2 ways of proving "2 = 1"

Forums: Riddles
Email this Topic • Print this Page

Method 1: Differentiation Method

Let a^b mean a "raised to" b

1^2 = 1
2^2 = 2 + 2
3^2 = 3 + 3 + 3

so for any X > 0

X^2 = X + X + X + ..... (X times)

Differentiate both sides with resepct to X

2X = 1 + 1 + 1 + ... (X times)

2X = X

2 = 1 .... Cancelling X from both sides as X <> 0

Method 2: Logarithm Method

We know that natural log of 2 i.e. Ln(2) = 0.6931471

As per log series transformation ....

Ln(2) = 1 - (1/2) + (1/3) - (1/4) + (1/5) .....

Ln(2) = (1 + 1/3 + 1/5 ....) - (1/2 + 1/4 + 1/6 ....)

Ln(2) = ((1 + 1/3 + 1/5 ....) + (1/2 + 1/4 + 1/6 ....)) - 2(1/2 + 1/4 + 1/6 + 1/8 ....)

Ln(2) = (1 + 1/2 + 1/3 + 1/4 ...) - (1 + 1/2 + 1/3 + 1/4 ....)

Ln(2) = 0

Now we also know Ln(1) = 0 ........ as e^0=1

So Ln(2) = Ln(1) = 0

e^Ln(2) = e^Ln(1) ......... taking the exponent of both sides

2 = 1 .............. as e^Ln(a) = a

Topic Stats
Top Replies
Link to this Topic

Type: Discussion • Score: 1 • Views: 3,574 • Replies: 12

No top replies

1. No - The sum rule cannot, in general, be applied to summations with variable number of terms.

2. Still no - Rearranging the terms of a conditionally convergent series may change it's limit, as it does in your example.

0 Replies

Explanation

x wrote:

1. No - The sum rule cannot, in general, be applied to summations with variable number of terms.

2. Still no - Rearranging the terms of a conditionally convergent series may change it's limit, as it does in your example.

Hay Can you explain the 2nd explanation in detail or suggest any link that explains this ...

0 Replies

Let me know if any of my notation confuses you. Anyway, an infinite series can be said to either diverge or converge. For example, sum(n, 1, infinity, n) (or 1+2+3+...) grows without bound while sum(n, 1, infinity, 1/2^n) (or 1/2+1/4+1/8+...) converges to 1 as you add more and more terms.

If the sum(n, 1, infinity, a_n) converges but sum(n, 1, infinity, abs(a_n)) does not, it is said to be conditionally convergent. The problem with conditionally convergent series is that their limit can be changed to anything (including making them diverge) by rearranging the terms. This result is known as the Riemann series theorem. This is what you are doing in the following step:

Quote:

Ln(2) = 1 - (1/2) + (1/3) - (1/4) + (1/5) .....

Ln(2) = (1 + 1/3 + 1/5 ....) - (1/2 + 1/4 + 1/6 ....)

The first series is conditionally convergent, but you still split it. Furthermore, neither of the two resulting series even converge. It is meaningless to take their difference in classical mathematics.

0 Replies

There is nothing wrong with vinsan's use of summation. The math is fine up to the point where Vinsan gets the equation

2X = X. This is a perfectly good equation which has the solution set (x = 0).

The only mistake is the way Vinsan solves the equation. Dividing both sides of an equation by a variable can change change the solution set. (since the only solution is X = 0, this is dividing by zero which is undefined).

A proper way to deal with the equation 2X = X is to subtract X from both sides yielding the equation X = 0. No problem there.

0 Replies

ebrown_p, I have to disagree. Take this as an example:

x = sum(n, 1, x, 1)

d/dx(x) = 1 != 0 = sum(n, 1, x, 0) = sum(n, 1, x, d/dx(1))

You have to remember that derivatives are always taken on functions and are themselves functions - forgetting this is what led one person I know to claim that "the derivative of 2^2 is 4." The derivatives of two equivalent functions must be equal over the entire domain, not just at one value.

0 Replies

I respectfully disagree with your disgreement.

It works just fine as long as x = 0.

That is what visans calculation's proved.

2X = X is a perfectly reasonable mathematical equation with one solution. There is no problem with that.

0 Replies

The argument containing an intermediate step that makes sense in one context doesn't change the fact that you cannot use the sum rule like that (as my simple example illustrates).

I can come up with a true statement by misusing theorems, but that doesn't mean I did it correctly.

0 Replies

Thanks

x wrote:

Let me know if any of my notation confuses you. Anyway, an infinite series can be said to either diverge or converge. For example, sum(n, 1, infinity, n) (or 1+2+3+...) grows without bound while sum(n, 1, infinity, 1/2^n) (or 1/2+1/4+1/8+...) converges to 1 as you add more and more terms.

If the sum(n, 1, infinity, a_n) converges but sum(n, 1, infinity, abs(a_n)) does not, it is said to be conditionally convergent. The problem with conditionally convergent series is that their limit can be changed to anything (including making them diverge) by rearranging the terms. This result is known as the Riemann series theorem. This is what you are doing in the following step:

Quote:
Ln(2) = 1 - (1/2) + (1/3) - (1/4) + (1/5) .....

Ln(2) = (1 + 1/3 + 1/5 ....) - (1/2 + 1/4 + 1/6 ....)

The first series is conditionally convergent, but you still split it. Furthermore, neither of the two resulting series even converge. It is meaningless to take their difference in classical mathematics.

Gotcha!

0 Replies

Glad to help. Who tried to pass those two 'proofs' on you?

0 Replies

nobody

x wrote:

Glad to help. Who tried to pass those two 'proofs' on you?

Nobody just got them on the net. :wink:

0 Replies

It could be this easy, since sum actually should be closed for this field, or it will not be LVS.

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

left = 2 + {[(-1) + 1] + [(-1) + 1] + ... }

right = 1 + {[1 + (-1)] + [1 + (-1)] + ... }

then left = right, we will get 2 = 1

hehe, we also could get, 1 = 0 and -1 = 0 or any X+1 = X by the same method.

0 Replies

Of course Z is closed under addition, but that applies only to finite sums. The original series does not converge and neither does left or right.

0 Replies

2 ways of proving "2 = 1"

Related Topics

Quick Links

My Account

able2know