Thread Closed

Is that an algebraic number?

 
Share Thread Thread Tools
May31-09, 05:56 AM   #1
 

Is that an algebraic number?


is [tex]\sqrt{2}+\sqrt{5}[/tex] an algebraic number?
i used 2 and 5 arbitrarily, try any integers (as long as they are not the same integer, in which case it is algebraic)
I tried finding a polynomial with rational coefficients that zeros at this value, but haven't found any.
 
PhysOrg.com
PhysOrg
science news on PhysOrg.com

>> Hong Kong launches first electric taxis
>> Morocco to harness the wind in energy hunt
>> Galaxy's Ring of Fire
May31-09, 06:04 AM   #2
 
Recognitions:
Gold Membership Gold Member
Science Advisor Science Advisor
Retired Staff Staff Emeritus
If it was, its powers would span a finite dimensional vector space over Q.
 
May31-09, 07:22 AM   #3
 
Recognitions:
Homework Helper Homework Help
Science Advisor Science Advisor
http://www.dpmms.cam.ac.uk/~wtg10/galois.html

is a useful link to expand on Hurkyl's idea.

Or.

Define x to be that expression above, what is x^2? what is x^2 - 2 - 5?
 
May31-09, 11:22 AM   #4
 

Is that an algebraic number?


thanks matt grime!
 
May31-09, 12:51 PM   #5
 
Recognitions:
Gold Membership Gold Member
Science Advisor Science Advisor
Retired Staff Staff Emeritus
Yes, of course it is. If [itex]x= \sqrt{2}+ \sqrt{5}[/itex] then [itex]x- \sqrt{2}= \sqrt{5}[/itex] and [itex](x- \sqrt{2})^2= x^2- 2\sqrt{2}x+ 2= 5[/itex]. Then [itex]x^2- 3= 2\sqrt{2}x[/itex] so [itex](x^2- 3)^2= x^4- 6x^2+ 9= 8[/itex]. [itex]\sqrt{2}+ \sqrt{5}[/itex] satisfies the polynomial equation [itex]x^4- 6x^2+ 1= 0[/itex] and so is algebraic.
 
May31-09, 01:40 PM   #6
 
HallsofIvy,
[itex]
(\sqrt{5}+\sqrt{2})^4-6(\sqrt{5}+\sqrt{2})^2+1 = 98.596
[/itex]


i got a different result, for any [itex]\sqrt{a}, \sqrt{b}[/itex]
just use [itex](\sqrt{a}+ \sqrt{b})*(\sqrt{a}- \sqrt{b})*(-\sqrt{a}+ \sqrt{b})*(-\sqrt{a}- \sqrt{b})[/itex] and expand
i haven't read the whole article, just the start and deducted this (without proof) by factoring the polynomial they gave for 2 and 3
 
Jun2-09, 01:01 PM   #7
 
Recognitions:
Gold Membership Gold Member
Science Advisor Science Advisor
Retired Staff Staff Emeritus
Thanks for the correction. Here's my mistake:
instead of [itex](x^2- 3)^2= x^4- 6x^2+ 9= 8[/itex] is should have
[itex](x^2- 3)^2= x^4- 6x^2+ 9= 8x^2[/itex]. I dropped the "x" in "[itex]2\sqrt{2}x[/itex]" when I squared.

With that correction, we get [itex]x^4- 14x^2+ 9= 0[/itex] and this time I checked, with a calculator, that [itex]\sqrt{2}+ \sqrt{5}[/itex] satisfies that equation.

Since [itex]\sqrt{2}+ \sqrt{5}[/itex] satisfies [itex]x^4- 14x^2+ 9= 0[/itex], it is algebraic.
 
Jun8-09, 11:06 PM   #8
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
If you can prove that the algebraic numbers form a group additively, then you are done.
 
Jun9-09, 03:55 PM   #9
 
jason, what do you mean "form a group additively"? I dont get it, do you mean some sort of commutative property? Although I doubt it..
 
Jun9-09, 03:59 PM   #10
 
It means if you add or subtract algebraic numbers from each other, you get an algebraic number.
 
Jun9-09, 04:17 PM   #11
 
Thats a great question. I was working on a similar question, whether e+pi was transcendental.
 
Jun9-09, 11:24 PM   #12
 
Recognitions:
Homework Helper Homework Help
Science Advisor Science Advisor
Quote by camilus View Post
Thats a great question. I was working on a similar question, whether e+pi was transcendental.
Hah! good luck.
 
Jun10-09, 10:43 AM   #13
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Quote by camilus View Post
jason, what do you mean "form a group additively"? I dont get it, do you mean some sort of commutative property? Although I doubt it..
I never said anything about commutativity (even though in this case there is).
 
Jun10-09, 10:44 AM   #14
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Quote by camilus View Post
Thats a great question. I was working on a similar question, whether e+pi was transcendental.
Haha, yeah like CRGreathouse said, good luck.

This question is way beyond the calibre of question compared to the one in the OP.
 
Jun10-09, 01:15 PM   #15
 
Blog Entries: 1
Recognitions:
Homework Helper Homework Help
The algebraic numbers form a field even, but that's a little tricky to prove (specifically if a and b are algebraic numbers, then a*b is too)
 
Thread Closed
Thread Tools


Similar Threads for: Is that an algebraic number?
Thread Forum Replies
Question of algebraic flavor in algebraic topolgy Differential Geometry 2
Number of ways to make change for dollar with even number of coins Calculus & Beyond Homework 1
Algebraic Number Theory Linear & Abstract Algebra 12
numerical solutions of system of nonlinear algebraic equations nonlinear algebraic eq General Math 6