# Please check my proof I used a “proof by contradiction” and have no idea that if it's good.

I have a proof to make in my homework and I wanted to ask if this proof is a good one as I used a logical explanation which I never used before. Please let me know if such thing works , if not please explain why and how should I prove...

$Let(\mathbb{F}\cdot+)$ be a field and $a,b,c,d\in\mathbb{F}$

Need to prove that if $a\cdot b=0_{\mathbb{F}}$ then $a=0_{\mathbb{F}}$ or $b=0_{\mathbb{F}}$

suppose that $a\cdot b=0_{\mathbb{F}}$ and $a\neq0_{\mathbb{F}}$ $\land$ $b\neq0_{\mathbb{F}}$

then $a\cdot b+a\cdot b=0_{\mathbb{F}}$ $\rightarrow$ $2(ab)=0_{\mathbb{F}}\rightarrow2=0$ we reached a contradiction, the statement $a\neq0_{\mathbb{F}}$ $\land$ $b\neq0_{\mathbb{F}}$ is false.

which means that the oppsite logical statement $a=0_{\mathbb{F}} \lor b=0_{\mathbb{F}}$

must be true

You cannot assume that if $2(ab)=0$ that $2=0$. This is only valid if $ab\neq 0$, but you don't know that.

To maybe shed more light about what is wrong with your proof, it should ring an alarm when you notice that the only thing you really used was the fact that $ab=0$. However, there are clearly pairs of $a,b$ such that $ab=0$ in actual fields, like $\mathbb R$. Just because $ab=0$, that cannot lead you to a contradiction.

• Thank you very much !!! Do you have any slight hint/idea on how this proof should be done then ? I am very puzzles.. – Pavel Penshin Feb 23 '16 at 12:57
• @user313448 It depends on what your definition of a field is. For example, if your definition includes that $\mathbb F\setminus \{0\}$ must be a multiplicative group, then the fact that $a\neq 0\land b\neq 0\implies ab\neq 0$ is a direct consequence of the fact that a group is closed under its operation – 5xum Feb 23 '16 at 13:36