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Normally in algebra, you can simplify an equation such as $a=a*b$ by dividing both sides by a common coefficient. For example, $\frac {a}{a}=\frac {ab}{a} \rightarrow1=b$. Obviously $b = 1$ is a correct solution to the equation, but clearly also is $a = 0$. Both solutions can be found inuitively, but it occurs to me that there should be a procedure that yields both, and I don't even know one that yields $0$ at all.

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$a-ab=0 \Rightarrow a(1-b)=0$ – pedja Nov 24 '11 at 10:05
When you divided by $a$ you implicitly assumed $a\ne 0$. It remains to consider the case $a=0$. – Bill Dubuque Nov 24 '11 at 17:06
up vote 7 down vote accepted

$$ a = a \ast b \Rightarrow a \cdot (1-b) =0$$ which says either $b=1$ or $a=0$ because when the product of $2$ numbers is $0$, then either one of them has to be $0$.

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This only happens in an Integral domain. – Hassan Muhammad Nov 24 '11 at 19:56
Note the algebra-precalculus tag - in this setting, we can assume integral domains. – Johannes Kloos Sep 19 '12 at 9:19

This question has a good application of the zero product property.$$\begin{aligned}a &=ab & \\ a - a b &= 0\\a(1 - b) & = 0 \\ \therefore a = 0 \vee b = 1\end{aligned} $$

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$a=0$ OR $b=1$, not necessarily both at the same time. – fretty Sep 19 '12 at 8:58
Good point... editing now. – Parth Kohli Sep 19 '12 at 9:06

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