Is ab = 1 ? (1) aba = a (2) bab = b

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15 Jul 2004, 21:49

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19. Is ab = 1 ?
(1) aba = a
(2) bab = b

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15 Jul 2004, 23:48

Answer is C

aba = a

From 1)
aba - a = 0 ----. a(ab-1)=0 ---- a = 0 or ab=1

From 2)

bab -b = = b = = or ab=1

So from 1 & 2

ab = 1

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16 Jul 2004, 03:46

Cant be C (which i thought of

)
aba=a & bab=b
hence, a=b=a^3=b^3 -> a=b=0 or a=b=-1 or a=b=1
cant determine whether ab=1 or not

E is the answer.
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16 Jul 2004, 05:22

E it is
Both combined, you can have all elements = 0 or all elements = 1 and the stem could or could not be true
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16 Jul 2004, 05:43

I think itz E.
a=b is all we get .
And a=0 or b=0 .Whatz the OA?

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16 Jul 2004, 05:44

K_s_r wrote:

Answer is C

aba = a

From 1)
aba - a = 0 ----. a(ab-1)=0 ---- a = 0 or ab=1

From 2)

bab -b = = b = = or ab=1

So from 1 & 2

ab = 1

If answer is indeed E, can someone point out a flaw in this solution?

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16 Jul 2004, 13:37

Let me see if I understand this correctly

Q: is ab=1

Method#1
know 1) aba=a which can be rewritten as a(ab-1)=0
know 2) bab=b which can be rewritten as b(ab-1)=0

From this:
a or b could equal 0
and
ab=1

Therefore C would be insufficient.

Alternative method #2
But plugging in a and b into ab=1, we get
(aba)(bab)=1 which rearranged is (b^3)(a^3)=1
The only way for this formula to be true would be if
b and a were 1
or
b and a were -1

Conclusion,

Method #1 tells us ab=1 which is true, but that a and b could also be 0. Thus inconclusive

Method #2 tells us a and b must both be 1 or both be -1.
Thus inconclusive

Let me know if I'm off here please.
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16 Jul 2004, 23:36

Paul wrote:

E it is
Both combined, you can have all elements = 0 or all elements = 1 and the stem could or could not be true

Paul can exaplin littel more,

waht was wrong with my method, mathematically or logicaly

[#permalink] 16 Jul 2004, 23:36

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Is ab = 1 ? (1) aba = a (2) bab = b

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