# Which propositions are true?

I'm practicing exam questions for the exam in january and would like to know if my reasoning is sound and the answers correct. I've formatted the correct answers in bold.

Which propositions are true? Recall that the symbol | means “divides”

• Svar 1.a: ∀n ∈ Z: 2n > n + 2 : wrong as 2*-2 is not greater than -2 + 2 = 0.

• Svar 1.b: ∃n ∈ Z: 2 | (3n + 1) : 2 does divide n = 3 in the expression 2 | (3n + 1)

• Svar 1.c: ∃k ∈ Z: ∀n ∈ Z: n = kn : Taking k = 0 makes the expression true.

• Svar 1.d: ∃k ∈ Z: ∀n ∈ Z: 2 | (n + k) : No because adding 1 would turn an even n into an uneven which 2 does not divide etc.
• Svar 1.e: ∀n ∈ Z: ∀k ∈ Z: (n > k ∨ k ≥ n) : Is false for all n.

## 1 Answer

For part $$b$$, you mean $$2$$ does divide $$3n+1$$ if $$n=3$$.

For part $$c$$, you mean taking $$k=1$$.

For part $$d$$. Suppose $$k$$ is odd, then if $$n$$ is even, then $$n+k$$ is odd, hence $$n$$ can't be odd. Similarly, if $$k$$ is even, then if $$n$$ is odd, then $$n+k$$ is odd. Hence no such $$k$$ exists.

For part $$e$$. We have $$n> k$$ or $$n \le k$$. It is true.

• "For part b, you mean 2 does divide 3n+1 if n=3." - Yes. "For part c, you mean taking k=1." - Right, yes. It can't be 0. "For part d. Suppose k is odd, then if n is even, then n+k is odd, hence n can't be odd. Similarly, if k is even, then if n is odd, then n+k is odd. Hence no such k exists." - Exactly what I was trying to communicate. Thanks. "For part e. We have n>k or n≤k. It is true." - Right. Because a proposition actually says that "It is true that n>k or n≤k for all n and all k."? – Travitrinket Sep 22 '18 at 14:41
• yup, I think so . – Siong Thye Goh Sep 22 '18 at 14:45
• I did my best formatting it. :) Makes sense. Thanks! – Travitrinket Sep 22 '18 at 14:46