# Groups - discrete math

• Jun 15th 2012, 01:13 PM
kljoki
Groups - discrete math
In G={x∈R | x>1} is defined operation * with x*y = xy-x-y+2. Show that {G,*} is Albelian group
thanks
• Jun 15th 2012, 01:29 PM
Plato
Re: Groups - discrete math
Quote:

Originally Posted by kljoki
In G={x∈R | x>1} is defined operation * with x*y = xy-x-y+2. Show that {G,*} is Albelian group thanks

What is the identity for this operation?
• Jun 15th 2012, 02:35 PM
Deveno
Re: Groups - discrete math
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?
• Jun 15th 2012, 02:46 PM
Plato
Re: Groups - discrete math
Oh how I wish we could allow the student to show some effort.
Giving a premature solution in my opinion is counterproductive.
• Jun 15th 2012, 04:34 PM
kljoki
Re: Groups - discrete math
x*e=(xe - x - e + 2) (xe = x)
=x - x - e + 2 = -e + 2 + x - x = -e + 2 + ex - x = ex - e - x + 2 = e*x ??????
• Jun 15th 2012, 04:55 PM
Plato
Re: Groups - discrete math
Quote:

Originally Posted by kljoki
x*e=(xe - x - e + 2) (xe = x)
=x - x - e + 2 = -e + 2 + x - x = -e + 2 + ex - x = ex - e - x + 2 = e*x ??????

You are making my points!
Do you even know what the identity for this operation is?
Am I wrong?
• Jun 15th 2012, 05:01 PM
emakarov
Re: Groups - discrete math
Quote:

Originally Posted by kljoki
x*e=(xe - x - e + 2) (xe = x)
=x - x - e + 2 = -e + 2 + x - x = -e + 2 + ex - x = ex - e - x + 2 = e*x ??????

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.
• Jun 15th 2012, 05:26 PM
kljoki
Re: Groups - discrete math
Quote:

Originally Posted by Plato
You are making my points!
Do you even know what the identity for this operation is?
Am I wrong?

I don't have idea, only this was given to the exam :/
• Jun 15th 2012, 09:25 PM
Deveno
Re: Groups - discrete math
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).
• Jun 16th 2012, 02:43 AM
kljoki
Re: Groups - discrete math
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)

• Jun 16th 2012, 12:01 PM
Deveno
Re: Groups - discrete math
e = (2x-2)/(x-1)

you're fine up to here.

but.....

2x-2 = 2(x-1)

do you think you might be able to simplify a little?
• Jun 16th 2012, 02:05 PM
kljoki
Re: Groups - discrete math
ups :)))
so e=2
and should i leave y in next proof like this y = (e + x - 2)/(x - 1) or change for e=2 so y = x/(x - 1)
• Jun 16th 2012, 10:04 PM
Deveno
Re: Groups - discrete math

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".