Can any one prove 1=2?

Question

show 1 = 2 any how.

Answer 1

There are many ways to do this, some of which rely on advanced bistromathics. Among the easiest, and most convincing, is the following proof using induction.

Lemma: all natural numbers are equal

Proof: Suppose a and b are natural numbers such that a≤1 and b≤1. There is only one such natural number, namely 1, and so a=1 and b=1. By transitivity, a=b.

Now suppose that it is true for some n that if a and b are natural numbers such that a≤n and b≤n, a=b. Let a and b be natural numbers such that a≤n+1 and b≤n+1. Then clearly a-1≤n and b-1≤n, so by our induction hypothesis, a-1=b-1. But that implies a=b. Therefore, by induction, if a and b are natural numbers less than or equal to any natural number, a=b. Since all natural numbers are less than or equal to some natural number, it follows that all natural numbers are equal. In particular, 1=2. Q.E.D.

Another proof uses the well established properties of infinite series:

0=0+0+0+0...
0=(1-1) + (1-1) + (1-1) + (1-1)...
Using the associative law:
0=1 + (-1+1) + (-1+1) + (-1+1) + (-1+1)...
0=1
Adding 1 to both sides:
1=2

Of course, sometimes algebra is fun too:

-2=-2
1-3=4-6
1-3+9/4=4-6+9/4
(1-3/2)²=(2-3/2)²
1-3/2=2-3/2
1=2

There is a well known proof that uses 1*0=2*0, therefore 1=2, but unfortunately I can't present it here because the last time I divided by zero, I recieved a message saying I had performed an illegal operation and must be shut down.

Here's one using calculus:

Consider ∫sin x cos x dx:
let u=sin x, du=cos x dx:
∫u du
u²/2
sin² x/2

However:

∫sin x cos x dx
let u=cos x, du=-sin x dx
-∫u du
-u²/2
-cos² x/2

So:
sin² x/2 = -cos² x/2

Now let x=π/2
1/2=0
Multiplying both sides by 2:
1=0
And adding 1 to both sides:
2=1

A fascinating proof using complex logarithms:

e^(iπ)=-1
ln (-1)=iπ
2 ln (-1) = 2iπ
ln ((-1)²) = 2iπ
ln 1 = 2iπ
0=2iπ
Dividing by 2iπ
0=1
Adding 1:
1=2

Another proof that all numbers are the same:

Let S be a group of n numbers. Suppose n=1. Then S contains only one number, so clearly all the numbers are the same. Now suppose that for some n any group of n numbers is all the same. Let S be a group of n+1 numbers. Remove some number a from S - this is then a group of n numbers, so all the numbers are the same. Similarly remove some number b from S - then that is also a group of n numbers, all of which are the same. Let c be a number in both S\a and S\b. Clearly, since all the numbers in S\a are the same, b=c. By similar logic, a=c. Therefore, a=b. Since all the other numbers in S have already been shown to be the same, we have all numbers in S are the same. By induction, this applies to sets of any size, so all numbers are the same. Q.E.D.

And finally:

x=0.[9] (the brackets indicate a repeating decimal)
10x=9.[9]
10x-x=9
9x=9
x=1
By transitivity of equality:
1=0.[9]
1-0.[9]=0
Multiply both sides by (1/(1-0.[9])):
1=0
2=1

Warning: The preceding proofs are for entertainment purposes only. Every last one of them has as much validity as a load of dingoes kidneys and do not constitute mathematical advise. If you are seeking assistance on an actual mathematical problem, consult a mathematician liscensed to practice in your area.

Answer 2

The answer to your question is a mathematical fallacy because it employs a serious violation of mathematical principles. The supposed proof runs like this: If a = b and b = 1 then; a^2 = a^2 :identity ab = a^2 : substitution ab - b^2 = a^2 - b^2 : axiom of subtraction b(a - b) = (a + b)(a - b) : factoring both sides b = (a + b) : cancelling same terms on both sides 1 = 1 + 1 : substitution 1 = 2 The fallacy lies in the cancellation of (a - b) from both sides. The action is tantamount to dividing (a - b) by itself and assuming that the quotient is 1. However the value of (a - b) is 0 and 0/0 is indeterminate and not 1. the proof therefore is a fallacy.

Answer 3

a = b
ab = b^2
ab-a^2 = b^2-a^2
a(b-a) = (b-a)(b+a)
a = b+a [dividing by a-b]
1=2 [since a = b]
alternate proof!
log2 = 1-1/2 +1/3-1/4+1/5....
= (1+1/3+1/5..) - (1/2+1/4.........)
= (1+1/2+1/3+1/4+1/5+1/6......) - 2(1/2+1/4+1/6...)
= (1+1/2+1/3+1/4+1/5+1/6......) - (1+1/2+1/3+1/4..........)
= 0
log2 = 0 = log1
2 = 1
alternate proof
one has three alphabets so has two.
hence 1 = 2
one is alone and when one gets married becomes two, but they are referred as together and one.
hence 1 = 2
Letters in alphabets
O - N + E = T - W + O
15 - 14 + 5 = 20 - 23 + 15
6 = 12
1 = 2

Answer 4

one has three alphabets so has two.
hence 1 = 2
one is alone and when one gets married becomes two, but they are referred as together and one.
hence 1 = 2
Letters in alphabets
O - N + E = T - W + O
15 - 14 + 5 = 20 - 23 + 15
6 = 12
1 = 2

Answer 5

Uh, I don't think that anybody can prove this since this is not true (at least for complex numbers). I am sure some sort of a number system can be generated where this is true.

BTW, both of the proofs above me are wrong. For the first one, you can't compare the fractions because the denominators are different.

For the second one, STOP DIVIDING BY ZERO.

Answer 6

Well, clearly this is untrue for all complex numbers.
And most of the solutions given say the following:

a=b
=>a^2=ab
subtracting b^2 from both sides
=>a^2-b^2=ab-b^2
=>(a+b)(a-b)=b(a-b)
but here (a-b) on the LHS and RHS can't be cancelled as division by zero is meaningless.

Answer 7

Given a=b=1

1. a = a
2. a*a = a*a
3. a^2 = a^2
4. a^2 - ab = a^2 - ab
By substitution, ab = b^2
5. a^2 - ab = a^2 - b^2
6. a(a-b) = (a+b)(a-b)
Dividing both sides by (a-b)
7. a = a + b
From the given, a = b
8. a = a + a
9. 1 = 2

*This, of course, is fallacious. Division by zero, which is illegal, has taken place.

Answer 8

fractions 1/3=2/6

Answer 9

1=2
1(1=2)
1=2
divide 1 both sides
1=2

or

1=2
2(1=2)
2=4
divide 2 both sides
1=2

Answer 10

1=2 apparently thats the only way
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