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For a problem such as:$$\forall k\in\mathbb{N},\exists j\in \mathbb{Z},k=j+1.$$I can immediately see that this is false as it does not hold for all $k$, for instance when $j=-10$, $k$ becomes negative and that is not a natural number. However how do I write like a proper explanation on my answers if this was to be answered in an exam?

commie trivial
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Aurora Borealis
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    You are misreading the claim. In English, it says "For every natural number $k$ we can find an integer $j$ such that $k=j+1$". Indeed, just take $j=k-1$. The example you give would be a counterexample for the claim $\forall j\in \mathbb Z,,\exists k\in \mathbb N$ such that $k=j+1$" but that's a very different claim. – lulu Oct 23 '20 at 10:14
  • As to the logic, a single counterexample suffices to disprove a claim. So, for example, your example instantly disproves the claim I wrote in my prior comment. As mentioned, though, it does not disprove the claim you wrote. – lulu Oct 23 '20 at 10:20
  • So technically speaking, the above statement is true because for every $k$, we can find a $j$ such that the statement holds true. But how do we show that mathematically? – Aurora Borealis Oct 23 '20 at 10:57
  • Not "technically speaking" . It's just plain true. To prove it, take $j=k-1$. I don't know how you have defined the integers, but I assume you have shown (or assumed) that the integers are closed under subtraction. – lulu Oct 23 '20 at 12:28

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Lets assume $k \in \mathbb{N}$. We know that $\forall k \in \mathbb{N}, k - 1 \in \mathbb{Z}$, so:

Take $j = k - 1$

$$ k = j + 1 \iff k = (k-1) + 1 \iff k = k - 1 + 1 \iff k = k $$

By induction (seems overkill though), take $j = k - 1 \in \mathbb{Z}$.

Base case, $k=0$:

$$ 0 = (0 - 1) + 1 \iff 0 = 0 $$

Inductive step:

Assuming $k = n$, and that $n = (n - 1) + 1$ holds, we want to check if $n + 1 = ((n+1) -1) + 1$ holds

$$ n + 1 = ((n+1) -1) + 1 \iff n + 1 = (n + 1 -1) + 1 \iff n + 1 = n + 1 $$

Since both the base case and the inductive step are true, then the original statement holds for every natural number $k$.

fire-bee
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