# rational numbers

• Sep 6th 2008, 11:54 AM
rmpatel5
rational numbers
Find two rational numbers with denominators 11 and 13, respectively, and a sum of 7/143.

I got x/11 + y/13=7/143.

Two unknowns and one equation. Kind of stuck.
• Sep 6th 2008, 12:05 PM
Matt Westwood
Might help to multiply through by 143 and clear the fractions. Note that 11 x 13 = 143.

Then you're into a conventional linear Diophantine equation for which there are techniques (which I'd need to look up). What's your level?
• Sep 6th 2008, 12:11 PM
Laurent
Don't forget $x$ and $y$ are integers. The equation they satisfy is $13x+11y=7$.
You should first find one particular solution $(x_0,y_0)$ (you may start by finding $x,y\in\mathbb{Z}$ such that $13x+11y=1$ (why is it possible ?)), and then look for the others: they satisfy $13x+11y=13x_0+11y_0$, hence $13(x-x_0)=11(y_0-y)$ and you should then be able to conclude that $x=x_0+11k$ and $y=y_0-13k$ for some integer $k$.

Laurent.
• Sep 6th 2008, 12:16 PM
rmpatel5
Quote:

Originally Posted by Matt Westwood
Might help to multiply through by 143 and clear the fractions. Note that 11 x 13 = 143.

Then you're into a conventional linear Diophantine equation for which there are techniques (which I'd need to look up). What's your level?

What do u mean level? I am taking Elem. Number Theory and this is some of the recm. hw problems to do for the section of euclidean algorithm
• Sep 6th 2008, 12:18 PM
Laurent
Quote:

Originally Posted by rmpatel5
What do u mean level? I am taking Elem. Number Theory and this is some of the recm. hw problems to do for the section of euclidean algorithm

• Sep 6th 2008, 12:20 PM
rmpatel5
Quote:

Originally Posted by Laurent
Don't forget $x$ and $y$ are integers. The equation they satisfy is $13x+11y=7$.
You should first find one particular solution $(x_0,y_0)$ (you may start by finding $x,y\in\mathbb{Z}$ such that $13x+11y=1$ (why is it possible ?)), and then look for the others: they satisfy $13x+11y=13x_0+11y_0$, hence $13(x-x_0)=11(y_0-y)$ and you should then be able to conclude that $x=x_0+11k$ and $y=y_0-13k$ for some integer $k$.

Laurent.

I am really lost with what you do. I know why i can say 13x + 11y=1 because the gcd(11,13)=1 and there is a thm that says ax+by=1. Should i just slove the two system of equations ?
• Sep 6th 2008, 12:23 PM
Laurent
Quote:

Originally Posted by rmpatel5
gcd(11,13)=1 and there is a thm that says ax+by=1.

There is a theorem that says that there exists x and y such that 11x+13y=1. Now, you should find an example of such a couple (x,y). Either you guess it, or you make use of the Euclidean algorithm as you may have done in class.
• Sep 6th 2008, 12:24 PM
Moo
Quote:

Originally Posted by rmpatel5
I am really lost with what you do. I know why i can say 13x + 11y=1 because the gcd(11,13)=1 and there is a thm that says ax+by=1. Should i just slove the two system of equations ?

Ok, so solve for x and y in 13x+11y=1, using the Euclidean algorithm (http://en.wikipedia.org/wiki/Euclidean_algorithm). Actually, do the Euclidian algorithm over 13 and 11, and observe...

Then, multiply by 7 :
13*(7x)+11*(7y)=7

Uh.
• Sep 6th 2008, 12:26 PM
Matt Westwood
Use the Euclidean Algorithm to get the gcd of 11 and 13 (yes, you know this is 1, but the calculations you did in the algorithm help you find out what values of a and b give you 11a + 13b = 1.

You now want two numbers x and y such that 11x + 13y = 7.

Well, you just got 11a + 13b = 1, so multiply everything by 7:

$11 \times 7a + 13 \times 7b = 7$

So 7a and 7b are the numbers you want for x and y. Job done.
• Sep 6th 2008, 04:59 PM
ThePerfectHacker
You do not need to use Euclidean algorithm.
It is quite obvious that $13(6) + 11(-7) = 1$.
Thus, $13(42) + 11(-49) = 7$.

This means the solutions to $13x+11y=7$ are: $x=42 - 11t \mbox{ and }y=-49 + 42t$.
• Sep 6th 2008, 05:15 PM
rmpatel5
6 and -5 work as well, so are there many solutions to this problem. I got them by doing the EA