A teacher says $x = \frac 1y $ is indeed a function, because it's the same as $y = \frac 1x$ But I don't think so, because no restrictions for $x$ , so $x$ could be zero but there is no solution for zero, so there's an $x$ that has no $y$ My dear teacher says 0 should be out of the domain in the first place Please help me know what's correct?
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Babak's answer is fine, so I won't add one of my own, but I will make a point. Since there is no $y$ that would make $x$ zero, then $x$ cannot be zero. Indeed, $y=\frac1x$ is a function where both $x$ and $y$ can take on all real values except zero, and so is $x=\frac1y$. – Cameron Buie Oct 21 '13 at 12:40
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"$x={1\over y}$" is a pixel heap, unless accompanied by some text indicating the intended interpretation. One interpretation could be that $x$ takes the value ${1\over y}$ whenever $y\ne0$. – Christian Blatter Oct 21 '13 at 12:51
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The relation $y=f(x)$ is indeed a function with respect to $x$ iff $$x_1=x_2\Longrightarrow f(x_1)=f(x_2)$$ Here, you have $xy=1$ or $x=1/y$ or $y=1/x$. If we get the later relation then, it is not hard seeing that $y=1/x$ or $y=f(x)=1/x$ is a well-defined function. Of course, $1/0$ is a forbidden state so, we cannot select $x=0$ to put in $f$. If fact we cannot let $x$ to stand on $0$ on the real line $\mathbb R$. This means that, your dear teacher told you the very right fact. $$0\notin D_{f(x)=1/x}$$
Mikasa
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