C
cantaloup63
Does 2 + 2 = 4?
A
Alarm
Not if it is 2 threes and 2 fives.
C
cantaloup63
But that would be 16, which is 4 to the power of 2, which would leave a spare 2
depends what it is two of?
eg. 2 apples + 2 oranges wont add up to either 4 apples or 4 oranges but will go a long way to your 5 a day.
eg. 2 apples + 2 oranges wont add up to either 4 apples or 4 oranges but will go a long way to your 5 a day.
T
toasty
I seem to recall a (flawed) mathematical proof from my degree days that proves 2+2 is 5 but I think the error is in assuming the square root of a number is always positive.
So, I concur, 2 + 2 is 4
So, I concur, 2 + 2 is 4
I seem to remember from my school days when putting 2 and 2 together, it often resulted in a clip around the ear!
So in summary 2+2=earache
So in summary 2+2=earache
C
cantaloup63
Don't recall that one. If you can remember how it goes, I'd appreciate it
T
toasty
I'll see if I can remember it. It was based around a convoluted Pythagorean quadratic if memory serves
Are we buying or selling?
T
toasty
Got it!
Right...
Let a & b each be equal to each other and both have a vaule of 1.
Since a & b are equal:
b^2 = ab - equation 1 (for non mathsy people, 'ab' means 'a' multiplied by 'b' and b^2 means 'b squared' or 'b' multiplied by 'b')
Since 'a' equals itself, it is obvious that
a^2 = a^2 - equation 2
Subtract equation 1 from equation 2. This gives
(a^2) - (b^2) = (a^2) - ab - equation 3
We can factor both sides of the equation:
(a^2)-ab equals a(a-b)
Likewise:
(a^2)-(b^2) equals (a + b)(a - b) - non mathsy people might struggle with this jump, but it is true.. (try plugging in numbers if you aren't sure)
Substituting into the equation 3 , we get
(a+b)(a-b) = a (a-b) - equation 4
So far, so good....?
Now divide both sides of the equation by (a-b) and we get
a + b = a - equation 5
Which means that:
b = 0 - equation 6
But we set b to 1 at the very beginning of this proof, so this means that
1 = 0 - equation 7 (oh dear!!)
So if 1 = 0 then we can say that 2 + 2 + 0 = 5 i.e 2 + 2 = 5 (as we have already proved that 0=1)
So, conceptually, any number is equal to zero.
That's also the reason why when you divide by zero, it is 'Undefined.' Which is what is happening in this equation...can you spot where?
So, as I say, it's a flawed proof, but kind of interesting I suppose...
Right...
Let a & b each be equal to each other and both have a vaule of 1.
Since a & b are equal:
b^2 = ab - equation 1 (for non mathsy people, 'ab' means 'a' multiplied by 'b' and b^2 means 'b squared' or 'b' multiplied by 'b')
Since 'a' equals itself, it is obvious that
a^2 = a^2 - equation 2
Subtract equation 1 from equation 2. This gives
(a^2) - (b^2) = (a^2) - ab - equation 3
We can factor both sides of the equation:
(a^2)-ab equals a(a-b)
Likewise:
(a^2)-(b^2) equals (a + b)(a - b) - non mathsy people might struggle with this jump, but it is true.. (try plugging in numbers if you aren't sure)
Substituting into the equation 3 , we get
(a+b)(a-b) = a (a-b) - equation 4
So far, so good....?
Now divide both sides of the equation by (a-b) and we get
a + b = a - equation 5
Which means that:
b = 0 - equation 6
But we set b to 1 at the very beginning of this proof, so this means that
1 = 0 - equation 7 (oh dear!!)
So if 1 = 0 then we can say that 2 + 2 + 0 = 5 i.e 2 + 2 = 5 (as we have already proved that 0=1)
So, conceptually, any number is equal to zero.
That's also the reason why when you divide by zero, it is 'Undefined.' Which is what is happening in this equation...can you spot where?
So, as I say, it's a flawed proof, but kind of interesting I suppose...
I
imamartian
if you're working in base 4, then 2+ 2 = 10
C
cantaloup63
Thanks toasty - by flawed proof you mean paradox 
Also by undefined you mean ill defined (as in 0/0 which is the flaw in the proof). But we'll let that pass 
Also by undefined you mean ill defined (as in 0/0 which is the flaw in the proof). But we'll let that pass 
