 # We conclude that 1 = 2 – Second part

Six is a number perfect in itself, and not because God created the world in six days; rather the contrary is true. God created the world in six days because this number is perfect, and it would remain perfect, even if the work of the six days did not exist.

Saint Augustine

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Let us prove 1 = 2! Is that sounds ridiculous? There are mathematical situations, from where we can easily conclude that 1 = 2. It’s unfortunate, to have such instances. This is the second part and the first part is available here (https://math1089.in/2021/01/03/we-conclude-that-1-2/).

They are mathematical fallacy, an assumption (or series of steps) which is seemingly correct, though actually specious reasoning. In this blogpost, we will present a few such results, where due to some bugs, we can easily arrive at the conclusion 1 = 2.

It is definitely against our known result, due to some error. But what is (are) that error(s)? Question is “Is the division by 0 allowed?”. Of course, the answer is “NO”. Then how we can do the same here? In step (f), while cancelling the factor x2xy, probably we ignored the fact that x2xy = xyxy = 0, by (a). Since division by 0 is undefined and we have performed the same here, so the absurd result 1 = 2.

If two numbers are equal, their squares are also equal. However, the reverse form of such a statement does not hold. Recall that, the result of a square root is not unique. In fact, one should keep in mind that, x2 = y2 means x = ± y and not that x = y. Using this result religiously, we find that the error occurred while travelling from step (f) to (g).

Revisiting 1 = 2

• (a) 10 = 1
• (b) 20 = 1
• (c) 10 = 20
• (d) 1 = 2

It’s worth mentioning that ax = bx means a = b only when a, b are positive and x ≠ 0.

Just one more time, 1 = 2

Consider the following infinite series:

1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + ∙ ∙ ∙

Let S be the sum of this infinite series. We now calculate the value of S in the following two ways. We can write

• (a) S = 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + ∙ ∙ ∙
• (b) S = (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) + ∙ ∙ ∙
• (c) S = 0 + 0 + 0 + 0 + ∙ ∙ ∙
• (d) S = 0

OR

• (a) S = 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + ∙ ∙ ∙
• (b) S = 1 – (1 – 1) – (1 – 1) – (1 – 1) – ∙ ∙ ∙
• (c) S = 1 – 0 – 0 – 0 – ∙ ∙ ∙
• (d) S = 1

Therefore 0 = 1. Now add 1 to both sides.

We conclude that 1 = 2.

Finally, a geometrical reasoning for 1 = 2

Let us draw an equilateral triangle with side length 1 unit. The ratio of the sum of the lengths of the two sides to the base is 2 : 1.

Bring the top vertex (here, A) of the triangle down to meet its base (here, BC) to create two equilateral triangles side by side.

We now repeat the procedure as shown below.

The length ratio of the sum of the two sides to the base will always be 2 : 1. We can continue folding down the flaps of each triangle to produce a doubled generation of increasingly shorter and shorter triangles.

Then optically we would reach a point where the peaks of triangles visually collapse into the base of the triangle we started with – giving the appearance of 1 = 2!

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1. Subhajit says:
1. Math1089 says: