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I must say that the last time I have done math (badly) at school was 20 years ago, and I have been given this function but even with the answer I am struggling

The anwer is:

$\{(x, y) \in \mathbb Q \times \mathbb Q^+ \mid y = \frac1x\}$ is not a function, because nothing corresponds to a negative x.

I even struggle with the notation... I assume that $\mathbb Q$ is the rational numbers (that inlcudes the negative $\mathbb Q$) and $\mathbb Q^+$ inlcudes the rational numbers but only the positives.

But why ther is nothing to correspond to a negative $x$? I am sure that starts with an intutive example but I lack of, can someone help?

MH.Lee
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Emilia Delizia
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  • So: is there a pair $(x,y)\in \Bbb Q\times\Bbb Q^+$ such that $x<0$ and $y=1/x$? –  Oct 28 '21 at 07:45
  • it still makes little sense to me, because I dont fully understand the notation. does this mean that x belongs to Q and y belongs to q+? – Emilia Delizia Oct 28 '21 at 07:52
  • It would seem that you do understand the notation. –  Oct 28 '21 at 07:53
  • if I substitute 2 to x becuase it has to be a positive rational, I get 2<0 and y =1/2. But 2 is not less than 0, so I deduct that it is not a function, but maybe I am just confusing myself – Emilia Delizia Oct 28 '21 at 08:00
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    The relation - call it $R$ - is defined as the set of pairs $(x,y)$ such that $x$ is rational, $y$ is a positive rational and $y=\frac 1 x$. – Mauro ALLEGRANZA Oct 28 '21 at 08:14
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    Thus, e.g. $(2,1) \notin R$ because $1 \ne \frac 1 2$ and also $(1,2) \notin R$ because $2 \ne \frac 1 1$ – Mauro ALLEGRANZA Oct 28 '21 at 08:16
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    At the same time, no pair with a non-positive first component will be in $R$ because if $x$ is negative we have that also $\frac 1 x$ will be and thus the condition that $y \in \mathbb Q^+$ is not satisfied. – Mauro ALLEGRANZA Oct 28 '21 at 08:18
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    Having said that, we have only one number $q$ that is equal to its reciprocal and it is $1$: $(1,1) \in R$ because $1= \frac 1 1$. In conclusion: $R$ is a function. – Mauro ALLEGRANZA Oct 28 '21 at 08:20
  • some of it makes sense, but it break down quickly, if Q are rational numbers why you are using 1 and 2, is 1 also a rational number ? 1=1/1. Sorry it is silly but it does no make sense for untrained – Emilia Delizia Oct 28 '21 at 08:25
  • Yes, $1$ is a rational number since it can be expressed as the quotient of two integers. – N. F. Taussig Oct 28 '21 at 09:11

1 Answers1

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For case where $x=0$: Because $\lim_{x\rightarrow 0^+}{\frac{1}{x}} = \infty$, the pair $(x,1/x)$ does not have a value when $x=0$, so the list of pairs $\{(x,y) \in Q\times Q^+| y=1/x\}$ is not a function because each value of $x$ does not have corresponding $y$ value ($\in Q^+$). QED, it's not a function.

For case where $x<0$: the $y=1/x$ would fall outside of $Q^+$ because y is negative, thus it's not a function of $Q \rightarrow Q^+$

For case where $x>0$: the $y=1/x$ would go inside $Q^+$. This alone cannot make it a function when the other cases claim otherwise.

tp1
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  • You have to evaluate whether or not there is a pair $(x,y)\in \Bbb Q\times \Bbb Q^+$ such that $x=0\land y=1/x$. The fact $\lim_{x\to 0^+}\frac1x=\infty$ is completely irrelevant and misleading. –  Oct 28 '21 at 08:15
  • The actual rule that says something is a function is based on "whether there is exactly one value for f(x) for each possible value of x". In this case it fails because 1/x is going outside of $Q^+$ range, and then it requires knowing that a list of pairs is another representation for a function when $(x,f(x))$. – tp1 Oct 28 '21 at 08:22
  • No, no, a hundred times no. It fails because $x$ is not in the domain. We could agree right now that the expression $1/0$ means $37$, and suddenly we'd have the pair $(0,37)$ in that set. Still, $\lim_{x\to 0^+}\frac1x =\infty$ would be true. You have to argue by using the defining property for what it means. –  Oct 28 '21 at 08:25
  • what you mean by $x$ is not in the domain? I would think $0 \in Q$ which would mean that x=0 is in the domain. – tp1 Oct 28 '21 at 08:33
  • I guess you call active domain what I call domain. –  Oct 28 '21 at 08:36
  • But you're right that the lim is kinda strange with $Q$, given that $\frac{a}{b}$ makes the lim jump from one rational to another in unpredictable manner. – tp1 Oct 28 '21 at 08:49
  • No, I've never said that the limit was "strange": the limit is fine in $\Bbb Q$. The point is that what $\lim_{x\to 0^+}\frac1x$ is has nothing to do, on a material level, with whether or not there is some $y\in\Bbb Q^+$ such that $(0,y)\in {(x,y)\in\Bbb Q\times\Bbb Q^+,:, y=1/x}$. –  Oct 28 '21 at 08:57