Quick help please: Prove that * defined on the rationals is a group by letting a*b=ab.

Question

fibonaccichick666 · Answer

I have group axioms 1,2 and closed covered, but for some reason I can't figure out how to show the inverse in the rationals.

fibonaccichick666 · Answer

@amistre64 , could you take a quick look?

amistre64 · Answer

inverses are defined as:  x x' = e

amistre64 · Answer

group properties (in my text) were defined as:

identity, inverses, associativity;  for a closed binary operation

fibonaccichick666 · Answer

right ooh ok, so let's say a=c/d and b=f/g so ab=(cf)/(dg) which is an element of the rationals, so (ab)^-1=(dg)/(cf) right?

fibonaccichick666 · Answer

yea, it's just inverses I seem to always get stuck on proving but, same

amistre64 · Answer

yes

fibonaccichick666 · Answer

thank you!!

amistre64 · Answer

let a = n/m  for some integers m,n;  m not 0
$$\frac nma'=e$$
$$\frac{m}{1}\frac nma'=\frac{m}{1}$$
$$\frac n1a'=\frac{m}{1}$$
$$\frac 1n\frac n1a'=\frac{m}{1}\frac{1}{n}$$
$$a'=\frac{m}{n}\in Q$$

amistre64 · Answer

or some such reasoning :)  yours was fine for me

fibonaccichick666 · Answer

thanks, ya know sometimes you just have to rubber duck with someone haha  thanks again!