## anonymous one year ago 2 + 2 =1, can anyone help to show me? #mathfallacy

1. anonymous

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

2. mathstudent55

$$a = b$$ $$a^2 = ab$$ $$a^2 - b^2 = ab-b^2$$ $$(a-b)(a+b) = b(a-b)$$ How do you go from the step just above this line to the step just below this line? $$a+b = b$$ $$b+b = b$$ $$2b = b$$ $$2 = 1$$

3. anonymous

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

4. mathstudent55

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."

5. anonymous

Thats correct

6. anonymous

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

7. mathstudent55

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

8. anonymous

thanks! mathstudent55.

9. anonymous

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

10. anonymous

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.

11. zzr0ck3r

$$2 +_3 2=1$$

12. mathstudent55

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

13. zzr0ck3r

silly

14. zzr0ck3r

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

15. geerky42

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.

16. geerky42

What does "$$+_3$$" mean? @zzr0ck3r

17. zzr0ck3r

18. anonymous

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

19. geerky42

I just did... lol.

20. zzr0ck3r

you cant show it

21. zzr0ck3r

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

22. zzr0ck3r

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

23. geerky42

Read hashtag. OP wants fallacy that shows $$2+2=1$$

24. geerky42

You can "prove" that $$m=n$$, when actually $$m\neq n$$. All you need is one fallacy.

25. zzr0ck3r

you can prove anything with one fallacy :)

26. geerky42

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

27. zzr0ck3r

here is a set theory proof that it can't happen $$2+2= \{\emptyset, \{\emptyset\}\}\cup\{(\emptyset,\emptyset), \{(\emptyset,\emptyset)\}\}=\{\emptyset, \{\emptyset\}, (\emptyset,\emptyset),\{(\emptyset, \emptyset)\}\}\ne \{\emptyset\}$$

28. zzr0ck3r

where we define the natural numbers $$1=\{\emptyset\}\\ 2=\{\emptyset, \{\emptyset\}\}\\3= \{\emptyset, \{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}\\.\\.\\.$$

29. zzr0ck3r

all you need is ZFC axioms

30. zzr0ck3r

actually just ZF