Question

Rational number question

Answer 1

If I give you a rational number reduced to lowest terms,

\[n=\frac{a}{b}\]

and I don't tell you what a and b are, is it possible just using n alone to find an expression \(f(n)\) for

\[f(n)=a+b\]

Answer 2

Your notation suggests that \(f\) is a function.

By definition of function, it must satisfy :
\[p=q\implies f(p) = f(q)\]

Answer 3

Easy to find a counterexample
let p = 1/2, q = 2/4

Answer 4

I don't think that's fair since n is always known to be reduced to lowest terms

Answer 5

Oh wait, then it requires some thinking...

Answer 6

then \[f(1/2) = f(2/4)\]

Answer 7

you should be able to use the first equation to solve for a and b in terms of n
substitute those into the second equation' and then, if we're lucky there might be a way to get rid of a and b altogether so we have just an equations containing n

Answer 8

if a is 1 and b is 3 then...

Answer 9

@Joshtray f(2/4) is invalid since \(n \ne \frac{2}{4}\)

Maybe this helps, this is always true:

\[\gcd(a,b)=1\]

Answer 10

but n can be 1/3

Answer 11

alright, I can see how you might get an expression for f(n) in terms of n, a, and b

Answer 12

but as far as getting rid of a and b, I'm not sure what they would be looking for there

Answer 13

f(a/b) = a+b

Answer 14

you're going backwards to the direction you're supposed to go to m8 @Joshtray

Answer 15

if n = a/b, then a = bn, and b = a/n

Answer 16

yeah

Answer 17

are you inputting n in decimal form ?

Answer 18

so take the second equation, and replace a with bn, and replace b with a/n

Answer 19

n right now is an unknown rational number

Answer 20

so no, we're not using any decimals at this point @ganeshie8

Answer 21

that is bn + a/n

Answer 22

just trying to get an expression for f(n) that includes n

Answer 23

and yes correct @Joshtray

Answer 24

and then you could combine that into one rational expression