# Next step to reach the contradiction?

This is a problem from Discrete Mathematics and its Applications Here are my notes and my current work so far for this problem. I started with an assumption that what i am trying to prove is false, that is there is a rational number r for which $r^3 + r + 1 = 0$. Under that assumption, I am trying to reach a contradiction(goes against the assumption, that this cannot happen under any circumstance). I first used to definition of rational number to break down $r$ into $\frac{p}{q}$. Following the hint and my natural intuition, I broke the equation down and reasoned that this will be true only if the numerator $p^3 + 2pq^2 + q^3 = 0$I am stuck at this point. How can I show that that expression cannot be equivalent to zero? I tried cubic factoring from http://www.purplemath.com/modules/specfact2.htm but that didn't really get me anywhere either(at the bottom of the page). The hint says to look at whether each a and b is odd or even but I don't see how that would help you prove that that expression will never be equivalent to zero.

From $p^3+pq^2+q^3=0$, discuss the four possible cases:

1. $p$ odd, $q$ odd.

In this case the three terms are all odd. They cannot sum up to $0$.

1. $p$ odd, $q$ even.

In this case, we have one odd term and two even terms. The sum must be odd, so they can not add up to $0$.

I guess you can now continue with the other cases now and conclude that they have to be both even, which is a contradiction.

• thanks but if they're both even, how does that contradict the original statement? the original statement says there is rational number r. r can be composed of one even divided by another Jan 12, 2015 at 2:18
• Remember your assumption is $r=p/q$ in lowest terms, which means $p$ and $q$ don't have common divisor. Jan 12, 2015 at 2:19
• I don't i included that part in my assumption. The book definition(the one I am using) of rational is "The real number r is rational if there exist integers p and q with q!=0 such that r = p/q." So technically 8 = 16/2 work with that definition of rational? Jan 12, 2015 at 2:23
• Well, in your first post, it says "a/b is in lowest terms". Even without thinking of that hint, you will have to make that assumption so that the proof would work. This is a very typical assumption when you proof things about rational numbers. Jan 12, 2015 at 2:28
• ohhh lowest terms, cause for evens, that denom can always get broken down to a 1. Jan 12, 2015 at 2:33