How would I go about turning english sentences into predicate logic?

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I am taking a Discrete Mathematics for Computer Science course and was having some issues figuring this out. I would rather not be given a specific answer, but the steps on how I could solve this. If you happen to have another example that would be great!

The equation $x^2 + 2x + 1 = 0$ has no solutions over the natural numbers.

Thanks!!!

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To turn a statement expressed in natural language into mathematical language, you have to transform it using a series of steps, at every step only perform a transformation that makes the statement more concise. (obviously ensuring to keep the exact same meaning step after step) Example:

  1. The equation $x^2+2x+1=0$ has no solutions over the natural numbers.
  2. The equation $x^2+2x+1=0$ has no solutions in $\mathbb{N}$.
  3. no $x$ in $\mathbb{N}$ is a solution of $ x^2+2x+1 = 0$
  4. if $x \in \mathbb{N}$ then $ x^2+2x+1 \neq 0$
  5. $(\forall x \in \mathbb{N})(x^2+2x+1 \neq 0)$

another example: $U$ (stack exchange humans set), $E$ (english speaking umans)

  1. (it is false to say that) (no one among us can talk english well.)
  2. $\lnot (\ (\forall p \in U)( p \notin E)\ )$
  3. $(\exists p \in U)(p \in E)$
  4. $U \cap E \neq \phi$

so in the first step i turned the "it is false to say that" into a $\lnot$ and i used sets and quantifiers to transform the rest then it's just algebra.

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I am assuming you are working in a first order language of arithmetic. The desired sentence would be

$$(\forall x)(x^2 + 2x + 1 \neq 0)$$

In your intended model of of arithmetic $\mathbb{N}$, you would have the the above is true in $\mathbb{N}$.

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There are no values $x$ such that $x \in \mathbb N$ and such that $x^2 + 2x + 1 = 0$...

This readily translates into the following translation: $$\lnot \exists x\big((x \in \mathbb N) \land (x^2 + 2x + 1 = 0)\big)$$

Alternatively, we can write $$\forall x\big(x \in \mathbb N \rightarrow x^2 + 2x + 1 \neq 0\big)$$

You can check for yourself that the two statements are equivalent.