# show the expressions are rational numbers

• March 5th 2008, 09:39 PM
rcmango
show the expressions are rational numbers
give a, b and c are rational numbers, show the expressions are rational.

i.) a^2 + b^2 + c ^2

ii.) (5a^3 +6b^4)/(a^2+b^2)

iii.) a + a^2 + a^3 + ... +a^10

I had alot of trouble proving this, and it was not enough work.

• March 6th 2008, 10:25 AM
CaptainBlack
Quote:

Originally Posted by rcmango
give a, b and c are rational numbers, show the expressions are rational.

i.) a^2 + b^2 + c ^2

ii.) (5a^3 +6b^4)/(a^2+b^2)

iii.) a + a^2 + a^3 + ... +a^10

I had alot of trouble proving this, and it was not enough work.

Write a, b and c as ratio of integers, then substitute these into the expressions and rearrange to show that they are the ratios of integers.

For example let $a=e/f$, $b=g/h$, $c=i/j$, where $e,\ f,\ g,\ i$ and $j$ are integers.

Then:

$a^2 + b^2 + c ^2=\frac{e^2}{f^2}+\frac{g^2}{h^2}+\frac{i^2}{j^2} =\frac{e^2h^2j^2+g^2f^2j^2+i^2f^2h^2}{f^2h^2j^2}$

RonL
• March 6th 2008, 02:34 PM
iknowone
Speaking generally there are a lot of different operations one can do to rational numbers without leaving the rational world.

The product of any finite number of rational numbers is rational:
let $q_1, q_2, ... q_k$ be rational numbers.
say $q_i=\frac{n_i}{m_i}, \forall i, \text{ and } m_i, n_i$ are integers.
then $n_1n_2...n_k \text{ and } m_1m_2...m_k$ are integers
therefore $q_1 q_2 ... q_k$ is rational.

Same kind of proofs can be used to show the sum of 2 rationals is rational, which extends inductively to the finite sum of rationals. The division of rationals remaining rational is a direct consequence of the multiplication of rationals remaining rational. So it is not hard to see that any finite composition of such operations on rationals results in a rational. Proving these general statements might be easier than specific examples which can get very messy.
• March 9th 2008, 03:44 PM
rcmango
iii.) a + a^2 + a^3 + ... +a^10

i could let a = f/e

then = (f/e)^2

then no matter what to the 10th power it is rational.

is that correct?