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1. Hi

I am in need of some help when working out a maths question in order to determine whether a set is a group using the four group axioms, that is;

G1 Closure,
G2 Identity,
G3 Inverses and,
G4 Associativity.

The set is (R , o) where x o y = x + y + e

I am having real trouble even identifying the mathematics behind each of the group axioms in order to determine whether or not this forms a group! Any help in order to determine this example question would be greatly appreciated! Many thanks all!

p.s the 'R' in the set of all real numbers and also o is the binary option.
2. (Original post by alex.kundert)
Hi

I am in need of some help when working out a maths question in order to determine whether a set is a group using the four group axioms, that is;

G1 Closure,
G2 Identity,
G3 Inverses and,
G4 Associativity.

The set is (R , o) where x o y = x + y + e

I am having real trouble even identifying the mathematics behind each of the group axioms in order to determine whether or not this forms a group! Any help in order to determine this example question would be greatly appreciated! Many thanks all!

p.s the 'R' in the set of all real numbers and also o is the binary option.
The group is (R , o), the set is R.

G1: Is x o y in R? If so, prove it.
G2: Does the identity exist such that x o e = e o x = e? Prove it.
G3: Does every element have an inverse? That is, for all x do we have x o x^(-1) = x^(-1) o x = e. Prove it.
G4: If (x o y) o z = x o (y o z)? If so, prove it.
3. (Original post by Zacken)
The group is (R , o), the set is R.

G1: Is x o y in R? If so, prove it.
G2: Does the identity exist such that x o e = e o x = e? Prove it.
G3: Does every element have an inverse? That is, for all x do we have x o x^(-1) = x^(-1) o x = e. Prove it.
G4: If (x o y) o z = x o (y o z)? If so, prove it.
Oh thanks a bunch for that.

For G2, the e is different to the 'e' I have posted in the question, right? Because the e in the question is relating to the irrational number e and the test in G2 is relating to the identity element?

I think my specific question was around determining whether or not each one of these axioms holds, such that they do actually show that this set constitutes a group based on the binary option. I think I am having trouble putting the proverbial 'pen to paper' and actually providing evidence against each one of these - effectively I am not particularly sure how you actually show that this is the case.
4. (Original post by alex.kundert)
Oh thanks a bunch for that.

For G2, the e is different to the 'e' I have posted in the question, right? Because the e in the question is relating to the irrational number e and the test in G2 is relating to the identity element?

I think my specific question was around determining whether or not each one of these axioms holds, such that they do actually show that this set constitutes a group based on the binary option. I think I am having trouble putting the proverbial 'pen to paper' and actually providing evidence against each one of these - effectively I am not particularly sure how you actually show that this is the case.
Yes, they're different. It's very common to use e for the identity of an arbitrary group.

As for proving the group properties.. well, you just do it. The defining property of an identity is that id o x = x o id = x. So, if this group has an identity, then x = x o id = x + id + e. So we must have x = x + id + e for all real x. Does a number id exist which makes this true for all x? If so - you've found your identity.

Similarly for inverses. Compute x o inv(x) in symbols - can you find what inv(x) will be?

Associativity, (G4), again just do it. Compute x o (y o z) and compute (x o y) o z. Are they the same or not?

Closure's just obvious really; adding real numbers gives real numbers .
5. (Original post by FireGarden)
Yes, they're different. It's very common to use e for the identity of an arbitrary group.

As for proving the group properties.. well, you just do it. The defining property of an identity is that id o x = x o id = x. So, if this group has an identity, then x = x o id = x + id + e. So we must have x = x + id + e for all real x. Does a number id exist which makes this true for all x? If so - you've found your identity.

Similarly for inverses. Compute x o inv(x) in symbols - can you find what inv(x) will be?

Associativity, (G4), again just do it. Compute x o (y o z) and compute (x o y) o z. Are they the same or not?

Closure's just obvious really; adding real numbers gives real numbers .

A question regarding Closure- can you define it as addition when the set operation is + e? Is this not some form of multiplication of elements?

For the new term introduced as d ( for identity) I am not sure what you mean when you are putting id for my specific example? I dont see how that follows?

Also for the inverse, is there a quick way to find out what this would be? I cant think as to how you would find that?

Regarding associativity - how do I determine what x , y and z are in order to compute this calculation?

I honestly really appreciate all of your help guys as I am self-teaching and finding group theory in paticular a big struggle!! I miss integration lol.
Posted from TSR Mobile
6. Let's work our way through the list:
1) Closure: this is pretty obvious: since e is a real number, and adding any three real numbers together gives you a real number (easy corollary of the fact that the real numbers form a group under addition), you're done.
2) Identity: Is there some number x such that for every other number, adding x to y then adding e gives you y again? (Yes, there is: it should be pretty obvious what x should be if you just write this equation down)
3) Inverses: If we call that identity that we've just found i (or just call it what it is, once you know what that is), and given any number x, is there something that we can add to x to get to i? (Again, this should be pretty obvious, and it should be pretty obvious what the inverse should be).
4) Distributivity: This should be easy to prove from the same for the real numbers under ordinary addition.
7. (Original post by alex7892)

A question regarding Closure- can you define it as addition when the set operation is + e? Is this not some form of multiplication of elements?

For the new term introduced as d ( for identity) I am not sure what you mean when you are putting id for my specific example? I dont see how that follows?

Also for the inverse, is there a quick way to find out what this would be? I cant think as to how you would find that?

Regarding associativity - how do I determine what x , y and z are in order to compute this calculation?

I honestly really appreciate all of your help guys as I am self-teaching and finding group theory in paticular a big struggle!! I miss integration lol.
Posted from TSR Mobile
For closure, we know adding real numbers gives a real number. x o y, the operation of the group, evaluates as x+y+e, which is the addition of real numbers. So the group operation takes two real numbers, and gives out a real number. It's closed.

Suppose d is the identity element. To prove it exists it is enough to just find out what it is. If d exists, then x = x o d = x + d + e.

So then x = x + d + e, and so further 0 = d + e. Obviously then, d = -e, and that's the identity.

Inverses are done the same as the identity. Let x be any element, and write X for its inverse. Then x o X = x + X + e = -e. The first equality is how the operation works, the second is because x and X are inverse to each other - their product has to be the identity, which we found earlier to be -e. Just rearranging we find that X = -(x+2e). So there's inverses.

Now associativity. You don't need to know anything other than the group operation to do this for this group, since we know it's about real numbers.

x o ( y o z) = x o (y+z+e) = x+y+z+e+e = x+y+z+2e.

(x o y) o z = (x+y+e) o z = x+y+e+z+e = x+y+z+2e.

They match; x, y, z were arbitrary so it matches always. Thus associative.

Perhaps now with this example it's more clear what is meant by just checking the axioms.
8. (Original post by FireGarden)
Inverses are done the same as the identity. Let x be any element, and write X for its inverse. Then x o X = x + X + e = -e. The first equality is how the operation works, the second is because x and X are inverse to each other - their product has to be the identity, which we found earlier to be -e. Just rearranging we find that X = -(x+e). So there's inverses.

x o ( y o z) = x o (y+z+e) = x+y+z+e+e = x+y+z+2e.

(x o y) o z = (x+y+e) o z = x+y+e+z+e = x+y+z+2e.
You should probably have a quick note saying that addition over the reals is commutative when simplifying the associative thing into x+ y + z + 2e (like how you justified closure by using the closure of real numbers under addition).

Also, I might be going crazy here, but is it not - probably just a typo on your part, but I'm pointing it out in case the OP gets confused.
9. (Original post by Zacken)
You should probably have a quick note saying that addition over the reals is commutative when simplifying the associative thing into x+ y + z + 2e (like how you justified closure by using the closure of real numbers under addition).

Also, I might be going crazy here, but is it not - probably just a typo on your part, but I'm pointing it out in case the OP gets confused.
You're right, it's -(x+2e). Silly slip.

When I wrote the final =x+y+z+2e on both lines, I intended that to be the statement "because real number addition lets us do this (i.e., they're commutative); see that the results are the same". I guess it may be too terse, but now your post explains it .
10. (Original post by FireGarden)
You're right, it's -(x+2e). Silly slip.

When I wrote the final =x+y+z+2e on both lines, I intended that to be the statement "because real number addition lets us do this (i.e., they're commutative); see that the results are the same". I guess it may be too terse, but now your post explains it .
Great! Hopefully this is enough for the OP. :-)
11. (Original post by Zacken)
Great! Hopefully this is enough for the OP. :-)
Yes, this really is. Thanks for investing your time above Zacken and FireGarden into helping me understand this concept and specific example! Really appreciated! Have a nice evening both,

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