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(where $n>0$)

I have been taught that an equation with the highest power $n$ will always have $n$ solutions. This does not appear to be the case with: $$x^n=1 \implies x=\pm1$$ Where $n$ is even, the solutions are $1$ and $-1$; where $n$ is odd, the only solution is $1$.

Why? Is it a special case, or is there something I'm missing?

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    $\begingroup$ It has $n$ solutions, but only one or two of them are real. The rest are complex numbers. The full set of solutions can be written $x = \cos(2\pi k/n) + i \sin(2\pi k/n)$ for $k=0,1,2,3,\ldots, n-1$. $\endgroup$ – Winther Aug 15 '15 at 19:59
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    $\begingroup$ Polynomial equations on the real numbers have at most as many solutions as the degree. But they can have none, for instance $x^2+1=0$. $\endgroup$ – egreg Aug 15 '15 at 20:00
  • $\begingroup$ You've gotten it backwards - there is only one real solution when $n$ is odd, and two if even. $\endgroup$ – Thomas Andrews Aug 15 '15 at 20:07
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    $\begingroup$ If OP is asking this question, I think he has still to learn about rings, fields, etc. I think mentioning stuff like that will confuse him more than help him. $\endgroup$ – YoTengoUnLCD Aug 15 '15 at 20:08
  • $\begingroup$ Thanks, @ThomasAndrews. Will edit. $\endgroup$ – Mutantoe Aug 15 '15 at 20:22
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When you speek about solutions of an equation you must always be careful about the "algebraic structure" (In a specialistic language Fields/Rings) where you are searching a solution.

For example, searching a solution in $\mathbb{Z}$ is not equal to searching a solution over $\mathbb{Q}$ even in very simple cases.

An interesting example of this phenomenon is the polynomial $2x-1$.

The related equation has no solution in the ring $\mathbb{Z}$ but a unique solution in $\mathbb{Q}$ (that is in specialistic language $\mathbb{Z}$'s fraction field).

Looking more closely to your case, given a polynomial with just one variable (a so-called "univariate polynomial") of degree $n$, its associated equation has always $n$ solution in (the field) $\mathbb{C}$ but it can have no solution in -(the field) $\mathbb{R}$

Look for example to the polynomial $x^2+1$.

Algebra in some sense is the study of this differences, but something sure are the following theorems:

Theorem Let $p(x)$ be a univariate polynomial of degree $n$. Then there always exist a "bigger" algebraic structure (called the splitting field of the polynomial p) where the equation $p(x)=0$ has exactely $n$ solutions (counted with multiplicity)

For example, in a case above, the field $\mathbb{C}$ is the splitting field of $x^2+1$ over $\mathbb{R}$

Well, to conclude, let's speack about $\mathbb{C}$. It is a very "beautifull" algebraic structure because has a very nice property:

Theorem Let $p(x)$ an univariate polynomial of degree $n$ with coefficients in $\mathbb{C}$. Then $p(x)=0$ has exactely $n$ solution over $\mathbb{C}$ (counted with multiplicity)

In particular, since $\mathbb{R} \subset \mathbb{C}$, a real polynomial is a complex polinomial and its associated equation has always $n$ complex solution but cannot have $n$ real solutions. This is precisely the case of $x^2+1$.

There are "not many" algebraic structure with the same property of $\mathbb{C}$. This kind of structures are called "algebrically closed fields" and are one of the most important objects of Algebra.

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    $\begingroup$ You just used the terms "field", "ring", and "univariate" casually in a response to somebody presumably in high school algebra $\endgroup$ – Alex Mathers Aug 15 '15 at 20:23
  • $\begingroup$ You're right. I hope now it's more clear to a high scool student $\endgroup$ – Sabino Di Trani Aug 16 '15 at 11:16
  • $\begingroup$ @JosephCurwen I can now understand quite a lot of that, thanks! $\endgroup$ – Mutantoe Aug 16 '15 at 19:20
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It has $n$ roots, but only if you consider roots $x_i\in \Bbb C$. These "special" ones are calleds the roots of unity.

As an example, consider $X^4=1$; try to check that $X_1=i$ is a root of said equation.

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The statement that there are always $n$ solutions is only true if you are working in the field of complex numbers. In the field of the real numbers this is not true, as you have shown.

Please note, that even in the field of complex numbers, a solution can occur multiple times, i.e. $(x-1)^2$ has a double zero at $1$. In that case, you need to define how you count the number of solutions.

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    $\begingroup$ So, you can have multiple solutions that are actually the same number? $\endgroup$ – Mutantoe Aug 15 '15 at 20:42
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    $\begingroup$ @Mutantoe well yes, that depends on the definition - if we take the multiplicity into account, then we have always $n$ complex solutions for a polynomial of degree $n$. If we do not care about multiplicity, then there can be less. $\endgroup$ – supinf Aug 16 '15 at 3:37
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Notice, we have $$x^n=1$$ $$x=(1)^{1/n}$$ $$=(\cos 0+i\sin 0)^{1/n}$$ $$=(\cos 2k\pi+i\sin2k\pi)^{1/n}$$ $$=\left(\cos \left(\frac{2k\pi}{n}\right)+i\sin\left(\frac{2k\pi}{n}\right)\right)$$ Thus there are $n$ roots which can be determined by setting $k=0, 1, 2, \ldots (n-1)$

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    $\begingroup$ I think this is wrong. By definition, $(1)^{1/n}$ is always $1$ and not something else $\endgroup$ – supinf Aug 15 '15 at 20:14
  • $\begingroup$ Here, we are considering complex roots. Could you please tell me why $1=\cos 0+i\sin 0$ not true in complex numbers ? $\endgroup$ – Harish Chandra Rajpoot Aug 15 '15 at 20:22
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    $\begingroup$ You made two mistakes. the first is from the first to the second row: counterexample: $x=-1,n=2$, then $-1 = x = (1)^{1/2} =1$, which is obviously wrong. The second mistake cancels the other out and is in the last row: there is no mathematical law that you can take roots like that $\endgroup$ – supinf Aug 16 '15 at 3:46
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    $\begingroup$ yes, i do know de moivre's theorem. do you apply it here? i don't think you do. And what is wrong with my proof $-1 = 1$ ? $\endgroup$ – supinf Aug 16 '15 at 4:14
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    $\begingroup$ Supinf is right, $1^{1/n}=1$ always so the second and third equality are not equivalent to the first one. Like for example, $\sqrt{4}=2$ and not $\pm 2$. But if $x^n=1$, then $x=e^{\frac{2ik\pi}{n}}$, $k=0,...,n-1$. $\endgroup$ – Surb Aug 16 '15 at 11:24

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