Mathematics paradoxes

Here is the version offered by Augustus De Morgan:

Paradox no:1

Proving that 2 = 1

Let x = 1.

Then x² = x.

So x² - 1 = x -1.

Dividing both sides by x -1, we conclude that x + 1 = 1;

that is, since x = 1, 2 = 1.






Paradox no: 2

Assume that

a = b. (1)

Multiplying both sides by a,

a² = ab. (2)

Subtracting b² from both sides,

a² - b² = ab - b² . (3)

Factorizing both sides,

(a + b)(a - b) = b(a - b). (4)

Dividing both sides by (a - b),

a + b = b. (5)

If now we take a = b = 1, we conclude that 2 = 1.

Or we can subtract b from both sides and conclude that a, which can be taken as any number, must be equal to zero.

Or we can substitute b for a and conclude that any number is double itself.

Our result can thus be interpreted in a number of ways, all equally ridiculous.

Paradox no: 3


Proving that 3 + 2 = 0

Assume A + B = C, and assume A = 3 and B = 2.

Multiply both sides of the equation A + B = C by (A + B).

We obtain A² + 2AB + B² = C(A + B)

Rearranging the terms we have

A² + AB - AC = - AB - B² + BC

Factoring out (A + B - C), we have

A(A + B - C) = - B(A + B - C)

Dividing both sides by (A + B - C), that is, dividing by zero,

we get A = - B, or A + B = 0, which is evidently absurd.

Paradox no: 4


Proving that n = n + 1

(a) (n + 1)² = n² + 2n + 1

(b) (n + 1)² - (2n + 1) = n²

(c) Subtracting n(2n + 1) from both sides and factoring, we have

(d) (n + 1)² - (n + 1)(2n + 1) = n² - n(2n +1)

(e) Adding ¼(2n + 1)² to both sides of (d) yields

(n + 1)² - (n + 1)(2n + 1) + ¼(2n + 1)² = n² - n(2n + 1) + ¼(2n + 1)²

This may be written:

(f) [(n + 1) - ½(2n + 1)]² = [(n - ½(2n + 1)]²

Taking square roots of both sides,

(g) n + 1 - ½(2n + 1) = n - ½(2n + 1)

and, therefore,

(h) n = n + 1