so 1=0 am i right or math is wrong
2006-08-10 09:22:26 · 8 answers · asked by omar g 2 in Science & Mathematics ➔ Mathematics
sorry DDK it's not my fault that u don't know this law
2006-08-10 09:34:22 · update #1
wrong! If F(x) == F(y), that doesn't mean X == Y,
take the function:
F(x) = 4
so a straight horizontal line
F(0) == F(1) == F(2)
Does that mean 0 == 1== 2?
2006-08-10 10:04:04 · answer #1 · answered by leatherbiker040 4 · 2⤊ 0⤋
Sorry, but you are wrong. If a=b (1! = 1) and b=c (1=0!) then a=c, aka 1! = 0!. There is no law that I know of in mathematics that states if a!=b! then a=b.
2006-08-10 09:30:33 · answer #2 · answered by DDK 2 · 2⤊ 0⤋
The math is wrong because 0! does not equal 0.
2006-08-10 09:27:11 · answer #3 · answered by bpiguy 7 · 0⤊ 2⤋
DDK is correct.
"If x!=y!, than x=y" is not a true statement.
The reason DDK does not "know" this law is because it does not exist.
2006-08-10 11:08:06 · answer #4 · answered by MsMath 7 · 1⤊ 0⤋
1! does not mean 1.
The math was right until then.
2006-08-10 09:28:50 · answer #5 · answered by billyandgaby 7 · 0⤊ 2⤋
You need to look at the definition of a function again.
Only functions that are one-to-one would have this property.
The gamma function fails the horizontal line test, making this argument invalid.
2006-08-10 09:31:14 · answer #6 · answered by Anonymous · 2⤊ 0⤋
You yourself have posted an example of where x! = y!, but x is not equal to y. Your assumption is invalid. This only holds for injective functions, which the factorial is not.
2006-08-10 12:40:17 · answer #7 · answered by AnyMouse 3 · 0⤊ 0⤋
No.
Your statement about x! = y! => x=y is WRONG
1! = 0! is proof of that
2006-08-10 09:31:47 · answer #8 · answered by Anonymous · 2⤊ 0⤋