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I'm currently starting a number theory book. On its exercise, there's:

Prove: If $a,b \in \mathbb{N}$ and $ab=1$, then $a=1$ and $b=1$.

Here's one proof I just did:

Since $ab=1$, $a=\frac{1}{b}$. But note that if $b \neq 1$, $b > 1$, and thus $\frac{1}{b} \notin \mathbb{N}$, which contradicts $a \in \mathbb{N}$. Similarly for $b$.

Is it safe to assume if $b \neq 1$, $b > 1$, and thus $\frac{1}{b} \notin \mathbb{N}$?


Also, here's another proof:

Since $ab=1$, $\forall c \in \mathbb{N}$, $abc=1c=c$.

Note that since $a \in \mathbb{N}$ and $c \in \mathbb{N}$, $ac \in \mathbb{N}$.

Thus, $\frac{abc}{b} = \frac{c}{b} = ac \in \mathbb{N}$.

Since $\frac{c}{b}$ for all $c \in \mathbb{N}$, $b=1$.

Are those 2 proofs OK? I'm wondering how far can I assume things to be "axioms"?

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3 Answers 3

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It depends. What is ${1 \over b}$ if you know only about $\mathbb N$?

Probably you should use something like $a>b \Rightarrow ac>ac$ for $a,b,c\in \mathbb N$ if this was proved (or assumed) in your book.

If $ab=1$ and $a>1 \Rightarrow 1=ab>b \Rightarrow 1>b$ which is impossible.

If $ab=1$ and $b>1 \Rightarrow 1=ab>a \Rightarrow 1>a$ which is impossible.

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You should probably use monotonicity of multiplication: it is safe to assume $a,b>0$ (otherwise the product would be zero) so $a\leq ab =1$. But since $a\geq 1$ we have $a=1$. Similarly, $b=1$.

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Since $ab=1$, therefore $a$ divides $1$(as $a,b\in \mathbb {N} $). As the only positive factor of $1$ is $1$ itself, therefore $a=1$. Similarly $b=1.$