Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am currently learning about Direct Proofs. I am struggling trying to find a starting point to prove the Statement: For all real numbers $c$, if $c$ is a root of a polynomial with rational coefficients, then c is a root of a polynomial with integer coefficients. Based on a definition given by the book and the professor: A number is $c$ is called a root of a polynomial $p(x)$ if, and only if, $p(c) = 0$. But how can I prove the veracity of this statement using the given definition?

share|cite|improve this question
Hint: use the lcm of the denominators of the coefficients of the polynominal – Stefan Feb 20 '13 at 19:31
up vote 4 down vote accepted

Example (as a hint):

$$P(x) = \frac{3}{2}x^2 + 2x - \frac{1}{3}$$

Multiply by 6:

$$ 6P(x) = \frac{18}{2}x^2 + 12x - \frac{6}{3}$$ $$ 6P(x) = 9x^2 + 12x - 2$$

If $c$ is a zero of $P(x)$, then $6P(c) = 0$, so $0 = 9c^2+12c-2$, and so $c$ is a zero of of a polynomial with integer coefficients.

share|cite|improve this answer

Any polynomial with rational coefficients can be made into a polynomial with integer coefficients by multiplying through by the product of the denominators of the coefficients (or the LCM of those denominators). All that remains, then, is for you to show that if a number $c$ is the root of a polynomial, then it is also the root of any polynomial formed by multiplying that polynomial by a constant.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.