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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2$\begingroup$ Suppose that $f(a,b)=f(c,d)$. Then $a+b\sqrt{11}=c+d\sqrt{11}$. Can you prove that $a=c$ and $b=d$? $\endgroup$– Ian ColeyDec 10, 2013 at 9:34
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2 Answers
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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.
answered Dec 10, 2013 at 9:40
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$\begingroup$ some care is due here: it may be that $\;(a,b)\neq (x,y)\;$ yet $\;b=y\;$ and thus you cannot divide by $\;b-y\;$ ... $\endgroup$ Dec 10, 2013 at 9:41
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1$\begingroup$ If $b=y$, then also $a=x$, so we can assume $b\ne y$. $\endgroup$– egregDec 10, 2013 at 9:56
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$\begingroup$ Exactly my point, @egreg : some care is due here. If $\;b=y\;$ and $\;a+b\sqrt{11}=x+y\sqrt{11}\;$ then...etc. $\endgroup$ Dec 10, 2013 at 9:59
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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 $v$?
answered Dec 10, 2013 at 9:43