Question

Let S = {x ER:[x]<1}=(-1,1). We will refer to E as hyperbolic relativity space. Now a+b define a binary operation by: if a,beR and ab +-1, then aob= 1+ ab Proposition 1. (5,0) is a group. Remark. This is the kind of problem that every student should become competent at doing. Perhaps some of the details here are more challenging than normally but understanding what are the steps to follow in such a problem is basic, and everyone should understand this. Proof. Here are the steps we must prove: 1. We must show that the operation on the set s, which means that if the operation is applied to any two elements of E, the result is also in E. Note: this property is also described by saying that E is closed under the operation 2. We must show that the operation is associative. That is: for any a,b,c es ao(bºc)=(a oboc 3. We must show that there is an identity element ees. That is: there is an element e e such that for any a E, aoereoa=a. 4. Finally, we must prove that every element of a € 5 has an inverse a-' €3. That is: for every a € 5 there is an a-e such that aoa'ra 'oare

Answer 1

ANSWER:

& ab ER but abf1 of (9,6) ER = = LAER: 141214 = (-1,1) {ab 7-1, a ob = atb 1 tab closure property A, LER =) ather atb ER tab a ob ER- i.e. a ob satisfies callosure property m=(-4,1) Anociative property! to any a, b, CEE 6+4 aolboc) = 1 tbc at (br a tabctbtc btc It bct abtac 1+ a. (btc) (1+bc) atb atb tc I tab 1t/ats (doblo c = It Cortesi) atbt (tabe I tabtactbc aolboc) (aoboc = " is associative

Exintence of Identity, let e EE be any identity elepest & at such that aoe = a coa 1 ate a lta e a te=atare e = aze e=a²e - e-a²e=0 -) ell-44) te=0 g a2=1 at/ but at (-1,1) eo af ti प Inverse, then there enis b=a 'GE a EE such that a obe ath = 0 Itab atba 6 = -a inverse ed has a 1-a. "(Ey isagroup.