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Prove or disprove the claim, and prove or disprove the converse:

Claim 1:

∀n ∈ ℕ, (Ǝk ∈ ℕ, n = 5k + 2) ⇒ (Ǝj ∈ ℕ, n^2 = 5j + 4)

Claim 2:

∀m,n ∈ ℕ, (Ǝk ∈ ℕ, m = 7k + 3) ∧ (Ǝj ∈ ℕ, n = 7j + 4) ⇒ (Ǝi ∈ ℕ, mn = 7i + 5)

I know how to prove claim 1 since it is fairly simple but I'm stuck on claim 2 as well as how to prove both their converse. I've tried several approaches but don't seem to reach anywhere. I do know both claims are true and both converses are false.

yus_m
  • 93

2 Answers2

1

$\forall m,n \in \mathbb{N}, \exists k \in \mathbb{N}, m = 7k + 3, \exists j \in \mathbb{N}, n = 7j + 4$ then

$mn=(7k+3)(7j+4)=7(7kj)+7(3j+4k)+12=7(7kj+3j+4k+1)+5$

let $i=(7kj+3j+4k+1) \in \mathbb{N}$

For converse, you can just give a counterexample.

For claim 1, let $n=3$, then $n^2=5+4$, but no $k \in \mathbb{N}$ fulfills the requirement.

For claim 2, let $m=4, n=3$, then $mn=7+5$, then use similar argument.

John
  • 13,204
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by hypothesis you have that $\exists k \in \mathbb{N}, m = 7k + 3$ and $\exists j \in \mathbb{N}, m = 7j + 4$

So if you multiply side by side , you will have that

$$m\cdot n = (7k + 3)( 7j + 4) = 49kj+ 28k +21j+12 =49kj+ 28k +21j+7 + 5 $$

then you take out the 7 that they have in common and have this:

$$m \cdot n = 7(7kj+4k+3j + 1) +5 $$

let $i = (7kj+4k+3j + 1) $ with $i\in \mathbb{N}$

So from here you have that $\exists i \in \mathbb{N}, m \cdot n = 7i +5$ and its done.

Jearson Narvaez Rojas
  • 629