- #1

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In other words,

I need to show: an*x^n + an1*x^(n-1) + ... + a1 * x^1 + a0 * x^0 =

where the a terms are not ALL 0 but some of them can be.

Like for root 2 by itself,

I have 1 * (root 2) ^ 2 + 0 * (root 2)^1 + -2 * (root 2) ^ 0

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- Thread starter lokisapocalypse
- Start date

- #1

- 32

- 0

In other words,

I need to show: an*x^n + an1*x^(n-1) + ... + a1 * x^1 + a0 * x^0 =

where the a terms are not ALL 0 but some of them can be.

Like for root 2 by itself,

I have 1 * (root 2) ^ 2 + 0 * (root 2)^1 + -2 * (root 2) ^ 0

- #2

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How about a5=1, a3=-10, a1=1, all others zero?

- #3

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According to Maple, that gives -.30e-7.

- #4

- 740

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then [tex](x - \sqrt{3})^2 = x^2 - 2\sqrt{3}x + 3 = 2[/tex]

... something like that. put one square root one one side & the rest on the other, square both sides, etc etc then you'll get a 4th-degree polynomial that has [tex]\sqrt{3} + \sqrt{2}[/tex] as a root. end of proof

- #5

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lokisapocalypse said:According to Maple, that gives -.30e-7.

Rounding error.

--J

- #6

Curious3141

Homework Helper

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The easiest way is just to find an equation with integral coefficients that has the number as a root.

What is the square of [tex]\sqrt{2} + \sqrt{3}[/tex] ? Express that in the form : integer plus a square root. Let's call this [tex]n_1 = a + \sqrt{b}[/tex]

Next, find the conjugate of the number. [tex]n_2 = a - \sqrt{b}[/tex]

Set up the equation [tex](y - n_1)(y - n_2) = 0[/tex] and expand it out. You know for certain that this equation is going to have integral coefficients because that's a property of conjugate surds.

Once you have that equation in y, just put [tex]x^2 = y[/tex] and get a quartic in x. That equation will have [tex]\sqrt{2} + \sqrt{3}[/tex] as one of the roots, and you're done.

What is the square of [tex]\sqrt{2} + \sqrt{3}[/tex] ? Express that in the form : integer plus a square root. Let's call this [tex]n_1 = a + \sqrt{b}[/tex]

Next, find the conjugate of the number. [tex]n_2 = a - \sqrt{b}[/tex]

Set up the equation [tex](y - n_1)(y - n_2) = 0[/tex] and expand it out. You know for certain that this equation is going to have integral coefficients because that's a property of conjugate surds.

Once you have that equation in y, just put [tex]x^2 = y[/tex] and get a quartic in x. That equation will have [tex]\sqrt{2} + \sqrt{3}[/tex] as one of the roots, and you're done.

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- #7

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You can look at this from the standpoint of Galois theory. Along with one surd we are going to get the conjugate of: [tex]\pm\sqrt2[/tex]. So putting both roots in a quadratic makes the coefficients rational. [tex]x^2-2=0[/tex].

The same idea can be applied to this expression: [tex]\pm\sqrt{2}[/tex][tex]\pm\sqrt{3}[/tex]. We don't want the surds to appear in the coefficients.

Maybe not as efficient as described in previous postings, but does offer some insight.

The same idea can be applied to this expression: [tex]\pm\sqrt{2}[/tex][tex]\pm\sqrt{3}[/tex]. We don't want the surds to appear in the coefficients.

Maybe not as efficient as described in previous postings, but does offer some insight.

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- #8

Hurkyl

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According to Maple, that gives -.30e-7.

It shouldn't; Maple is capable of doing symbolic arithmetic. Don't ask it to approximate. :tongue2:

By the way, please don't multiple post.

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- #9

HallsofIvy

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Standard method: Use the fact that (a+ b)(a- b)= a^{2}- b^{2} to get rid of the square roots.

[tex](x- \sqrt{3}-\sqrt{2})(x-\sqrt{3}+ \sqrt{2})= (x- \sqrt{3})^2- 2= x^2- 2\sqrt{3}x+1[/tex]

[tex](x^2+1-2\sqrt{3}x)(x^2+1+2\sqrt{3}x)= (x^2+1)^2- 12x^2= x^4+ 2x^2+ 1- 12x^2= x^4- 10x^2+ 1[/tex]

[tex]\sqrt{3}+ \sqrt{2}[/tex]

is algebraic because it satisfies

[tex]x^4-10x^2+ 1[/tex]

[tex](x- \sqrt{3}-\sqrt{2})(x-\sqrt{3}+ \sqrt{2})= (x- \sqrt{3})^2- 2= x^2- 2\sqrt{3}x+1[/tex]

[tex](x^2+1-2\sqrt{3}x)(x^2+1+2\sqrt{3}x)= (x^2+1)^2- 12x^2= x^4+ 2x^2+ 1- 12x^2= x^4- 10x^2+ 1[/tex]

[tex]\sqrt{3}+ \sqrt{2}[/tex]

is algebraic because it satisfies

[tex]x^4-10x^2+ 1[/tex]

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- #10

Hurkyl

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You made a typo in there somewhere.

- #11

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I apologize for the multiple post...I was in a panic. Won't happen again.

- #12

HallsofIvy

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Hurkyl said:You made a typo in there somewhere.

Yep, I dropped an "x" early on. I've edited it. Thanks, Hurkyl.

- #13

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Wow, I hope I never have to do these kind of problems. What course is this for?

- #14

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adc85 said:Wow, I hope I never have to do these kind of problems. What course is this for?

Modern Algebra course of some kind. I'm doing a similar one myself this semester. Actually, I solved that very exercise a couple of hours ago.

Very nice course, by the way. Don't hope without knowing.

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