The real (short) proof that 0=1 (or why you should never trust a mathmatician)

[Jothaxe]

<< When i was in high school, my algebra teacher should us a forumla, to make 0=1. The formula involved a variable, and the only problem with it was if you used a certain number, then you ended up dividing by 0, which we all know is a no no, but for the laymen when you should it to them, they all thought you were the devil and wanted to burn you on a stake?

Does anyone know the formula I speak of, cause I can't remember it. >>




That proof is no good at all, mikef208, because it is false. Here is a better proof that is convincing because it is real:


proof that 1 = -1

... and since -1 = 1

we can add 1 to each side to get 0 = 2

Then divide both sides by 2 to get:

0 = 1

QED

I guess this is the end of science and math as we know it. If they cant get this straight, what's the use??

 

[GL]

hmmm.....

-GL

 

[Jothaxe]

Ok GL, you definately hit it right on the head.

Thought it would take longer, but I guess not...

If you feel like deleting the answer so other people can get perplexed, feel free!

 

[GL]

NM

-GL

 

[Jothaxe]

<< not sure if you were serious or if this was just an amusement post >>



No, of course I'm not serious. The picture of the proof that I posted was actually a problem from a really old assignment we had in an applied math course I took.

The point is that you have to remember to deal with multiple roots carefully in complex variable problems... or else you get absurd statements like 1 = -1.

 

[Jothaxe]

By the way GL, thanks for deleting the answer. Maybe it can stump a few more people!

What kind of educational background do you have?

*spelling edit*

 

[cdan]

Heh heh my friend confused his math teacher with this. Gotta remember &quot;Aunt Sally&quot;

 

[Bloodybrain]

Wrong!
You can't add 1 to each side of the equation while keeping the equality. If you add 1 to the left side, you must substract 1 to the right side.

Nice try.

 

[Jokeram]

<< Wrong!
You can't add 1 to each side of the equation while keeping the equality. If you add 1 to the left side, you must substract 1 to the right side.

Nice try. >>



ok.. lets try that out...
2=2 then with your version
2+1 = 2-1
3=1?????

Is this what you are saying??? WOW now we got something else to think off...

 

[Homer_Simpson]

<<Wrong!
You can't add 1 to each side of the equation while keeping the equality. If you add 1 to the left side, you must substract 1 to the right side.

Nice try.>>

Bloodybrain:

Did you not pass pre-algebra? or you just haven't taken it yet...

You are a dumbass...

OF COURSE YOU CAN ADD 1 TO BOTH SIDES AND STILL KEEP THE EQUALITY!!! THAT'S WHY THEY ARE CALLED EQUATIONS!!!

-1=1
-1+1=1+1
0=2

You can't add one to the left side and subtract one on the right side...that's not keeping the equality moron.

if you add something to one side of an equation, you add the exact thing to the other side. Geez...some people...

Homer

 

[GoldenBear]

I think making 1=-1 opens the door to an unlimited amount of possibilities.

 

[Jothaxe]

<< I think making 1=-1 opens the door to an unlimited amount of possibilities. >>



Right you are GoldenBear, if 1 = -1 then you can prove that any number equals any other.

I just show that 0 = 1 because mikef208 asked for this proof...

 

[GoldenBear]

Actually now that I look at the proof again..one of the steps is in error:

((-1)(-1))^(1/2)

You have to do the parenthesis first before you raise it to the 1/2 power, which would make it 1^(1/2).

Is that right?

I realize ((x)(y)^(1/2) is x^(1/2)*y(1/2), but ((x)(x)^1/2 is just x, ((-x)(-x)^1/2 is x as well..or it very well should be.

 

[Pretender]

erasing post, don't wanna spoil the surprise

 

[Buddhist]

whole numbers don't really exist, only in conceptual theory.

chew on that concept.

-M.T.O

 

[Jothaxe]

You hit the target Pretender. Nice!

You can erase your post now if you feel like it. Or leave it if you want... whatever!

 

[Jothaxe]

<< I realize ((x)(y)^(1/2) is x^(1/2)*y(1/2), but ((x)(x)^1/2 is just x, ((-x)(-x)^1/2 is x as well..or it very well should be. >>



You have given proof here for why it is ok to take ((-1)(-1))^(1/2) and break it into (-1)^(1/2)*(-1)^(1/2).

Thats not the &quot;illegal&quot; part of the proof. Actually the same error is used twice in the proof - once at the end and once at the beginning...

 

[Maetryx]

I think numbers do exist. The philosophers that deny numbers are called particularists, I think. They deny any property that can exist in multiple places at the same time, such as redness, yellowness and numbers.

Those properties are called universals. Personally, I think universals exist.

 

[Static911]

too many smart ppl here!

ahh!...

static911 runs away!

 

[Buddhist]

Maetryx,
i'm not challenging wether or not numbers exist, because on a conceptual scale they are the &quot;form&quot; of the value associated with a number.
However, when this number is transposed into reality, a slight flaw makes it no longer whole. Therefore, whole numbers can only exist in conceptual theory and not in reality. Thus the whole foundation of mathematics is conceptually perfect, in practice flawed. Chaos theory adds to this fundamental error and thus slight assumptions can over time add up quite significantly.

-M.T.O

 

[Bloodybrain]

Doh! I wonder how I managed to brain-fart such a nonsense. I guess I was just too stoned. In fact I passed calculus II and diff equations in college, and it made me unable to solve the most simple algebra problems. Thanks for pointing out Homer_Simpson, even if you did it in the most childish and immature manner.

Buddhist -
That's true for measurement of time, distances, mass, and so on, but if you take in consideration discrete quantities (say, there's two apples on the table), you can only use real numbers. It is still a concept in our minds, but fits really well with reality. Now the question is whether the universe is based on strict physical laws (numbers are behind the reality) or is it just a flawed and approximate way to interpret the physical reality? Could we run a model of the universe on an infinitely powerful computer, supposing we knew the exact initial state before the big bang, and obtain the same results as in our &quot;real&quot; world?

 

[Buddhist]

&quot;That's true for measurement of time, distances, mass, and so on, but if you take in consideration discrete quantities (say, there's two apples on the table), you can only use real numbers&quot;


&quot;two&quot; apples as we say, refer to something that approaches the &quot;Concept&quot; of apple, or the &quot;form&quot; of an apple. Thus even &quot;two&quot; apples as we see before us in fact may not constitute &quot;two&quot; apples, but rather approach the sum of two apples. However since its such a pain in the ass to think like this and we really couldn't think straight if we did, we assume it is two whole apples.


&quot;Could we run a model of the universe on an infinitely powerful computer, supposing we knew the exact initial state before the big bang, and obtain the same results as in our &quot;real&quot; world?&quot;

No. Chaos theory would intervine. That is to say UNLESS we knew EVERY SINGLE variable which would easily approach infinity. Which would theoretically be impossible.

-M.T.O

 

[bigdog1218]

this is off topic, but since this is a math relate thread, why does 0! = 1, it makes no sense, my math prof just said it equals one because it makes a lot of equations work out nicer, he can't find a proof for it, he thinks it was basically made up, i mean 2! = 2x1=2, and 1!=1 so 0!=1?

 

[Belegost]

BigDog: Because we define it as such. A proof of it is not needed, since we define 0!=1, there is nothing to prove, it simply is what we have decided to use as the convention for factorials.

I personally like to mess with conditionally convergent series. It's just fun to make something equal 1/2 of itself.

 

[Deeko]

q^r=L

so there