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
Posts
