There aren't any two *distinct* numbers that are equal to each other. The definition of equality necessarily says that if x = y, they are the same number. So, I suspect this may be a trick question.
(Oh, and to the respondents above me: there is such a thing as - 0; it happens to be equal to 0. The negative of a given number x is its additive inverse, ie. the unique number -x, which, when added to x, yields zero. And as 0 + 0 = 0, -0 = 0.)
2007-03-27 22:44:30
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answer #1
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answered by MHW 5
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What kind of question is that? You might as well ask how much does something have to weight to weigh 2 ounces? And there is NO SUCH THING AS -0! lol help yourself to a math book dude
2007-03-28 05:13:34
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answer #2
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answered by Anonymous
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it is a tricky question indeed, but try this:
if 1! = 1, and 0! = 1, then 1 = 0
2007-03-28 06:38:33
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answer #3
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answered by The Pretender 2
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1=1, 2=2,3=3, 4=4, 5=5, 6=6, 7=7, 8=8, 9=9, 10=10, 11=11, 12=12, 13=13, 14=14, 15=15..................................................
2007-03-28 05:47:03
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answer #4
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answered by Lexu 1
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know wat ???
1=2 !!!
want the proof ?? look into this...
Consider that a=b,
multiply by 'a' on both sides,
-> a^2=ab
subtract b^2 on both sides,
-> a^2-b^2=ab-b^2
we know by identity that a^2-b^2=(a+b)(a-b),
->(a+b)(a-)=ab-b^2
taking 'b' as common out from the RHS,
-> (a+b)(a-b)=b(a-b)
cancelling (a-b) from both sides,
-> (a+b)=b
but since a=b as per our assumption,
-> b+b=b
that becomes,
-> 2b=b
cancelling 'b' on both sides,
-> 2=1 !!!
got the proof ?!?!?
now if u r clever enuf, find out the flaw in the above solution !!!
2007-03-28 08:22:39
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answer #5
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answered by j 2
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Minus zero and zero
2007-03-28 05:12:43
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answer #6
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answered by M00ND0CT0R 6
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yeh it is a trick question
minus zero lol where do u do ur maths? lol
2007-03-28 05:40:42
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answer #7
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answered by Allison 1
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