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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"?

Thanks :D

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Division is probably not a good way to go, since you are almost certain not to have even defined it for natural numbers. Do you have the notion of order for natural numbers? What have you proven about them? We cannot tell you what you can assume as axioms, because it depends on what your presentation has been. –  Arturo Magidin Mar 15 '12 at 17:23
    
This is eerily reminiscent of yesterday's "In which of following stuctures is valid implication". –  Henning Makholm Mar 15 '12 at 17:27
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Over $\mathbb{N}$: if $ab = 1,$ then both $a$ and $b$ are units. What are the units in $\mathbb{N}$? –  user2468 Mar 15 '12 at 17:35
    
Is the distributive law available? –  Mark Bennet Mar 15 '12 at 19:17

2 Answers 2

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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