1

$P(x): x ≤ 0,$

$Q(x): $x^2$ = 1,$

$R(x): x $ is odd,

$S(x): x = x + 1.$

Statement:

$∀x ∈ Z, S(x) → R(x) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$

$∃x ∈ Z$ such that $Q(x) ∧ ∼ R(x) \;\;(2)$

$∃x ∈ Z$ such that $P(x) → S(x) \;\;\;\;(3)$

  1. I try to let $x$ in $S(x)$ is even and odd to prove $S(x) → R(x)$, $x = (2n+1) + 1 =2n +2$ this is an even, so this statement is false?

  2. The true set of $Q(x)$ is $\{-1, 1\}$, the set is not satisfactory for $~R(x)$, because $~R(x)$ is an even. Thus this is false.

  3. I have no idea about this one, I find $P(x)\leq 0$, but there no result is satisfactory for $S(x)$, so this is false..?

TomAsh
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  • It would be really helpful if you could set your question in MathJax. Is x 2 supposed to be $x$ squared? If so, write it as $x^2$. – David Aug 20 '18 at 00:33
  • @David Sorry about the negligence, x 2 is x^2. – TomAsh Aug 20 '18 at 00:35
  • So can you change it, and see if you can improve the rest too? Making it difficult for people to help you is not a good idea. – David Aug 20 '18 at 00:36
  • As for answering the question, take a close look at S(x) and see if you can figure out which values of x make it true. – ConMan Aug 20 '18 at 00:41
  • @ConMan This is what I am considering now, I want to find a value from P(x) is satisfactory for S(x), but no value is suitable. – TomAsh Aug 20 '18 at 00:45
  • Assuming you're operating in the natural numbers, integers or real numbers, there is no satisfactory value for $S(x)$. How can a number be equal to one more than itself? – ConMan Aug 20 '18 at 00:46
  • @David I have already change it, thank you for advice, I will be more careful in the future question. – TomAsh Aug 20 '18 at 00:46
  • @TomAsh You can typeset math by enclosing it in dollar signs: for example, $x^2$ gives $x^2$. See the link that David provided for a more extensive guide. – Théophile Aug 20 '18 at 00:47
  • @ConMan So S(x) is false. Ah, thank you! – TomAsh Aug 20 '18 at 00:52
  • @TomAsh Thanks for making the edits. FYI - set of integers $\mathbb{Z}$ can be obtained with \mathbb{Z}. – David Aug 20 '18 at 02:00
  • Additionally, $P(x) \leq 0$ is a meaningless combination of symbols. $P(x)$ is the statement that $x \leq 0$, so it is satisfied by values of $x$ that are less than or equal to 0. But the statement itself is either true or false, and doesn't have a numeric value. – ConMan Aug 20 '18 at 06:30

1 Answers1

1

The first statement, $$∀x ∈ Z, S(x) → R(x)$$ is true because $S(x)$ is always false.

You are correct on the second one.

The third statement is also true. You may pick $x=2$ for example.

Mohammad Riazi-Kermani
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