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Is there an easy way to justify: $$x(x-1)(x+1) \equiv x(x^2-1) \Rightarrow (x-1)(x+1) \equiv x^2-1,$$ even for $x=0$? I seemingly have to divide by $x$ which should place the restriction $x \neq 0$ on the final result. Does this work only for polynomials?

EDIT: thank you for the comments, in light of the suggestions to do case work I'll update with a more involved example to demonstrate why I am not looking for this approach. I'm sorry to move the goal posts a bit, let me know if this should be a new question.

Let's suppose that $w=e^{2\pi i/n}$ where $n$ is an integer. Let's say I've deduced that $$(z-1)(z-w)(z-w^2)...(z-w^{n-1}) \equiv (z-1)(1+z+z^2+...+z^{n-1}).$$

I want to conclude here that $(z-w)(z-w^2)...(z-w^{n-1}) \equiv 1+z+z^2+...+z^{n-1}$ including $z=1$ - it's not easy to verify by cases anymore since I am actually trying to use this factorisation to show that $(1-w)(1-w^2)...(1-w^{n-1})=n$.

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    $\begingroup$ What do you mean by $\equiv$? $\endgroup$ – José Carlos Santos May 27 '19 at 22:55
  • $\begingroup$ Equal for all complex values of $x$ let's say. $\endgroup$ – SquareCircle May 27 '19 at 22:56
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    $\begingroup$ You don't need the implication since both equalities are demonstrably true by the usual properties of multiplication/addition/subtraction $\endgroup$ – Henry May 27 '19 at 22:57
  • $\begingroup$ If $\equiv$ means identically equal, then you could first deduce the result for $x\neq0$ and then invoke the theorem that two polynomials that agree at infinitely many inputs are identically equal. Another theorem that would do the job is that a quadratic polynomial is determined by its values at any three inputs. $\endgroup$ – Andreas Blass May 27 '19 at 22:58
  • $\begingroup$ Ok, to check the implication, you have to check that $(x-1)(x+1)= x^2-1$ for any real $x$. You divide it in two cases. The first one: $x=0$. In this case it is trivial that $(0-1)(0+1)= 0^2-1$. The second one: x$\neq0$. In this case, you use the equality $x(x-1)(x+1) = x(x^2-1)$ and divide by $x$ (which you are supposing not zero) and conclude that $(x-1)(x+1)= x^2-1$. $\endgroup$ – André Porto May 27 '19 at 23:00
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This is a consequence of the fact that a nonzero polynomial whose coefficients are complex numbers (or in any integral domain) has no more roots than its degree. This quickly yields what we seek, viz.

Theorem $ $ If $\,f,g,h\,$ are polynomials with coefficients in $\Bbb C$ (or any infinite field C) and $\,f\neq 0\,$ then

$$\begin{align} f(x) g(x) &= f(x) h(x)\ \ \text{for all }\, x\in C\\ \Rightarrow\ \ g(x) &= h(x)\qquad\ \ \text{for all }\, x\in C\end{align}$$

Proof $\ $ Since $\,f\neq 0\,$ it has only finitely many roots (at most $\deg f).\,$ Thus there are infinitely many nonroots $\,c\in C\,$ where $\,0\neq f(c)\,$ so it is cancellable, thus

$$f(c)\,(g(c)-h(c)) = 0\,\Rightarrow\, g(c)-h(c) = 0$$

Thus the polynomial $\,g(x)-h(x)\,$ has infinitely many roots $\,x = c\,$ so it is identically zero.

Remark $ $ It fails for finite fields, e.g. over $\,\Bbb Z_3 = $ integers $\!\bmod 3\,$ we have $ x(x^2) = x(1)\,$ for all $\,x,\,$ but $\, x^2 = 1\,$ is false at $\,x = 0$.

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  • $\begingroup$ Thank you. The idea that g(x)-h(x) has infinitely many roots implies g(x)=h(x) - is there a proof of this somewhere? $\endgroup$ – SquareCircle May 28 '19 at 0:33
  • $\begingroup$ @SquareCircle If it were nonzero then it would have only finitely many roots (no more than its degree) by the linked proof. $\endgroup$ – Bill Dubuque May 28 '19 at 0:39
  • $\begingroup$ Ah silly me yes, you're right. Great answer this is exactly what I was looking for. $\endgroup$ – SquareCircle May 28 '19 at 0:44
  • $\begingroup$ @BillDubuque Shouldn't it not be for any field of characteristic 0? There are many infinite fields with characteristic p. $\endgroup$ – sqtrat May 28 '19 at 12:25
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    $\begingroup$ @sqtrat Polynomials still have only finitely many roots in characteristic p, so it works in an infinite field of any characteristic. $\endgroup$ – Carmeister May 28 '19 at 14:43
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You can first note that the implication is true for $x=0$ by checking it. Then you can say you still have to prove it for $x\neq 0$, but then you can divide by $x$.

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For an alternative perspective, consider that the polynomials you are looking at can be thought of as purely formal objects without analytical meaning. Don't think of the polynomials as polynomial functions that you need to worry about being $0$ when you plug in different values of $x$, but as algebraic objects belonging to the ring of polynomials $\mathbb C[x]$.

Then your problem disappears because the ring of polynomials over $\mathbb C$ is an integral domain, meaning that for $f(x),g(x)\in\mathbb C[x]$, if $f(x)g(x)=0$ (importantly: here $=$ is a polynomial equality, not an equality of complex numbers), then either $f(x)=0$ or $g(x)=0$. You can prove this property in many ways. In fact, it is true that $R[x]$ is a domain whenever $R$ is, and $R$ does not need to be a field like $\mathbb C$. One nice way is to consider the leading coefficients of two nonzero $f(x),g(x)$ in the ring of polynomials: since the leading coefficients are nonzero, then their product is nonzero (because the ring of coefficients is a domain) so the polynomial product is nonzero too.

How does this tie in with the usual concept of polynomial functions? Well, for any $\alpha\in\mathbb C$, there exists a unique homomorphism $\phi_\alpha:\mathbb C[x]\to\mathbb C$ that sends $x$ to $\alpha$, called the evaluation homomorphism. This homomorphism essentially gives you a way to "plug in" the value of $\alpha$ into a formal polynomial, recovering the idea of a function. But since we have dealt with the issue of division by zero within the ring of polynomials where division by zero is a non-issue, we are now safe, because two polynomials that are formally equal define the same function over $\mathbb C$.

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  • $\begingroup$ Note that the OP mentioned in a comment that "I am only working with complex numbers, not come across rings yet (pre-university)". Also, when explaining the difference between formal and functional polynomials it is important to explain what goes awry over finite fields (no hint of that above). $\endgroup$ – Bill Dubuque May 28 '19 at 13:20
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    $\begingroup$ Hi YiFan as Bill says I don't really understand this, but that's okay I think it's good to post this answer for the sake of completeness so other interested viewers with sufficient background can see it from another perspective. Maybe I'll look at this answer again later when I learn a bit more maths. $\endgroup$ – SquareCircle May 28 '19 at 13:33
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Here's a way we could go about it, using proof by contrapositive.

Suppose that there is some $x\in\Bbb C$ such that $$(x-1)(x+1)\neq x^2-1.$$ Noting that $$(0-1)(0+1)=(-1)(1)=-1=0-1=0^2-1,$$ we must have $x\ne 0.$ Thus, since $(x-1)(x+1)\neq x^2-1,$ then we have $$x(x-1)(x+1)\neq x\left(x^2-1\right).$$


Added: Bearing in mind that, for any statements $p$ and $q,$ we have that $p\implies q$ is equivalent to $(\neg p)\vee q,$ the "implication" is true trivially, merely because $(x-1)(x+1)\equiv x^2-1$ is true. By the same token, $$1\neq 1\implies(x-1)(x+1)\equiv x^2-1.$$ Clearly, no division by $0$ takes place in this implication, yes?

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You're not dividing by 0 though.

You're saying $x(x-1)(x+1)=x(x^2-1)$ implies $(x-1)(x+1)=x^2-1$ for any $x \in \mathbb{R}$. That is a true statement. I understand where your confusion is coming from. You're dividing both sides by $x$, and since $x \in \mathbb{R}$, then $x=0$ at some point. Think of the statement "$x(x-1)(x+1)=x(x^2-1)$ implies $(x-1)(x+1)=x^2-1$" as taking a sort of "leap of logic". It's true, even if $x=0$, the statement is true. Now what happens when $x=0$ and you have the equation $x(x-1)(x+1)=x(x^2-1)$? You can't divide both sides by $x$ anymore. However, this does not imply that $(x-1)(x+1)=x^2-1$ is false, it just means you can't logically jump from $x(x-1)(x+1)=x(x^2-1)$ to $(x-1)(x+1)=x^2-1$ when $x=0$.

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