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Proving that sum of Rational #s is Rational

 
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Sep14-08, 02:07 PM   #1
 
Question

Proving that sum of Rational #s is Rational


1. Let a and b be rational numbers. Prove or provide counterexample that

A) a+b is a rational number.
B) Is ab necessarily a rational number?




2. How can you prove that the sume of two rational number is rational? Well I am not really good at math



3. This is what I've tryed to do for part a. But I'm stuck proving the sum (product) of two integer is an integer.

Since a and b are rational numbers they can be written as a=x/y and b=w/z where x,y,w,z are all integers. Then a+b= x/y+w/z = (xz+wy)/(yz)

Now I'm stuck, and I have no clue how to do part B...
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Sep15-08, 09:29 AM   #2
 
You might want to look at this thread

http://www.physicsforums.com/showthread.php?t=166106

it's similar
Sep15-08, 09:36 AM   #3
 
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Quote by rs21867 View Post
1. Let a and b be rational numbers. Prove or provide counterexample that

A) a+b is a rational number.
B) Is ab necessarily a rational number?




2. How can you prove that the sume of two rational number is rational? Well I am not really good at math



3. This is what I've tryed to do for part a. But I'm stuck proving the sum (product) of two integer is an integer.

Since a and b are rational numbers they can be written as a=x/y and b=w/z where x,y,w,z are all integers. Then a+b= x/y+w/z = (xz+wy)/(yz)

Now I'm stuck, and I have no clue how to do part B...
You are exactly right for (a): you have shown that a+ b= (xz+ wy)/(yz) and so is the ratio of two integers (and the denominator yz is not 0 because neither y nor z is 0).

To do B, do exactly the same thing! If a= x/y and b= w/z, what is ab? Is it a ratio of two integers? Could the denominator be 0? If anything, B is easier than A because it is easier to multiply two fractions than to add!
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