I am allowed to use Prime$(x)$ and Even$(x)$, and quantifiers. I just wanted to make sure if I'm on the right track.

  1. There is no greatest number: $$\forall x \exists y(x<y)$$

  2. Any number added to itself is even: $$\forall x \mathrm{Even}(x+x)$$

  3. Every even number is the sum of two primes numbers: $$\forall x (\mathrm{Even}(x)) ⟹ \exists y \exists z(\mathrm{Prime}(y) \land \mathrm{Prime}(z) \land x = y+z)$$

  4. No square number is prime: $$\lnot \exists x \mathrm{Prime}(x \cdot x)$$

  5. The result of multiplying an odd number by itself is always odd.

I have no exact idea of how to approach this one. Would it be "for all the $x$'s that are not even, the result is not even when $x$ is multiplied by itself"?

  • $\begingroup$ What exactly do you mean when you say $\forall x ⟶ R$? $\endgroup$
    – Nika
    Nov 24 '19 at 22:06
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    – Caffeine
    Nov 24 '19 at 22:18
  • $\begingroup$ I meant real numbers. I should have used the symbol ℝ ! $\endgroup$ Nov 24 '19 at 22:30
  • $\begingroup$ Ah, ok. I assumed by $R$ you meant $\mathbb{R}$. But $\forall x ⟶ \mathbb{R}$ doesn't quite mean 'for all real numbers'. Typically the way to say that is $\forall x \in \mathbb{R}$, does that seem familiar? $\endgroup$
    – Nika
    Nov 24 '19 at 22:36

The consequent in ($3$) has a free variable, namely $x$. This can be fixed by moving the parenthesis at the end of the antecedent to the end of the entire expression, so:

$$\forall x (\mathrm{Even}(x)) ⟹ \exists y \exists z(\mathrm{Prime}(y) \land \mathrm{Prime}(z) \land x = y+z)$$


$$\forall x (\mathrm{Even}(x) ⟹ \exists y \exists z(\mathrm{Prime}(y) \land \mathrm{Prime}(z) \land x = y+z))$$ So that the universal quantifier can quantify over the whole expression.

For ($5$), the way you phrase it in English makes it somewhat unclear what you mean for it to say. Particularly with the word 'result'. Try:

$$\forall x (\lnot \mathrm{Even}(x) \implies \lnot \mathrm{Even}(x \cdot x))$$


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