Disproved these relationships...

[Qacer]

Can someone help me point out the errors?


******************************************************
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

 

[darthsidious]

Divide by zero problem. because a+b = c, a+b-c =0. So essentially you are dividing 0/0, which is what casues 3=4.

Again in the second 'theorem', you divide out by a-b, which is zero.

Almost all such wrong 'theorems' come from dividing by zero.

The last one is a little more interesting. it has an issue with the squareoot.

 

[blackllotus]

Square roots have two real number results. a^2 = b^2 does not mean that a = b, it means that a = b OR a = -b.

 

[AStar617]

Do your own homework!

OK here's one simple one.

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

AFAIK, you can't just slide the decimal in there when doing the c/$ switchover without making the exponent negative.

 

[jacob0401]

these are old...and pretty simple...

 

[mchammer187]

first 2 are wrong due to division by zero

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

second line is wrong = $1= 10^2 * c not 10^2 *c^2

if you wanted to rewrite it as a square root than $1 = (10 * sqrt (c)) ^2)

fourth one

every number has positive or negative roots

sqrt 1 = +/- 1

sqrt (-1) = +/- i

i*i = -1

all the last line is saying is +/- 1 * +/- 1 = +/- (i) * +/- (i)


last one

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 (incorrectly factored) (b^2 - 4ac + c ^2) = (b-c)^2 in this case the right side is not correctly factored because b= 4 c= 9/2 than -4 *ac=72 and -81/4 is not equal to -72 for the right side b=5 and c = 9/2 than -4ac = -90 and 90!= 81/4
4 = 5

 

[Dumac]

Originally posted by: Qacer
Can someone help me point out the errors?


******************************************************


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

Someone already pointed out that the second line is wrong

 

[Qacer]

That's what I was missing... division by zero. Thanks!

 

[Qacer]

I didn't look at the $ and cents. I just looked at the first two. I encountered them before, but I forgot what was wrong. It's divide by zero all along. I used to get that a lot in my C class.