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A real number is said to be algebraic if it is root of a non-zero polynomial with integer coefficients.

Which of the following real numbers are algebraic?

(a) $\cos (\pi/5)$
(b) $e^{\frac{1}{2}\log2}$
(c) $5^{1/7}+7^{1/5}$

(a) I am not sure.
(b) it is algebraic since this is $\sqrt{2}$ and $x^2-2$ is the polynomial.
(c) not sure but I think this is also algebraic and a polynomial og order 35 may be the answer.but I cant prove this.

can I get some help please.

Community
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patla
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  • $(a)$ http://math.stackexchange.com/questions/49297/the-only-two-rational-values-for-cosine-and-their-connection-to-the-kummer-rings – lab bhattacharjee Dec 30 '13 at 05:46
  • Assuming natural logarithm, $$e^{\frac12\log 2}=e^{\log\sqrt2}=\sqrt2$$ – lab bhattacharjee Dec 30 '13 at 05:47
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    The sum of two algebraic numbers is algebraic. If these were questions in an exam you took, or are on a practice exam, then you should probably know this fact and how to prove it. (Also note $\cos\pi/5$ is a sum of two algebraic numbers.) One doesn't need to find an explicit polynomial. – anon Dec 30 '13 at 05:56
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    Chiming in with anon. $e^{i\pi/5}$ is a root of $x^5+1=0$, so it is algebraic. Same holds for $e^{-i\pi/5}$. Thus $\cos(\pi/5)$ which is their average is algebraic also (dividing by two gives another algebraic, the polynomial won't be monic this time). Note that replacing $5$ with a larger integer poses no problem, if you think about it in this way. – Jyrki Lahtonen Dec 30 '13 at 06:01

3 Answers3

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You are correct on (b).

For (a) and (c), the key facts are that a sum, quotient and product of algebraic numbers are algebraic. (You need to prove this if you haven't yet!)

$$\cos (\pi/5)=\frac{ \zeta_{10}+\zeta_{10}^{-1}}2$$ where $\zeta_{10}$ is a $10$th root of unity, and (c) is plainly obvious from the key facts.

PVAL-inactive
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(a) Since $$\alpha=\cos\left(\frac{\pi}{5}\right)=\frac{1+\sqrt 5}{4}\Rightarrow 4\alpha-1=\sqrt 5\Rightarrow (4\alpha-1)^2=5,$$ you'll know that $x=\alpha$ is a root of $$4x^2-2x-1=0.$$

(c) Wolfram alpha gives us this.

mathlove
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  • but how can I check in exam? – patla Dec 30 '13 at 05:52
  • You mean how we can get $\cos(\pi/5)=(1+\sqrt 5)/4$? – mathlove Dec 30 '13 at 05:53
  • for both.In (a) if 5 is repleced by any other large iteger then what should we do? – patla Dec 30 '13 at 05:59
  • for (c) in exam how to check this types of equations without any software? – patla Dec 30 '13 at 06:00
  • It is known that $\cosα$ is a transcendental number if $α\not=0$ is an algebraic number. For $(c)$, I don't know any good way to check it without using computer. By the way, if you want to get $\cos(\pi/5)=(1+\sqrt 5)/4$, then consider an isosceles triangle $ABC$ with $\angle{ABC}=\angle{ACB}=36^\circ,$ and take $D$ on $BC$ such that $\angle{BAD}=72^\circ$. – mathlove Dec 30 '13 at 06:07
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(c) this is a solution of 35x7 deg polynomial when we solve : x-5^(1/7)=7^(1/5) raising 35 power x^35 - 5^5 - 7^7 = 5^(1/7)*(polynomial of degree 34)

again raising 7th power we get desired polynomial.

(a) refer wolfram alpha.

Sry
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    Raising $x-5^{1/7}=7^{1/5}$ to $35$ does not give us $x^{35}-5^5-7^7=5^{1/7}$. Why would you think that? – anon Dec 30 '13 at 06:34
  • it seems that you didn't got my point.I have given a hint,its not complete sol and i don't see any reason to downvote – Sry Dec 30 '13 at 07:16
  • The downvote (not from me, because you are new) is because your hint is incorrect, not incomplete. That's what my comment was saying, so it looks like you didn't get my point. There's also the difficult-to-read formatting to consider. – anon Dec 30 '13 at 07:24