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Define a graph $\phi$ as;

$\phi(x)=|x|, \forall x\in [-1,1]$ and $\phi(x+2)=\phi(x)$.

I'm trying to prove that $\phi$ is a function, but it seems something is wrong in my argument.

It's not hard to see that "$\forall x\in \mathbb{R}, \exists y\in[0,1]$ such that $(x,y)\in \phi$"

I'm trying to prove that $x_1=x_2 \Rightarrow \phi(x_1)=\phi(x_2)$ and here's my argument below.

============= Suppose $x_1=x_2$.

Then there exists a unique $n\in\mathbb{Z}$ such that $n_i≦\frac{x_i + 1}{2}<n_i + 1$, and $n_1=n_2$.

Thus $\phi(x_1)=|x_1 - 2n_1|=\phi(x_2)$. Q.E.D

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I'm sure something's wrong in my argument, but don't know what it is..

(This argument works even when $\phi(1)≠1$, so something's wrong here..)

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Every real number is congruent to a unique real number in $(-1,1]$. (Note the open parenthesis on the left).

So if $\phi$ had been defined by $\phi(x)=|x|$ on $(-1,1]$, and $\phi(x+2)=\phi(x)$, there would be no issue whatsoever. (In principle one would need an easy induction to show that $\phi(x+2n)=\phi(x)$.)

But because $\phi(x)$ was defined as $|x|$ on the closed interval $[-1,1]$, we need to check that $\phi(x)$ is well-defined at odd integers.

The issue is that the definition, for example, simultaneously defines $\phi(5)$ as $\phi(-1)$ and as $\phi(1)$. But since $|-1|=|1|$, there is no problem.

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  • $\begingroup$ Would you explain what is wrong in my argument please? $\endgroup$
    – Katlus
    Dec 14, 2012 at 16:36
  • $\begingroup$ In principle there is a missing induction, which is also missing from mine, too trivial. Your assertion that $n_1=n_2$ is incorrect. We can think of $5$ as $-1+6$ or as $1+4$. $\endgroup$ Dec 14, 2012 at 16:45
  • $\begingroup$ It seems there's nothing wrong with $n_1=n_2$ since i assumed $x_1=x_2$. I just realized that taking $\phi$ as a function at the last step is incorrect,which is assuming what i was going to prove.. $\endgroup$
    – Katlus
    Dec 14, 2012 at 16:54

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