In G={x∈R | x>1} is defined operation * with x*y = xy-x-y+2. Show that {G,*} is Albelian group
thanks
we need to show 4 things:
1. * is an associative binary operation on G. this means two things:
a) x*y has to be in G
b) for any 3 elements x,y,z of G, (x*y)*z = x*(y*z)
2. there is some element e in G, with x*e = e*x, for ALL x in G.
3. for any x in G, there is some element y in G with x*y = y*x = e.
4. for any x,y in G, x*y = y*x.
so, let's look at #1 first. suppose x,y are in G. this means x,y are both real numbers greater than 1. we need to see if x*y is also a real number greater than 1.
well, x*y = xy - x - y + 2, and xy - x - y + 2 is certainly a real number. but is it greater than 1?
xy - x - y + 2 = (x - 1)(y - 1) + 1.
now x > 1, so x - 1 > 0.
similarly, y > 1, so y - 1 > 0. therefore (x - 1)(y - 1) > 0, so (x - 1)(y - 1) + 1 > 0 + 1 = 1. so, yes, x*y > 1. this show closure, so * is indeed a binary operation on G.
so now we can turn our attention to whether or not * is associative:
(x*y)*z = (xy - x - y + 2)*z = (xy - x - y + 2)(z) - (xy - x - y + 2) - z + 2
= xyz - xz - yz + 2z - xy + x + y - 2 - z + 2
= xyz - xz - yz - xy + x + y + z (2z - z = z, and -2 + 2 = 0)
= x(yz - y - z) - (yz - y - z) + x
= x(yz - y - z) - (yz - y - z) + 2x - x - 2 + 2
= x(yz - y - z + 2) - x - (yz - y - z + 2) + 2
= x*(yz - y - z + 2) = x*(y*z).
so * is indeed associative. what do you suppose e and x-inverse might be?
The important part of the identity group axiom is not that x * e = e * x, but that both of those expressions equal x. Also, note that x * e should equal x, but xe (with regular multiplication) is not necessarily x.
We have that x * e is a polynomial of x. Use the fact that two polynomials are equal for all values of x iff their corresponding coefficients are equal. From there you can find what e is.
we are looking for some real number e > 1, with the property that:
x*e = e*x = x for any real number x > 1. you should prove that x*y = y*x yourself, it's not hard, and you might learn something. but if we take that as given, we only need to find some e with x*e = x (since then e*x = x*e anyway, so will have to equal x).
to prove this, we need to use the definition of *.
x*e = xe - x - e + 2.
we want this to equal x, so we set it equal to x, and try to solve for e:
xe - x - e + 2 = x
now, solve for e, using ordinary algebra. if, for some strange reason, you need to divide by x - 1 at some point, rest assured that this is ok, since x > 1, x - 1 > 0, so in particular it never equals 0 (so we can divide by it).
So x*e = e*x = x
xe - x - e + 2 = x
xe - e = 2x -2
e(x - 1) = 2x - 2 => e = (2x - 2)/(x - 1)
and x*y = y*x = e
x*y = e
xy - x - y + 2 = e
xy - y = e + x - 2
y(x - 1) = e + x - 2 => y = (e + x - 2)/(x - 1)
and what about x*y=y*x ???
now start with:
x*y = 2 that is:
xy - x - y + 2 = 2
and solve for y in terms of x.
you should arrive at y = x^{-1} = x/(x-1) easily. note the x-1 in the denominator is ok, since x > 1. there *is* one thing you need to be sure of, however.
you need to check that x/(x-1) is *in* G, that is: that x/(x-1) > 1, when x > 1. first check that top and bottom are positive, then "cross-multiply".
then check your answer by computing:
x*(x/(x-1)). if you've done everything right up to here, you should get 2.