Showing an equation has no rational roots

  • Thread starter XYZeagle
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  • #1
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I've been working through "A Course of Pure Mathematics" and there is one problem I'm really stuck on. I'm wondering if anyone could help me out. To avoid typing it all out, I here's a link. http://books.google.com/books?id=a3...a=X&oi=book_result&resnum=1&ct=result#PPA7,M1

Here's the problem taken out of context. It's 6 and page 7.

Show that if Pn = 1 and p1+p2+...+pn[tex]\neq[/tex] -1, then p1xn+p2xn-1+p3xn-2+...+pn = 0 can not have a rational root.

I apologize if I put this in the wrong section.
 

Answers and Replies

  • #2
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There is a strategy to solve such problems. Let the roots be a1,a2 ...
now u know that pn=a1*a2...*an
and pn-1= [tex]\Sigma[/tex]a1*a2
and so on. Try to negate your proof...
 
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  • #3
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The author showed that the equations roots can only be integers and it's very obvious that pn must be divisible by the root. With pn being 1, the root can either be 1 or negative 1. If the root is 1, the two sides aren't equal because p1+p2+...+pn is not -1 and because of that the root can not be rational.

I hope that made sense. I'm really not sure how to convey that using mathematical symbols or what to do in the case that the root is -1.

I made a mistake in creating this thread here. Could someone move this to the homework forum? Thanks.
 
  • #4
gel
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The question doesn't make sense, unless you meant to write p0xn+p1xn-1+p2xn-2+...+pn = 0.

Even so, if n=1 and p0=p1=1 then -1 is a root. I can't see inside that book from your link, so I don't know what it says. Maybe they say positive root?
 
  • #5
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You're right it should be
xn+p1xn-1+p2xn-2+...+pn = 0. It says to show it can't have a rational root. The author showed the equation can't have a rational root other than -1 or 1. I showed its not 1 and I don't know what to do about -1.
 
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  • #6
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Clearly 1,p_1,..,p_n have no common factor (the polynomial is 'primative'). Call the polynomial f(x) and suppose that f(x)=g(x)(x-q) for some rational number q and some function g(x)=x^{n-1}+a_1x^{n-2}..+a_{n-1} with rational coefficients. Then there exist integers a,b such that ag(x) and b(x-q) are primative polynomials (proove this!) It is clear that their product ag(x)b(x-q)=abf(x) is also primative. But f(x) is a polynomial, so a and b must be 1 or -1. Hence g(x) is a polynomial and q is an integer. But we already know this is impossible.

This is very similar to Gauss's Lemma which should be in your book or on wikipedia.
 
  • #7
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Clearly 1,p_1,..,p_n have no common factor (the polynomial is 'primative'). Call the polynomial f(x) and suppose that f(x)=g(x)(x-q) for some rational number q and some function g(x)=x^{n-1}+a_1x^{n-2}..+a_{n-1} with rational coefficients. Then there exist integers a,b such that ag(x) and b(x-q) are primative polynomials (proove this!) It is clear that their product ag(x)b(x-q)=abf(x) is also primative. But f(x) is a polynomial, so a and b must be 1 or -1. Hence g(x) is a polynomial and q is an integer. But we already know this is impossible.

This is very similar to Gauss's Lemma which should be in your book or on wikipedia.
I can't see how that can show the equation's root isn't 1 or -1.

It's easy to see that -1 and 1 are the only possible roots and 1 isn't a root, but with -1 I'm hitting a brick wall.
 
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  • #8
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Yes, you're right; gel's comment still holds in the new version you posted (should've checked that!) My post shows that any root has to be an integer, so the only possibility is -1, hence there are no positive rational roots.

Perhaps the book meant x^{2n}+p_1x^{2n-2}+..+p_n?
 
  • #9
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Yes, what I copied is what the book says. What's weird is I can plug in numbers for the integers to that meet the requirements set and show that -1 can be a root.

The equation was xn+p1xn-1+p2xn-2+...+pn = 0.

Let n=2, x=-1, p1=2, p2=1 (because pn was required to be 1)
p1+p2=2+1[tex]\neq[/tex]-1
It meets both of the requirements set to not have a rational root.

x2+p1x1+p2=0

(-1)2+2(-1)1+1=0

1+(-2)+1=0

I really hope what the book said is a misprint, because I can not see how it can't have a negative rational root.
 
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