can any one prove 1=0?

Question

1/0=&(infinity)-a
0/1=0 -b
a*b
1/0*0/1=&*0
0/0=0 (multiplyin th denominators and th numerators)
1=0 (any no divided by that no is 1)
hence
1=0___

any comments?

Answer 1

Can someone explain to me the satisfaction gained in "proving" what is known to be untrue and can only be "proved" by disguising the deception involved. Is it the appeal of "Magic" conjuring tricks?

Dividing by zero is simply a sleight of hand which any half-educated person knows is not a legitimate move you can make. So all such "proofs" are at pains to disguise that is what they are doing. And once spotted, are revealed to be a hollow sham.

THIS DOESN'T SEEM A USEFUL WAY FOR ANYONE TO SPEND THEIR TIME, TO ME!

Answer 2

Yes. Crappy logic. Division is defined as 'the number of times one number can be subtracted from another'.

If you start with 1 (or any other number), and subtract 0, you can continue doing that an infinite number of times. So 'infinity' is NOT a number and to assume that you can subtract anything from infinity is false.

1/0 = ∞ is a true statement. 1/0 =∞-a is not merely untrue, it is invalid unless a=0

0/1 = 0 is also true, but 0/1 = 0-b is only true if b=0

In other words you are applying computaional practice to things which are NOT numbers. 0 is not a 'number' by definition, ∞ is not a number because it simply means 'no end'.

If you could prove that unity is equivalent to zero, then you would have proved that God does not exist (but that is entirely another argument.)

Answer 3

Take the following identity...
-12 = -12
Express it in an equivalent way...
9 - 21 = 16 - 28
Re-express them...
3² - 3*7 = 4² - 4*7
Add 49/4 to each side to complete the square...
3² - 3*7 + 49/4 = 4² - 4*7 + 49/4
Factor each side...
(3 - 7/2)² = (4 - 7/2)²
Take the square roots...
3 - 7/2 = 4 - 7/2
Now, add 1/2 to each side:
3 - 6/2 = 4 - 6/2
Which equals 0 = 1. Simple as that!

Try finding the conditional equality in my proof. Good luck!

BTW, this proof was derivated from another person's trying to figure out that 2 + 2 = 5.

EDIT: Here's another proof with an obvious error that should help you find the conditional equality:

1 = 1
square root both sides:
sqrt(1) = sqrt(1)
1 = -1
add 1 to both sides
2 = 0
divide both sides by 2:
1 = 0.

Answer 4

comments a lots

u are palying with maths like anything


u must know that

infinity multiplies by 0 is not 0->its not a defined number

secondly

a/a ==1 if and only if a bot equal to zero

if a =0 a/a is undefined

so ur proof is completely wrong


along with it i may give u 1 more solution


let

a=b=1

the a^2 = ab

a^2-ab=0

a(a-b)=0

dividing by (a-b)

a=0

but initially we took a=1

so 1=0

whats the pitfall

here u r dividing

in rhs 0 by (a-b) //but a-b is already zero so this step is wrong

Answer 5

2 power 0 =1 power 1 . bases not are equal then powers should not be equal. 1 is not 0. One disproof is enough to show. A foolish answer for a foolish question.;-)

Answer 6

1/0 is not defined
why u take infinity
how u multiply infinituy by zro
& * 0 not equal to zero
0/0 also not defined


for 0/0 u read srinivasa ramanjan story
if u not found i will tell the story

Answer 7

0=0+0+0+0.............

1-1=0

substitute 1-1 for 0
0= (1-1)+(1-1)+(1-1)+(1-1)+..........

by use of associative addition we can get

0=1+ (-1+1)+(-1+1)+(-1+1)+........

which is the same as 0=1+0+0+0+0........

ergo 0=1

Answer 8

in the last u said any no divided by that no is 1
then how u can get 1/0=infinity

Answer 9

well, good attempt but u failed to prove 1=0 bcz 1 is not equal to 0. accept it ! try n prove valid mathematical expressions n who knows one day u might put pythogoras to shame . all the best . have fun

Answer 10

it is not possible to divide anything by zero .. so your proof of 1=0 is incorrect .... There is only one case where 1 = 0 ... and that is by MISTAKE .. :)