Timeline for Proof that for all nonzero real numbers $a$, $\frac{1}{a}$ is nonzero

7 events

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Jul 9 at 9:28	comment	added	Peter		The reason is that for real numbers $a,b$ and $b\ne 0$ , we have $\frac{a}{b}=0\implies a=0$ by simply multiplying the equation with $b$. By division by $b$ , we get also the opposite direction and finally have $\frac{a}{b}=0\iff a=0$
Jul 9 at 9:26	comment	added	Peter		This simple fact can be extended and used to find the roots of some function of the form $\frac{f(x)}{g(x)}$. We only have to find out the roots of $f(x)$ and eliminate those who are also roots of $g(x)$. In particular , in the case of rational functions , this is very useful.
May 9 at 22:11	comment	added	hamam_Abdallah		If $\frac 1a=0$ then after multiplying by a^2, we find that $a=0$
May 9 at 22:08	comment	added	Paul Tanenbaum		Your proof is fine, and your concern is unfounded because it amounts to another (very similar) proof of your result. That’s because with $a\neq 0$, you can multiply both sides of any equation by $a$ and equality still holds. So from an assumption that $1/a=0$ it would follow that that $1=0$.
May 9 at 20:48	comment	added	user1318062		It is perfectly fine. The important part is that $1/a$ has a meaning, which it does since $a\neq 0$. So maybe note that as well somewhere. (Didn't check the image but some axiom will tell you that it exists.)
S May 9 at 20:42	review	First questions
May 9 at 21:11
S May 9 at 20:42	history	asked	user1320946	CC BY-SA 4.0