# Thread: More Parsonson - Rational Numbers

1. ## More Parsonson - Rational Numbers

Hi all. I'm trying to complete Ex. 11 (photos of the book pages below).
.would someone be able to give me a hint on how to approach this?

I have tried replacing (x,1) with (b/a,1) in the given expression to show the answer gives (a.b/a,1) and therefore the expression holds.

Not sure if this is satisfactory?

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2. ## Re: More Parsonson - Rational Numbers

When you say "I have tried replacing (x,1) with (b/a,1)" you are completely misunderstanding what is being said!

The whole point is that a rational number (which might be represented as b/a) can be written as the pair of integers (b, a). You can't replace x by b/a- x must be an integer. The definition of multiplication is given by $\displaystyle (p_1, q_1)\times (p_2, q_2)= (p_1p_2, q_1q_2)$ so that $\displaystyle (a, 1)\times (x, 1)= (ax, 1)$ so saying that $\displaystyle (x, 1)\times (a, 1)= (b, 1)$ means immediately that b= ax.

But then the definition of "equal" pairs, that $\displaystyle (a_1, b_1)= (a_2, b_2)$ means $\displaystyle a_1b_2= a_2b_1$, (b, a)= (x, 1) is the same as saying that b(1)= b= a(x), exactly what we had.

3. ## Re: More Parsonson - Rational Numbers

Originally Posted by HallsofIvy
When you say "I have tried replacing (x,1) with (b/a,1)" you are completely misunderstanding what is being said!

The whole point is that a rational number (which might be represented as b/a) can be written as the pair of integers (b, a). You can't replace x by b/a- x must be an integer. The definition of multiplication is given by $\displaystyle (p_1, q_1)\times (p_2, q_2)= (p_1p_2, q_1q_2)$ so that $\displaystyle (a, 1)\times (x, 1)= (ax, 1)$ so saying that $\displaystyle (x, 1)\times (a, 1)= (b, 1)$ means immediately that b= ax.

But then the definition of "equal" pairs, that $\displaystyle (a_1, b_1)= (a_2, b_2)$ means $\displaystyle a_1b_2= a_2b_1$, (b, a)= (x, 1) is the same as saying that b(1)= b= a(x), exactly what we had.
Oh dear. I'm not very good at this. I understand your explanation though. Thanks HallsofIvy.