I am learning about predicates and am somewhat a beginner in proofs. I would like your help on proving why statement 2 is false. These are the two statements:

$ \text{statement 1: }(\exists x_1 \in \mathbb{N}, x_1|165) \wedge(\exists x_2 \in \mathbb{N}, 7|x_2)\\ \\ \text{statement 2: }\exists x \in \mathbb{N}, x|165 \wedge 7|x \\ \text{Note: a|b is read as a divides b, and is defined as: } \exists k \in \mathbb{Z}, a.k = b\\ \text{I believe statement 1 is true, since, there exists a natural number that divides 165 (165 itself),}\\ \text{and there exists another number that is divisible by 7 (7 itself). So statement 1 is true.}\\ \text{Statement 2 is false, but I'm not sure why, and how can we prove it to be false?}\\ \text{So I started by trying to prove its negation to be true, i.e:}\\ \forall x \in \mathbb{N}, x \nmid165 \vee 7 \nmid x \\ \Leftrightarrow \forall x\in \mathbb{N} , x|165 \implies 7\nmid x $ Some discussion(using what I've learnt so far about proofs): $\text{Let x be a natural number, and assume x|165 , i.e } \\\exists k1 \in \mathbb{Z},x.k1 = 165 \text{ . Let k1 be this number.} \\ \text{I want to show } 7 \nmid x, \text{i.e } \neg(\exists k_2 \in \mathbb{Z}, 7.k2 = x) \Leftrightarrow (\forall k_2 \in \mathbb{Z}, 7.k_2 \neq x).... \\ \text{Now, how can I use my assumption to reach what I want to show? Thank you for your directions.} $

  • $\begingroup$ Do you have the fundamental theorem of arithmetic? $\endgroup$ – Eric Towers Apr 13 '18 at 23:26
  • $\begingroup$ In the first case, there exists $x_1 = 55$ such that $x_1\mid 165$, and there exists $x_2 = 49$ such that $7\mid x_2$. True. The second statement specifies the existence of some particular x such that both $x\mid 165$, and $7\mid x$. There is no such one $x$. $165 = 3\cdot 5\cdot 11$, which is divisible 1, 3, 5, 11, 15, 33, 55, 165$. None of these divisors is divisible by 7. $\endgroup$ – Namaste Apr 13 '18 at 23:33

We cannot find a number which simultaneously divides 165 and is divided by 7.

Proof Perhaps this is not a particularly elegant proof but the the numbers that divide 165 isn't all that big of a set: {1, 3, 5, 11, 15, 33, 55, 165} and indeed not one of them is a number which 7 divides.

Two proofs for the price of one: If a number was divided by 7 we would know that it is 7x for some x in the naturals. Then we would want this 7x to divide 165 but that would mean that 7 would divide 165. And that's a contradiction.

The first proof is just exhausts all the possibilities.

The second proof requires the fact that when ab divides c this implies that both a divides c and b divides c. I think this fact is the one to exploit for your proof.


Note that

$$7|x \text { and } x|165 \implies 7|165$$

We know that $7$ dose not divide $165$.

Thus statement $(2)$ is false.


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