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3=4
Posted:
Mon Oct 10, 2005 1:54 am
by Sotos
Theorem: 3=4
Proof:
Suppose:
a + b = c
This can also be written as:
4a - 3a + 4b - 3b = 4c - 3c
After reorganizing:
4a + 4b - 4c = 3a + 3b - 3c
Take the constants out of the brackets:
4 * (a+b-c) = 3 * (a+b-c)
Remove the same term left and right:
4 = 3
Posted:
Mon Oct 10, 2005 10:03 am
by devil
Much easier
3 x 0 = 0
4 x 0 = 0
3 x 0 = 4 x 0
Divide each side by 0
3 = 4
Posted:
Mon Oct 10, 2005 11:52 am
by garbitsch
devil, you cannot divide zero with zero. Or you cannot even divide any numbers with zero just fyi.
Posted:
Mon Oct 10, 2005 1:35 pm
by devil
garbitsch wrote:devil, you cannot divide zero with zero. Or you cannot even divide any numbers with zero just fyi.
DUH !!!! I wonder why no one ever told me that before, when I was doing degree maths????
Posted:
Mon Oct 10, 2005 1:57 pm
by garbitsch
Now you've learnt it
Posted:
Mon Oct 10, 2005 2:30 pm
by nasos007
reminds me of the Salary Theorem (the less you know, the more you make).
Proof:
Postulate 1: Knowledge is Power.
Postulate 2: Time is Money.
As every engineer knows: Power = Work / Time
And since Knowledge = Power and Time = Money
It is therefore true that Knowledge = Work / Money .
Solving for Money, we get:
Money = Work / Knowledge
Thus, as Knowledge approaches zero, Money approaches infinity, regardless of the amount of Work done.
Posted:
Mon Oct 10, 2005 4:17 pm
by Alexis
devil, you cannot divide zero with zero. Or you cannot even divide any numbers with zero just fyi.
Isn't that what Sotos did by dividing by (a+b-c), also zero,
or am I missing something?
Posted:
Mon Oct 10, 2005 4:36 pm
by dms007
for that matter its not just 3=4, any two consequtive natural numbers can be "proved" to be equal
Posted:
Mon Oct 10, 2005 4:45 pm
by dms007
but.... to the finer eye
4 * (a+b-c) = 3 * (a+b-c)
is where the flaw is......
a+b=c
so a+b-c=0
so effectively it becomes
4*0=3*0
since in the original post the expression is being divided by (a+b-c) which is actually 0 the division will fail, 0 divided by zero is undefined.
so all that we can prove is 0=0 which is true for all numbers.
Posted:
Mon Oct 10, 2005 5:07 pm
by devil
Which is why my version, which Garbitsch tried to trash, is much easier: it is exactly the same.