8 = 9 (?)

Obese
Posers and Puzzles 10 Feb '07 02:15
1. 10 Feb '07 02:151 edit
-72 = -72
64 - 136 = 81 - 153
(8)^2 - 2(8)(17/2) = (9)^2 - 2(9)(17/2)
(8)^2 - 2(8)(17/2) + (17/2)^2 = (9)^2 - 2(9)(17/2) + (17/2)^2
(8 - 17/2)^2 = (9 - 17/2)^2
8 - 17/2 = 9 - 17/2
8 = 9
2. 10 Feb '07 02:33
A^2 = B^2 does not imply A = B: it is possible that A = -B

In this case (8-17/2) = -0.5 and (9-17/2) = 0.5, so their squares are the same, but 8 and 9 remain distinct.

ðŸ˜€

Better than the proof that 1 = 2 because
Let a = b
(a - b) = 0
(a - b) = (a - a) + (b - b)
(a - b) = (a - b) + (b - a)
(a - b) - (b - a) = (a - b)
(a - b) + (a - b) = (a - b)
2*(a - b) = (a - b)
2 = 1
3. 10 Feb '07 07:31
Originally posted by GregM
Better than the proof that 1 = 2 because
Let a = b
(a - b) = 0
(a - b) = (a - a) + (b - b)
(a - b) = (a - b) + (b - a)
(a - b) - (b - a) = (a - b)
(a - b) + (a - b) = (a - b)
2*(a - b) = (a - b)
2 = 1
This is a classic! A nice one!
It gives you ants in your head until you are presented the solution.
Then it is crystal clear why this 'proof' is not valid.
4. PBE6
Bananarama
10 Feb '07 07:44
Originally posted by GregM
A^2 = B^2 does not imply A = B: it is possible that A = -B

In this case (8-17/2) = -0.5 and (9-17/2) = 0.5, so their squares are the same, but 8 and 9 remain distinct.

ðŸ˜€

Better than the proof that 1 = 2 because
Let a = b
(a - b) = 0
(a - b) = (a - a) + (b - b)
(a - b) = (a - b) + (b - a)
(a - b) - (b - a) = (a - b)
(a - b) + (a - b) = (a - b)
2*(a - b) = (a - b)
2 = 1
Aha...(b - a) = 0, but the trouble arises when you bring it from the RS to the LS and give it a negative sign. Giving (b - a) on the LS a positive sign alleviates the problem. Tricky!
5. 10 Feb '07 10:22
The problem is in the last line when both sides are divided by zero (a-b).

A simply example which shows why this doesn't work is:

2*0 = 1*0

(2*0)/0 = (1*0)/0

2*(0/0) = 1*(0/0)

2 = 1
6. 10 Feb '07 10:50
Sometimes one encounter people who thinks that the rule that division by zero is forbidden is a limiting rule. Once we had a discussion here at some Forum who stated that "Now there is a new mathematics where it is okay to divide by zero".

This so called paradox shows in a quite amusing way that all sorts of problems arises when one tries to divide by zero.

Some says we already divide by zero when using the l'Hôpital rule, but actually we aren't.
7. 15 Feb '07 06:10
In the fallacy of proving 8=9, the mistake lies in the penultimate step.
that is, when a^2=b^2, we cannot conclude a=b. The proper answer and right inference is PLUS OR MINUS a= PLUS OR MINUS b. In the so called proof the author has conveniently ignored this fact and facilitatingly assuming a=b to arrive at his favourable contradiction. Q.E.D. "QUOD ERAT DEMONSTRANDUM"
id: kesavan7777
Real Name: R.Kesavan.
8. PBE6
Bananarama
15 Feb '07 14:17
The problem is in the last line when both sides are divided by zero (a-b).

A simply example which shows why this doesn't work is:

2*0 = 1*0

(2*0)/0 = (1*0)/0

2*(0/0) = 1*(0/0)

2 = 1
Oops! You're right. ðŸ˜³
9. 17 Feb '07 19:40
Another square root fallacy occurs in this rather nice trig "proof".

Rearranging the identity cos^2(x) + sin^2(x) = 1,

cos^2(x) = 1 - sin^2(x)
cos(x) = sqrt(1 - sin^2(x))
1 + cos(x) = 1 + sqrt(1-sin^2(x))

This is true for all values of x, so let x=pi

so, 1 + cos(pi) = 1 + sqrt(1 - sin^2(pi))
1 + (-1) = 1 + sqrt(1 - 0^2)

hence 0 = 2. QED