Quantifier Proofs

**shilz222** · August 18th 2007, 09:28 PM

Are these quantifier proofs correct? You just have to take counterexamples to disprove, and examples to prove right?

**red_dog** · August 18th 2007, 11:41 PM

(i) Let $x\in\mathbf{R}$ . Take $y=-x+1\Rightarrow x+y=1>0$ , so the statement is true.
(ii) Let $x\in\mathbf{R}$ . Take $y=x-1\Rightarrow x-y=1>0$ , so the statement is true.
(iii) For $y=x-1\Rightarrow x+y=-1<0$ , so the statement is false.
(iv) For $x=0\Rightarrow 0\cdot y=0,\forall y\in\mathbf{R}$ , so the statement is false.
(v) If $x=0\Rightarrow 0\cdot y=0,\forall y\in\mathbf{R}$ .
If $x\neq 0$ , take $y=-x\Rightarrow xy=-x^2<0$ , so the statement is false.
(vi) Let $x\in\mathbf{R}$ . Take $y=x\Rightarrow xy=x^2\geq 0$ , so the statement is true.
(vii) Take $x=0\Rightarrow 0\cdot y=0\geq 0,\forall y\in\mathbf{R}$ , so the statement is true.
(viii) Take $y=-x$ and the statement is true.
(ix) $(x+y>0) \wedge (x+y=0)$ is false, so the statement is false.
(x) $(\forall x\in\mathbf{R}, \exists y\in\mathbf{R},x+y>0)$ is true (see (i)).
$(\forall x\in\mathbf{R},\exists y\in\mathbf{R},x+y=0)$ is true (take $y=-x$ ), so the statement is true.

**shilz222** · August 19th 2007, 08:13 AM

so the way i do it (choose numbers) is wrong?

**ThePerfectHacker** · August 19th 2007, 08:14 AM

Originally Posted by shilz222

so the way i do it (choose numbers) is wrong?

Yes. Look at how the Dog did it it. He pick any number and should you can find a corresponding number for that one. You picked a specific number and picked a corresponing number.

**shilz222** · August 20th 2007, 09:48 AM

For (3) do you mean $y = -x-1$ ?

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