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Consider the function $f: \mathbb N$ × $\mathbb N$ → $\mathbb R$, $f(a,b) = a+b \sqrt{11}$

How do I show this function is an injection (one to one)?

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Suppose that $f(a,b)=f(c,d)$. Then $a+b\sqrt{11}=c+d\sqrt{11}$. Can you prove that $a=c$ and $b=d$? –  Ian Coley Dec 10 '13 at 9:34
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2 Answers

up vote 5 down vote accepted

Suppose $(a,b) \neq (x,y)$ but $f(a,b) = f(x,y)$, then

$$ a + b\sqrt{11} = x + y \sqrt{11} \implies (b - y) \sqrt{11} = x - a \implies \sqrt{11} = \frac{x-a}{b-y} \in \mathbb{Q}$$

contradiction, so $(a,b) = (x,y) $.

Notice, we can assume $b \neq y$, otherwise we would have $a = x$. So the division by $b -y$ is allowed.

Notice, also: any $f: \mathbb{N} \times \mathbb{N} \to \mathbb{R} $ such that

$f(a, b) = a + b \sqrt{p} $ where $p$ is prime is always injective.

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some care is due here: it may be that $\;(a,b)\neq (x,y)\;$ yet $\;b=y\;$ and thus you cannot divide by $\;b-y\;$ ... –  DonAntonio Dec 10 '13 at 9:41
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If $b=y$, then also $a=x$, so we can assume $b\ne y$. –  egreg Dec 10 '13 at 9:56
    
Exactly my point, @egreg : some care is due here. If $\;b=y\;$ and $\;a+b\sqrt{11}=x+y\sqrt{11}\;$ then...etc. –  DonAntonio Dec 10 '13 at 9:59
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I corrected my solution Don! –  Euler Dec 10 '13 at 10:06
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One way showing that some function $f:A\to B$ is injective one, is by showing that

$$\forall x_1,x_2\in A: f(x_1)=f(x_2) \rightarrow x_1=x_2 $$.

In your question, $f:\mathbb{N}^2\to\mathbb{R}$ given by:

$$f((a,b))=a+b\sqrt{11}$$

$f$ is indeed injective:

Suppose that $f((x,y))=f((u,v))$, then $x+t\sqrt{11}=u+v\sqrt{11}$.

The last, we could write as:

$$(x-u)+\sqrt{11}(t-v)=0$$

Remember that $x,y,u,v\in\mathbb{N}$, what can you say about $x$ and $u$; and about $t$ and $v$?

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