Question

2 + 2 =1, can anyone help to show me? #mathfallacy

Answer 1

a = b
a squared = ab
a squared - b squared = ab-b squared
(a-b)(a+b) = b(a-b)
a+b = b
b+b = b
2b = b
2 = 1

Answer 2

\(a = b\)
\(a^2 = ab\)
\(a^2 - b^2 = ab-b^2\)
\((a-b)(a+b) = b(a-b)\)

How do you go from the step just above this line to the step just below this line?

\(a+b = b\)
\(b+b = b \)
\(2b = b \)
\(2 = 1\)

Answer 3

Look at the hashtag for the original poster #mathFALLACY, except he spelled it wrong

Answer 4

The missing step is:

\(\dfrac{\cancel{(a - b)}(a + b)}{\cancel{a - b}} = \dfrac{b\cancel{(a - b)}}{\cancel{a - b}}\)

The problem with this line is that since a = b, a - b = 0, and you can't divide by 0, so you can't divide by a - b. That is the fallacy with this "proof."

Answer 5

Thats correct

Answer 6

it's 2=1 i'm asking 2+2=1, thanks,kheath39 for error attention in hashtag

Answer 7

I've seen 2 + 2 = 5, but I've never seen 2 + 2 = 1.

Answer 8

thanks! mathstudent55.

Answer 9

mathstudent55! any logic in your mind to have 2+2=1?

Answer 10

a = b
a squared = ab
a squared - b squared = ab-b squared
(a-b)(a+b) = b(a-b)
a+b = b
b+b = b
2b = b
2 = 1

now add

x = 1.
x2 = 1
x2 – 1= 0
(x+1)(x-1) = 0
Divide both sides by (x-1): x+1 = 0
Substitute the value of x: 1 + 1 = 0
2 = 0.

Don't know if this counts.

Answer 11

\(2 +_3 2=1\)

Answer 12

See this for 2 + 2 = 5:
http://math.stackexchange.com/questions/457490/22-5-error-in-proof

Answer 13

silly

Answer 14

\(1=1^1=1^{\frac{1}{1}}=1^{\frac{2}{2}}=(1^2)^{\frac{1}{2}}=((-1)^2)^{\frac{1}{2}}=-1^{\frac{2}{2}}=-1^1=-1\)

Answer 15

Just use 2=1;

If we have 2=1, then 1+1=1;

So we can plug RHS into LHS;

\(1+1=1\quad\Longrightarrow\quad(1+1)+(1+1)=1\\~\\\phantom{1+1=1}\quad\Longrightarrow\quad2+2=1\)

Simple.

Answer 16

What does "\(+_3\)" mean? @zzr0ck3r

Answer 17

addition mod 3

Answer 18

no fallacy found to show it. thanks to all. closing it.

Answer 19

I just did... lol.

Answer 20

you cant show it

Answer 21

its not true for real numbers, anything else is a lie.

Answer 22

1=2 can only happen in a field of one element, i.e. the trivial ring, and then we are no longer in addition on the reals

Answer 23

Read hashtag. OP wants fallacy that shows \(2+2=1\)

Answer 24

You can "prove" that \(m=n\), when actually \(m\neq n\).

All you need is one fallacy.

Answer 25

you can prove anything with one fallacy :)

Answer 26

Exactly. and we need to show that to OP lol.

Answer 27

here is a set theory proof that it can't happen

\(2+2= \{\emptyset, \{\emptyset\}\}\cup\{(\emptyset,\emptyset), \{(\emptyset,\emptyset)\}\}=\{\emptyset, \{\emptyset\}, (\emptyset,\emptyset),\{(\emptyset, \emptyset)\}\}\ne \{\emptyset\}\)

Answer 28

where we define the natural numbers

\(1=\{\emptyset\}\\
2=\{\emptyset, \{\emptyset\}\}\\3= \{\emptyset, \{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}\\.\\.\\.\)

Answer 29

all you need is ZFC axioms

Answer 30

actually just ZF