AB operation

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AB operation

by joyseychow » Fri Aug 14, 2009 2:39 am
I know this is easy but I can't seem to see it....
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by truplayer256 » Fri Aug 14, 2009 6:24 am
Statement I: a+b-ab=b+a-ba. If you compare both sides, you can see that this is true for all integers a and b.

Statement II: a+0-0=a-->a=a. This is true for all integers a.

Statement III: (a*b)*c=(a+b-ab)*(c)=a+b-ab+c-(a+b-ab)(c)
a*(b*c)= a*(b+c-bc)=a+b+c-bc-(a)(b+c-bc)

Simplify both equations above to see if they're equal.

(a+b-ab)*(c)=a+b-ab+c-ca-cb+cab.

a*(b+c-bc)= a+b+c-bc-ba-ac+abc

Once again, if you compare both sides you can see that both expressions are equal for all integers a, b, and c.

* denotes the circle symbol.

E is the correct answer. Picking numbers for this problem would've saved you a lot of time. I just did it the long way so you can understand it better.
Last edited by truplayer256 on Fri Aug 14, 2009 6:35 am, edited 1 time in total.

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by ssmiles08 » Fri Aug 14, 2009 6:29 am
I am guessing you already figured out I and II and need to know why III is also the answer.

IMO picking numbers here help.

a = 1, b= 2, c = 3.

left side:
1+2 - 2 = 1

1 + 3 - 3 = 1

right side:
2+3 - 6 = -1

1 + (-1) - (-1*1) = 1

1 = 1; so it is a valid option.

(E)
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