Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Let (G,∗) be a binary structure that has the following properties: 1) The binary operation ∗ is associative. 2) There exists an element e∈G such that for all a∈G, e∗a=a (Existence of left Identity). 3) For all a∈G, there exists b∈G such that b∗a=e (existence of left inverses) Prove that (G,∗) is a group.

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

All you need to show, is that it also has a right identity, and a right inverse.
yea,,, I have hourssss trying to :(
That's true as well for non-Abelian groups, right?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Wait, yes it is. I was derping; I forgot the exact definition of Abelian for a second.
It is true for non-abelian groups too, but I'm starting to see why he was having problems. I keep getting that statements that use circular logic.
I know that in order to prove the right identity i use some how my left inverse... but i keep getting stuck
Man, I'm too sleepy for this. :(
bad reference... please... i have 3 nights with no sleep :(
I'm running in circles, and I can't tell if I'm just falling for a trap, or if I'm just derping because I'm sleepy.
:(
According to the always helpful wikipedia page http://en.wikipedia.org/wiki/Elementary_group_theory, Let \(y\) be the inverse of \(a*x\), and \(x\) be the inverse of \(a\). Then \[e=y*(a*x)\]\[e=y*(a*(e*x))\]\[e=y*(a*((x*a)*x))\]\[e=y*(a*(x*(a*x)))\]\[e=y*((a*x)*(a*x))\]\[e=(y*(a*x)*a)*x\]\[e=e*(a*x)\]\[e=a*x\]Thus we have a right inverse also.
As for right identity, let \(x\) be the inverse of \(a\). Then\[a*e=a*(x*a)\]\[a*e=(a*x)*a\]\[a*e=e*a\]\[a*e=a\]So we also have right identities.
That took some time to take apart and figure out exactly what was going on.
im still trying to understand the inverses
Any particular step in that proof?
well,, im trying to understand step by step... let me see if i get stuck
Woops, I tried proving it backwards using the right inverse. I feel dumb.
waittt
if x is the inverse of a and if we have left inverse the will be (x*a) nooo (a*x)
If that was unclear, it means that \(x*a=e\).
And we want to show that \(a*x=e\) as well.
yea... but u are using the right one
yea but you can apply that statement like its true.. u want to get that statement
In the proof, using meticulous use of associativity, it shows that if \(y*(a*x)=e\), and \(x*a=e\), then \(a*x=e\). Both \(y\) and \(x\) are left inverses of \(a*x\) and \(a\) respectively.
but my understanding is that u cant just say x*a=e=a*x.. u have to get the answer
We didn't just say that. By converting \(x\) into \(e*x\) and \(e\) into \(x*a\) (true by hypothesis), we can reorganize our terms to prove that \(x*a=a*x=e\)
ok let em check again
I need to go to bed now, but if you continue to have questions, feel free to ask and I'll get back to you tomorrow.
ok thanks you very much I kind of getting it..

Not the answer you are looking for?

Search for more explanations.

Ask your own question