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Field Axioms $M4$ and $M5$:

$M4$. There is a unique number $1$ such that $1 \ne 0$ and $(x)*(1)=x$ for all $x \in \mathbb{R}$

$M5$. For each $x \in \mathbb{R}$ with $x\ne 0$, there is a unique number $(\frac{1}{x})$ such that $(x)*(\frac{1}{x})=1$.

Prove if $x\ne 0$, then $(\frac{1}{x})\ne 0$.

proof.

Let $x$ $\ne$ $0$. Suppose $(\frac{1}{x})=0$, then by $M5$, $1=(x)*(\frac{1}{x})=(x)*(0)=0$. This leads to a contradiction. Therefore, $(\frac{1}{x})\ne 0$. QED

Apparently, this is a contradiction of Axiom $M4$. I'm having trouble seeing why this is so.

Pedro
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Skm
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1 Answers1

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Well, you have $1 = 0$ which contradicts that $1\ne 0$.

However, you do have to prove that $x*0 = 0$.

Which can be proven using $0 = 0 + 0$ (Addditive identity axiom. A4???)

So $x*0 = x*(0 + 0) = x*0 + x*0$ by Distributvie axiom.

So by Additive inverse axiom (A5????) $x*0 + (-x*0) = x*0 + x*0 +(-x*0)$

$0 = x*0$.

fleablood
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