a y". Is this correct? Are my answers correct? (my answers are the italics and the problem sets are to the left)

Domain: $\mathbb R$ (all real numbers)

a) ∀x∃y(x^2 = y) = True (for any x^2 there is a y that exists)

b) ∀x∃y(x = y^2) = False (x is negative no real number can be negative^2

c) ∃x∀y(xy=0) = True (x = 0 all y will create product of 0)

d) ∀x(x≠0 → ∃y(xy=1)) = True (x != 0 makes the statement valid in the domain of all real numbers)

e) ∃x∀y(y≠0 → xy=1) = False (no single x value that satisfies equation for all y

f) ∃x∃y(x+2y=2 ∧ 2x+4y=5) = False (doubling value through doubling variable coefficients without doubling sum value)

g) ∀x∃y(x+y=2 ∧ 2x−y=1) = False (really unsure about this one)

$\endgroup$
  • $\begingroup$ Not "there is a y that exists" but "there is a y" or "exists a y". $\endgroup$ – Mauro ALLEGRANZA Feb 10 '17 at 15:05
  • $\begingroup$ @MauroALLEGRANZA got it. That makes more sense "exists a y". Do my answers conform with that logic? $\endgroup$ – StormsEdge Feb 10 '17 at 15:12
  • $\begingroup$ @MauroALLEGRANZA and thank you for your help! $\endgroup$ – StormsEdge Feb 10 '17 at 15:12
  • $\begingroup$ g) consider it as a system of equations; adding them we get : $3x=3$ that is not true for every $x$. Proof: consider $x=2$; then we must have some $y$ such that $2+y=2$ and $4-y=1$, which is impossible. Conclusion: FALSE. $\endgroup$ – Mauro ALLEGRANZA Feb 10 '17 at 15:12
  • $\begingroup$ Other than slight roughness in the wording (which I expect will improve as you read more mathematics), the thinking appears to be good. For g), you just need one counterexample. An easy one is $x=0.$ (In fact you only needed one counterexample for b), although you found infinitely many.) $\endgroup$ – David K Feb 10 '17 at 15:14
0
$\begingroup$

$a) True$

$b) False$

$c) True$

$d) True$

$e) False$

$f) False$

$g) False$

Basically you got them all right. Regarding $g)$, you can either find an $x$ for which the statement isn't true or you solve the equation, obtaining specific (static, say) values for $x$ and $y$, thus proving it's falsity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.