castor28 said:$f(x)$ is a degree 10 polynomial such that $f(p)=q$, $f(q)=r$, $f(r)=p$, where $p$, $q$, $r$ are integers with $p<q<r$.
Show that not all the coefficients of $f(x)$ are integers.
That is quite correct (except for a small typo: $r-q\mid p-q$ should be $r-p\mid p-q$).kaliprasad said:Let all coefficients be integers
we have m-n divides $f(m)-f(n)$
so p- q | f(p) - f(q) | q-r
similarly
q -r | r-p
and r-q | p - q
from above as p - q | q-r | r- p | p- q (all nteger multiples) so all are same and hence a contradiction
so all coefficients cannot be integers
castor28 said:That is quite correct (except for a small typo: $r-q\mid p-q$ should be $r-p\mid p-q$).
Congratulations!:)
A polynomial is an algebraic expression consisting of variables and coefficients, typically combined using addition, subtraction, and multiplication. For example, f(x) = 3x^2 + 2x + 1 is a polynomial with variables x and coefficients 3, 2, and 1.
A polynomial with integer coefficients means that all the numbers in the expression (including variables raised to a power) are whole numbers, either positive, negative, or zero.
One way to show this is by providing a counterexample. For instance, you can find a polynomial with at least one coefficient that is not an integer, such as f(x) = 2.5x^2 + 1. This disproves the idea that all coefficients must be integers.
Proving that not all coefficients of a polynomial are integers can help us better understand the properties and limitations of polynomials. It also allows us to explore different types of numbers and their relationships, such as rational and irrational numbers.
Yes, a polynomial can have only non-integer coefficients. For example, f(x) = √2x^2 + πx + e is a polynomial with non-integer coefficients. This type of polynomial is called an irrational polynomial.