Prove that 1=2?

Question

I m giviu a hint, a=b
its goin to be a tough one

Answer 1

There is no proof that 1 = 2. If 1 were really equal to 2, the mathematical foundation would crumble.

Here is the false proof that 1 = 2.

Let x = 1 and y = 1. Then

x = y. Multiply both sides by x,

x^2 = xy. Subtract y^2 both sides,

x^2 - y^2 = xy - y^2. Factor both sides,

(x - y)(x + y) = y(x - y). Cancel the common term of (x - y) on both sides.

x + y = y

Resubstitute x = 1 and y = 1

1 + 1 = 1
2 = 1

The only reason why this appears to work is because of an invalid step. Nevertheless, this is the false proof.

Answer 2

It really is not tough at all. There are usually two equations that people use to "prove" that 1=2. See my comment below.

1=2: A Proof using Beginning Algebra
-----------------------------
Step 1: Let a=b.
Step 2: Then a^2 = ab,
Step 3: a^2 + a^2 = a^2 + ab,
Step 4: 2 a^2 = a^2 + ab,
Step 5: 2 a^2 - 2 ab = a^2 + ab - 2 ab,
Step 6: and 2 a^2 - 2 ab = a^2 - ab.

Step 7: This can be written as 2 (a^2 - a b) = 1 (a^2 - a b),
Step 8: and canceling the (a^2 - ab) from both sides gives 1=2.
================
1=2: A Proof using Complex Numbers
-----------------------------
Step 1: -1/1 = 1/-1

Step 2: Taking the square root of both sides: sqrt(-1/1) = sqrt(1/-1) (where "sqrt" denotes the square-root symbol which cannot be displayed on text-only browsers.)

Step 3: Simplifying: sqrt(-1) / sqrt(1) = sqrt(1) / sqrt(-1)

Step 4: In other words, i/1 = 1/i.

Step 5: Therefore, i / 2 = 1 / (2i),

Step 6: i/2 + 3/(2i) = 1/(2i) + 3/(2i),

Step 7: i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ),

Step 8: (i^2)/2 + (3i)/2i = i/(2i) + (3i)/(2i),

Step 9: (-1)/2 + 3/2 = 1/2 + 3/2,

Step 10: and this shows that 1=2.
--------------------------------------------------
Now that we have seen the "how to," let us consider that this question is regarding what are called "Classic Fallacies"

Each equation has a fallacy in one of it's steps. Said in a different manner, it simply is not true that 1=2, but then if one had common sense, they would not be trying to prove it.

Answer 3

Given a = b therefore (a – b) = 0
1 × (a – b) = 0
2 × (a – b) = 0

Equating both 1 = 2

Answer 4

let a = b

a² = ab
Multiply both sides by a
a² + a² - 2ab = ab + a² - 2ab
Add (a² - 2ab) to both sides
2(a² - ab) = a² - ab
Factor the left, and collect like terms on the right
2 = 1
Divide both sides by (a² - ab)

Answer 5

u've given so, a=b
multiplying by a, a^2=bxa
subtracting ,b^2, a^2-b^2=ab-b^2
applying formula, a^2-b^2=(a+b)(a-b)
&taking common b, we get,
(a+b)(a-b)=b(a-b)
cancelling both sides, (a+b)=b
now as a=b therefore,
a+a=a
2a=a
2=1


You think that this quest., is tough but it is not tough at all......

Answer 6

Let a = b = 1

a = b

a^2 = ab
(Multiplying both sides with a)

a^2 - b^2 = ab-b^2
(subtracting both side with b^2)

(a+b)(a-b) = b(a-b)
(factorizing LHS)

(a+b) = b
(by dividing both sides with (a-b)

(1+1) = 1

Ha ha ha - Maths on its head.

Forget the proof. In sanskrit it is called Kutarka - example If door is half closed it means it is half open. By same logic , if it is fully closed, is it fully open?

Answer 7

let a = b

a² = ab
Multiply both sides by a
a² + a² - 2ab = ab + a² - 2ab
Add (a² - 2ab) to both sides
2(a² - ab) = a² - ab
Factor the left, and collect like terms on the right
2 = 1
Divide both sides by (a² - ab)..

Answer 8

simple multiply by zero on both sides.
1*0=0
2*0=0
there fore 1=2

Answer 9

as we muiltiply 1X1 it is not having any changes so we multiply as really 0 but 1x1+1 =2. I have given you a star as you hve asked this question so you also give me as best answer

Answer 10

-(1/2) = -(1/2)

1/2-1=1/2-1
(1/2-1)^2=(1/2-1)^2
(1/2)^2-2.(1/2).1+1=(1/2)^2-2.1/2.1+1

= 1- 2.(1/2).1+(1/2)^2
=(1-1/2)^2

Now we have

(1/2-1)^2 = (1-1/2)^2

1/2-1= 1-1/2

There fore

1/2+1/2 = 1+1

or 1=2

Ofcourse this is a Mathematical fallacy