Chapter 1.2, Problem 85E

Solution

To prove: the graph of every odd function with domain all real numbers must pass through the origin.

Let $f$ represent an odd function defined for all real $x$ . Because $f$ is defined for all real $x$ , then $0$ is in the domain of $x$ . That is, there exists some $y$ in the range of $f$ such that $y=f\left(0\right)$ .

As $f$ is odd, then it must be true that

  $f\left(-x\right)-f\left(x\right)$

Let $x=0$ . 'Then

  $f\left(-0\right)=-f\left(0\right)$ and therefore

  $f\left(0\right)=-f\left(0\right)$

By the multiplication property of negative one,

  $f\left(0\right)=-1\cdot f\left(0\right)$

And by the multiplication property of zero,

  $f\left(0\right)=0$

from which conclude that $\left(0,0\right)$ is a point in the graph of $f$ , which means that that graph passes through the origin.