Question

let * be defined on Q(rational number) by letting a*b=ab.
how to show that it has no inverse?

Answer 1

@lgbasallote Can you show us how to do it .. ?

Answer 2

honestly...i have no idea what the question means o.O sorry

Answer 3

well i dont really get the qn but....
for inverse to exist it shud be 1-1 nd onto
may be u can find some c nd such that
c*d=a*b

Answer 4

me too,
but I just wanna me notified when some will do it ;)

Answer 5

well is it like ab is a number nd a*b = ab?
like 0*0=00 ?

Answer 6

actually the question ask whether the set is a group or not...

Answer 7

the set is a group....
Let a,b, and c be any elements of the real number
ok so first, we need to prove it is closed under that binary operation * defined as a*b=ab
so if we have two elements a and b which are rational numbers, definitely when you multiply them it will still be a rational number :)). Now if we have a,b and c which are elemnts of real number, then we know that (a*b)*c=(ab)c=a(bc)=a*(b*c) and is also in the set. Now we need to prove that there is an identity element which is also in the set. So in this case our identity is 1 which is an element of real numbers since a*1=ax1=a.
now we need to find an inverse for each element such that if b is an inverse of a, then a*b=identity element=1 . So in this case, the inverse for any element a in R will just be 1/a which is still in the set, and since a*(1/a)=a(1/a)=1.
Now since the set is closed under the binary operation *, an identity element exists which is in the set, and that for every element in the set there is an inverse for it which is also in the set, then the set of all real numbers with a binary operation of multiplication is a group.