Math Challenge For The Brave

in #math8 years ago

A monic polynomial has highest coefficient equal to 1. How do you prove that if a monic polynomial has a rational zero, then this zero must in fact be an integer?
I can explain the full logical proof. Maybe there are other methods too?

Here's some nice polynomial art:
serveimage-1.png

Sort:  

Do you mean a monic polynomial with 'rational' coefficients?
(By the way, I received 3SBD. Thank you:) )

I think the easiest way to prove that is the following:
Let f(x)=x^{n}+a_{n-1} x^{n-1} + ... a_{0} be a monic polynomial with rational coefficients and let p/q be a root of this polynomial. (Here, q is nonzero, p and q are coprime.)
Then we have
(p/q)^{n} + a_{n-1} (p/q)^{n-1} + ... + a_{0} = 0.
By multiplying q^{n} on the both sides, we get
p^{n} + a_{n-1} p^{n-1}q + ... a_{0} q^{n} = 0,
i.e., p^{n}=-( a_{n-1} p^{n-1}q + ... a_{0} q^{n} ) .
Since ( a_{n-1} p^{n-1}q + ... a_{0} q^{n} )is a multiple of q, q must divide p^{n}. Also, since p and q are coprime, q must divide p as well. Now we can conclude that q=1 and p/q is an integer .

Excellent, that's how I did it

I did not take that into account. It would seem that having rational coefficients is required, but I think it would work for real or complex coefficients as well, just as long as the zero is rational.

Coin Marketplace

STEEM 0.12
TRX 0.34
JST 0.033
BTC 122796.22
ETH 4599.52
BNB 1301.94
SBD 0.77