Thread: A question.

Will 4=5.

My maths teacher proved this in my class when I was in high school.
How come it is possible.

I guess you should ask your maths teacher because I see that you didn't learn from when the maths teacher taught you how its done.

Hmmm...

I can prove 1=2

Let x = y

x(x - y) = x^2 - xy

but x=y

x(x - y) = x^2 - y^2

x (x - y) = (x + y)(x - y)

Divide by (x - y)

x = x + y

but x=y

Therefore x = x + x

x = 2x

1=2

Wow happy days 1=2, now all sorts of strange stuff can happen....
I can't remember which mathematician did this but I'm sure someone will enlighten us.
He proved that if 1=2 you could prove anyone was the pope.
i.e. Fact 1: John and the pope are 2. Fact 2: 1=2 therefore we can deduce that John and the pope are 1.

Now where is the problem with our maths.....

Well the problem is when we divide by (x - y), as x - y = 0 and we cannot divide by 0

I'm not sure how your teacher proved that 4=5 but you can see that 4*(0)=5*(0) [i.e. 0=0] and you cannot divide by 0 so I'm sure he did it in a similar way to the 1=2.

1 will not ever equal 2 if you graph it the slops are both 0 so the lines will never intersect. Unless, you add a variable like X, but then it would not read 1=2.

How are you plotting 1=2?

If you are plotting x=1 and x=2 then you would have two lines with infinte slope, and of course they would never come into contact unless you folded the x-y plane to make it happen

In my last post I just pointed out that 1 * (0) = 2 * (0)

you can graph 1 and 2 on a graphing calculator like the TI-83 or TI-92 by puting y1=1 and y2=2. Also, I agree with you that 1 * (0) = 2 * (0).

Let a=b=1

Therfore a*a=ab
a*a-b*b=ab-bb
a*a-b*b=b(a-b)
(a+b)(a-b)=b(a-b)
a+b=b
2=1 (since a=b=1)
2+3=1+3
5=4

:wink:

As Mobius said, his version of the proof is dependent upon a stage where you divide by zero. This should stop the calculation at this point, but to one who isn't paying proper attention or may not be that educated in math, the proof seems valid.

I had a physics teacher that showed me an equation proving 1=-1. I always had these visions of antimatter in my head with that one. Imagine the dismay when I learned the vile trick used to perpetrate such a sham.

Meh.

Anyway.
Here's some other erroneous proofs I dug up for a similar thread in some long ago.

Theorem : 3=4
Proof:
Suppose:
a + b = c
This can also be written as:
4a - 3a + 4b - 3b = 4c - 3c
After reorganising:
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


Theorem : All numbers are equal to zero.
Proof: Suppose that a=b. Then
a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
a + b = b
a = 0


Theorem: 1$(dollar) = 1c(cent).
Proof:
And another that gives you a sense of
money disappearing...
1$ = 100c
= (10c)^2
= (0.1$)^2
= 0.01$
= 1c


Theorem: 1 = -1 .
Proof:
1/-1 = -1/1
sqrt[ 1/-1 ] = sqrt[ -1/1 ]
sqrt[1]*sqrt[1] = sqrt[-1]*sqrt[-1]
ie 1 = -1


Theorem: 4 = 5
Proof:
16 - 36 = 25 - 45
4^2 - 9*4 = 5^2 - 9*5
4^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 - 9/2 = 5 - 9/2
4 = 5