|
|
|
|
![]() ![]() |
Dec 9 2006, 07:23 PM
Post
#1
|
|
|
Incest is a game the whole family can play. ![]() Group: [MODERATOR] Posts: 1,205 Joined: 11-February 05 From: Heaven Member No.: 3,709 |
In light of the topic, 2+2=5 methought I'd create my own puzzle:
Let x = y Therefore x² = xy Therefore x² - y² = xy - y² If we factorise these equations we end up with (x + y)(x - y) = y(x - y) We now have a common term on both sides "(x - y)," which we will divide off to make our problem easier to solve x + y = y But x = y so y + y = y thus 2y = y Then, if we divide off the common value "y," 2 = 1 Simple! Where's the flaw? |
|
|
|
Dec 9 2006, 09:59 PM
Post
#2
|
|
|
EVIL CORN! ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 617 Joined: 7-May 06 From: USA Member No.: 23,230 ![]() |
I remember this from last year in my math class. It was in the step when you canceled (x-y) I believe, because (x-y) is 0, therefore you are multiplying by 0 which would be 0. You can't cancel (x-y). Good one!
This post has been edited by husker: Dec 9 2006, 11:28 PM |
|
|
|
Dec 10 2006, 12:03 AM
Post
#3
|
|
|
Trap Double Mocha Member ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: [HOSTED] Posts: 2,220 Joined: 5-November 05 From: That one place over there... Member No.: 13,830 |
Well like husker said (x-y) would equal zero and it's the fact that you can't divide by zero not multiply like husker said. So that would really mean that it ends up as 0 = 0 because if you plug in the 0 of (x - y) , that would multiply with the rest of the equation to make 0 because it is all multiplication, no addition or subtraction except between parentheses. There may be a way though so keep working at it.
|
|
|
|
Dec 10 2006, 05:37 AM
Post
#4
|
|
|
Privileged Member ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: [HOSTED] Posts: 537 Joined: 21-August 06 From: Ziya's Heart Member No.: 28,693 |
As I have said in similar topic 2+2=5, I would again say that this type of questions are raised by people with little knowledge of Mathematics.
As explained by Husker and Plenoptic as x and y are equal therefore x-y is zero and whenever zero comes it cant be cancelled from both sides, if this is possible then all numbers will be equal to each other, like this. 0=0 or, 0*a=0*b i.e. a=b where a and b are any number. therefore zeros from both sides cannot be cancelled. |
|
|
|
Dec 10 2006, 09:03 AM
Post
#5
|
|
|
Incest is a game the whole family can play. ![]() Group: [MODERATOR] Posts: 1,205 Joined: 11-February 05 From: Heaven Member No.: 3,709 |
Congrats! The answer was you can't divide by 0 or x - y. You can prove anything when you divide by 0.
|
|
|
|
Dec 20 2006, 04:36 PM
Post
#6
|
|
|
Newbie ![]() Group: Members Posts: 7 Joined: 19-December 06 Member No.: 35,637 |
Here is another way of proving that 1 = 2
QUOTE Everyone knows this: -1/1= 1/-1 Now we will square root both sides: √-1/1 = √1/-1 Now we break up the roots: √-1 √1 --- = --- √1 √-1 The square root of a negative 1 is i and the square root of 1 is 1. In other words: i/1 = 1/i Now we multiply the entire thing by 1/2: i/2 = 1/2i Now let's add 3/(2i) to this to make the math easier. i/2 + 3/2i = 1/2i + 3/2i Now we can multiply the entire thing by i: i(i/2 + 3/2i) = i(1/2i + 3/2i) So now we expand this beast: 1^2/2 + 3i/2i = i/2i + 3i/2i We know that the square root of -1 is i, so i^2 must be -1.: -1/2 + 3i/2i = i/2i + 3i/2i Now we simplify the i's -1/2 + 3/2 = 1/2 + 3/2 Let's calculate this thing: 2/2 = 4/2 And so 1 = 2 That does work as far as I can see. And with this proof you can prove that dividing by zero is possible, because anygivennumber^0 = 1 Source I know the source isnt reliable, but the 1-=2 thing I posted does work as far as I can see. This post has been edited by Terciel Silvi: Dec 20 2006, 06:37 PM |
|
|
|
Dec 20 2006, 08:09 PM
Post
#7
|
|
|
Member [Level 2] ![]() ![]() ![]() ![]() ![]() Group: Members Posts: 84 Joined: 24-November 06 Member No.: 33,967 |
1 will never equal 2 by definition. So I'm going to bother with the second proof because there's definately a flaw somewhere. If it was correct then there would be a flaw in our addition, multiplication, etc.
Remember you can't divide by variables unless you can prove they aren't 0 like y = x^2 + 1 where x is a real number (no i's |
|
|
|
![]() ![]() |
Similar Topics
|
Lo-Fi Version | Time is now: 25th July 2008 - 07:41 PM |