AB operation
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I know this is easy but I can't seem to see it....
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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.
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.
- ssmiles08
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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)
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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