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Show that $x-1=0$ has a solution in $\mathbb{N}$.

It seems too easy. Just use the induction process? I know $x$ is $1$, which is a natural number. May be that is all we have do that.

Also show that $x+1=0$ does not have a solution in $\mathbb{N}$.

$x=-1$, which is not a natural number.

I think my professor wants a formal proof of these two problems. Therefore, any help is appreciated. Thank you.

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    Do you know what additive inverses are? Does $\Bbb N$ have any? – 2017-01-24
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    Yes, -a+a=0 is an example of additive inverse. – 2017-01-24
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    Note that $n + 1 > 0$ for all $n \in \mathbb{N}$. – 2017-01-24
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    Therefore my 1st math induction step should be find an x where x+1>0 for all x$\in$N. – 2017-01-24

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For the first one, it is sufficient to note that $1$ is a solution - I don't know how rigorous your definition of "$-$" is, so I can't say whether you might need to prove that $1$ is a solution. If $a - b$ is just defined as "$a$ minus $b$", then just stating the example should be fine.

For the second one, $x = -1$ is not sufficient evidence - for example, $x = -1$ is also a solution to $x^2 = 1$, but that doesn't mean $x^2 = 1$ doesn't have any solutions in $\mathbb{N}$. You should prove by induction that $x + 1 \neq 0$ for all $x \in \mathbb{N}$.

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    To prove by math induction, I know I need to find an x$\in$N so that x+1=0. But there is no such number. So, I am already stuck at the first step of the induction. – 2017-01-24
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    @lily To prove by induction you first show that 0 is not a solution, and then that if $n$ is not a solution, $n+1$ is also a solution. – 2017-01-24
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    Step 1: Let x=1, then 1+1$\neq$0. Thus, x$\neq$1. – 2017-01-24
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    Step 2: let x=n+1 for n$\in$N. Then n+1+1$\neq$0, which is the same as n+2$\neq$0. We know this statement is true because n+2 is greater than 0. There, I have proved this statement. – 2017-01-24