# Theory Of Numbers question

Suppose we have the equation:

x=1 now, multiplying L.H.S and R.H.S with 'x' I get;

x2=x

This equation has the roots x=0 and x=1 now if I again multiply with 'x' I get;

x3=x2

This equation has the roots x=0 and x=0 and x=1.

I understand that if x is equal to 0 then repeatedly multiplying with x will give me solutions but why can't x be equal to 1?? for if I continue multiplying with 1 my equation remains correct ???

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$0\times x=0\times y$ does not mean that $x=y$... – J. M. Apr 19 '12 at 9:07
Well multiplying by 0 somehow mutitliates the equation – Rohan Apr 19 '12 at 9:09
The title does not reflect the question. – lhf Apr 19 '12 at 10:02

If you have an equality $\rm a=b$ and you apply some transformation to both sides correctly, the resulting equation will be true if the first one was. However, the converse is not true: there are irreversible transformations (or transformations that are only reversible under special conditions), and after applying these the resulting equation will not be logically equivalent to the original one.
Take, for example, $\rm x=1$. If you multiply both sides by $\rm 0$, you get $\rm 0=0$, which is of course true but does not say anything at all about the value of $\rm x$, let alone that it is equal to $\rm 1$. It is often the case that with polynomials and radicals, some of our transformations (like squaring) will introduce more solutions than the equation originally had, and these must be evaluated and discarded in the final analysis.
In your case, going from $\rm x=1$ to $\rm x^2=x$ introduces the false solution $\rm x=0$ (I say it is false because it is clearly incompatible with $\rm x=1$ in any familiar setting), and we must discard it in order to keep our analysis consistent with our original hypothesis.