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An odd function is symmetrical in the 1st and 3rd quadrants. Does this means that it always passes through the origin?

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Continuity is a key issue here. – David H Aug 9 '14 at 15:57
But I heard somewhere that it is not sure that it will always pass through origin. – user168467 Aug 9 '14 at 15:59
I might be making a fool of myself here, but isn't $y = 1/x$ an odd function? – Niet the Dark Absol Aug 9 '14 at 17:08
@DavidH Why so? Do you consider the discontinuous (though right-continuous) function $f(x)=1$ for $x\ge 0$ and $f(x)=-1$ for $x<0$ an odd function? The function $\mathrm{sign}(x)$ is not continuous, but it is odd and its graph contains the origin. – Jeppe Stig Nielsen Aug 9 '14 at 22:06
@DavidH No, it is not. See André Nicolas's answer for a proof that whenever $0$ is in the function's domain $f(0)=0$, which does not need or mention continuity at all. – Federico Poloni Aug 10 '14 at 10:58

As André Nicolas showed, under your conditions and if $f(0)$ exists, $f(0)=0$. However, nothing in your question implies that $f(0)$ must exist.

If you let $f(x)=\frac1x$ then $f$ is a symmetrical odd function, its graph is in quadrants I and III, but $f(0)$ is undefined.

So, you can say "$f(0)$ is either $0$ or undefined." Or, if you want to stick to terminology about graphs, "the graph of $f$ either passes through the origin or it does not intersect the $y$-axis at all."

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I think your conclusion is too strong. I have in mind some variation on $\sin \frac{1}{x}$ which is odd, discontinuous, and passes $0$ infinitely often. – nomen Aug 10 '14 at 3:34
@nomen: But $\sin \frac{1}{x}$ never intersects the $y$ axis. (It's undefined at $x=0$.) The statement holds. And if you have an equation where $y$ has multiple values at $x=0$, what you have by definition isn't a function of $x$. – cHao Aug 10 '14 at 4:41
You're right, I misunderstood. – nomen Aug 10 '14 at 4:57

Let $f(x)$ be odd, and defined at $x=0$. Then $f(-x)=-f(x)$ for all $x$. In particular, $f(-0)=-f(0)$, so $f(0)=-f(0)$. It follows that $f(0)=0$. The "curve" $y=f(x)$ passes through the origin.

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Well, it follows that the function equals 0 at the origin, but to say it "passes" through the origin is misleading language in my opinion. What if $f(x)=\operatorname{sgn}x$ for $x\neq 0$? – David H Aug 9 '14 at 16:03
What if the function is not defined around the origin? It can still be odd can't it? – I like Serena Aug 9 '14 at 16:16
@IlikeSerena: It certainly can be. Often unless a specification to the contrary is made, it is assumed that a function is defined everywhere. Thank you for your comment, I have made the requirement that $f$ is defined at $0$ explicit. – André Nicolas Aug 9 '14 at 16:21
I certainly don't mind the "nitpicking," which I would rather call being precise, and which is part of our DNA as mathematicians. "Passes through the origin" is an informal term whose formal definition, if one were given, perhaps should include the condition of being defined in an interval, and should perhaps include some smoothness condition. But I have not seen a formal definition of "passes through $P$," and took the broadest possible interpretation. – André Nicolas Aug 9 '14 at 16:35
The answer explicitly assumes that $f$ is defined at $0$. So probably "wrong" is not quite right, though "incomplete" certainly would be. – André Nicolas Aug 9 '14 at 18:32

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