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I was wondering if someone could check my proof that "For all $a\in\mathbb{R}$, if $a\neq 0$ then $\frac{1}{a}\neq 0$". The definitions/assumptions I am basing the proof off of come from "Intro to Real Analysis" by Bartle and Sherbert Fourth Edition. I have provided an image of what I am using in my proof:

2.1.1 Algebraic Properties of R

I am using (M3), (M4), and Theorem 2.1.2(c). Here is my proof:

Proof. Let $a\in\mathbb{R}$. Assume $a\neq 0$. Suppose for a contradiction $\dfrac{1}{a}=0$. Then by (M4), $1=a\cdot\dfrac{1}{a}=a\cdot0$. By Theorem 2.1.2(c), $a\cdot0=0$. Thus $1=0$. But by (M3), $1\neq0$, so we have a contradiction. Thus $\dfrac{1}{a}\neq 0$. QED

Is it awkward that I assume $\frac{1}{a}$ to be zero? I ask because if $\frac{1}{a}=0$, it seems that this implies $a=\frac{1}{0}$, which is undefined, so I cannot apply (M4) the way I did in my proof. However I am not sure if this reasoning is correct or my proof is fine as it is.

Any feedback would help. Thanks.

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    $\begingroup$ 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.) $\endgroup$
    – user1318062
    Commented May 9 at 20:48
  • $\begingroup$ 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 $1=0$. $\endgroup$
    – Paul Tanenbaum
    Commented May 9 at 22:08
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    $\begingroup$ If $\frac 1a=0$ then after multiplying by a^2, we find that $a=0$ $\endgroup$
    – hamam_Abdallah
    Commented May 9 at 22:11

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