Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
What am I doing wrong, because 1+1 DOES NOT equal 1...?
Anybody got an explanation for this?
Problem: A=1, B=1.
A=B.
A^2=AB (multiply both sides by A. so far, so good.))
A^2 - B^2 = AB - B^2 (Subtract B^2 from both sides... still is legitimate.)
(A+B)(A-B) = (B)(A-B) (factoring... part 1 expands to A^2 + AB - AB - B^2, so that's good factoring, part 2 expands to AB - B^2, so that's good factoring too.)
(A+B) = B (divide both sides by A-B, still good...)
1+1 = 1 (substitute values.)
How on earth do legitimate math procedures make this 1+1=1? Am I missing something?
6 Answers
- 1 decade ago
subtracting B^2 from both sides is probably your problem. if A^2 is 1 and B^2 is 1 and you take 1 from 1 you get a whopping ZERO, and once you get a zero in algebra you end up going in circles (as the girl above me beautifully demonstrated).
:).
- Anonymous1 decade ago
A - B = 0
You are dividing by 0 in the 4th step.
- How do you think about the answers? You can sign in to vote the answer.
- 1 decade ago
Let us depart from you at (A+B)(A-B)=B(A-B). The expression leads to
(A-B)(A+B-B)=0
(A-B)*A=0
A=1 is not zero; therefore, A-B=0, or A=B
That is the only legitimate result.
