2007-12-07 07:01:22 · 13 answers · asked by mody m 1 in Science & Mathematics ➔ Mathematics
Let a=b.
a^2 = a.b
a^2 + a^2 = a^2 + a.b
2(a^2) = a^2 + a.b
2(a^2) - 2ab = a^2 + a.b - 2ab
2(a^2) - 2ab = a^2 - a.b
2(a^2 - a.b) = 1.(a^2 - ab)
2 = 1
Q.E.D
Quad Erroneous Demonstrandum!!
2007-12-07 07:06:14 · answer #1 · answered by Emma Jean 7 · 4⤊ 1⤋
let 1 = 1 pair of shoes
let 2 = 2 shoes
then 1 = 2
2007-12-07 07:04:34 · answer #2 · answered by Anonymous · 0⤊ 0⤋
We know if : a^n=a^m,
then n=m
now 1^1=1^0
therefore 1=0
add 1 both sides
2=1 or 1=2
2007-12-07 07:06:53 · answer #3 · answered by shanu_gupta2003 2 · 0⤊ 1⤋
1 whole is 2 halves
2007-12-07 07:05:03 · answer #4 · answered by Anonymous · 0⤊ 0⤋
The usual middle-school-level "proof" is
a = b
multiply left by a, right by b (fine as they're equal)...
a² = ab
subtract b² from both sides...
a² - b² = ab - b²
factor both sides...
( a + b ) ( a - b ) = b ( a - b )
divide both sides by ( a - b )...
a + b = b
substitute a = b...
b + b = b
add like terms...
2b = b
divide both sides by b...
2 = 1
Now - why is this "proof" invalid?
2007-12-07 07:06:19 · answer #5 · answered by jgoulden 7 · 2⤊ 1⤋
Is this English related? 1 could be one pair or one dozen depends on the speaker
2007-12-07 09:14:43 · answer #6 · answered by Anonymous · 0⤊ 1⤋
if (a=b) is true, then (a-b) would equal 0
thats why the proof is incorrect and 1 does not equal 2
2007-12-07 07:08:57 · answer #7 · answered by Anonymous · 0⤊ 1⤋
Rise one finger. Now rise two fingers. Obviously it is not the same thing.
2007-12-07 07:08:57 · answer #8 · answered by vahucel 6 · 1⤊ 0⤋
Can't do it unless you allow me to divide by 0.
2007-12-07 07:08:10 · answer #9 · answered by historian 4 · 0⤊ 1⤋
you can verify it.
I give you $1 , you give me $2 ,
they are equal.
2007-12-07 07:14:34 · answer #10 · answered by Any day 6 · 0⤊ 1⤋