Does anyone have any guesses on who R.A.B is??? its driving me crazy!!!
2007-03-24 02:46:43 · 9 answers · asked by ♥miss liss♥ 3 in Arts & Humanities ➔ Books & Authors
My dad said it might be possible that it could be borgin... from the borgin and burkes shop but idk
2007-03-24 07:18:20 · update #1
The Regulus Black theory is backed up by the fact that in the Dutch translation the initials are 'R.A.Z.' and 'Black' happens to be translated to 'Zwarts'; no other names are spelled with a 'B' in English and a 'Z' in Dutch. Of course it's remotely possible the translator made a false assumption, but I know a few translators and it's unlikely... usually they check stuff like this out with the publisher and/or the author.
2007-03-25 11:30:27 · answer #1 · answered by Sheriam 7 · 0⤊ 0⤋
I agree, when I first saw the initials R.A.B. I thought it was Regulus, but knowing how Rowling writes she'll probably create a new character just to throw everybody off.
2007-03-24 10:59:12 · answer #2 · answered by Mizz Nay 2 · 1⤊ 0⤋
Well, most people are thinking its Regulus Black. The "A" part is his middle name. He was Sirius's brother and former Death Eater. We'll really just have to wait and see though.
2007-03-24 10:34:05 · answer #3 · answered by f.ireworks 2 · 0⤊ 0⤋
RAB stands for regulas black or sirius's brother. the A is supposedly a middle initial.
2007-03-24 11:09:29 · answer #4 · answered by Anonymous · 0⤊ 0⤋
I agree with Mizz Nay
2007-03-24 15:04:32 · answer #5 · answered by KU Fan 2 · 0⤊ 0⤋
It is Regulus Black, Sirius' brother.
2007-03-24 16:18:41 · answer #6 · answered by Anonymous · 0⤊ 0⤋
im goin with everyone else. regalus black. that makes sense, plus i cant think of ne else's name that starts with r except remus but his last name is lupin. so it has to be sirius' brother.
2007-03-24 13:27:03 · answer #7 · answered by Ay Pey Bay 4 · 0⤊ 0⤋
its the brother of sirius black
2007-03-24 10:39:16 · answer #8 · answered by ? 5 · 0⤊ 0⤋
everyone's already told ya :(
2007-03-24 12:05:06 · answer #9 · answered by Malfoy vs Potter 5 · 0⤊ 0⤋