I'm a Genius!! (1 Viewer)

[Dumbarse]

spent me alot of hardwork and time, but finally ive done it..

PROOF FOR 1=2

let a=b

ab = b*2

ab - a*2 = b*2 - a*2

a (b-a) = (b+a) (b-a)

a = b+a # but a=b #

so a = a + a

ie. a=2a

.'. 1=2
________________


Nobel prize worthy, dont u agree?

 

[Lazarus]

Perhaps it would be, if not for the error you made.

 

[Dumbarse]

what error?! no error!
100% perfect, heck i could prove it by induction its that foolproof!

(lazdog i'll cut u a deal, 50/50 in all prize money for my proof if u keep hushed)

 

[spice girl]

from

a (b-a) = (b+a) (b-a)

to

a = b+a # but a=b #

#div/0 error (b-a=0)

 

[mannnnndy]

I just have 2 say 1 thing, dumbarse you have a lot of time 2 on your hands shouldnt u be spending your time studying for that thing called the HSC which will determine what our lives will b like? well I cant really talk, but Ive never been so bored 2 come up with something like that.

 

[wogboy]

The error occurs after this line.

a (b-a) = (b+a) (b-a)

You can't just cancel out (b-a) from both sides, since 0/0 can equal any number not necessarily 1. Instead, subtract (b+a)(b-a) from both sides so that,

a(b-a) - (b+a)(b-a) = 0

factorising,

(b-a)(a-b-a)=0

therefore,

b-a = 0 OR a-b-a=0

therefore,

a = b OR a = b +a

You will note that there is an "OR" in cases like this and that means they don't both have to be correct. At least one must be correct. The correct statement is that a=b of course.

Your "proof" has the same logic as this:

let x = -2

therefore,

x^2 = 4

now we take the square roots of both sides and

x = 2.

Now I've proven 2 = -2

 

[Dumbarse]

manndy do u think i'd be that dumb?, (dont let the name fool ya), my maths teacher told me,
still i could study more...

 

[Lugia]

some one showed me this proof which proves -1 = 1. And I can't figure out whats wrong with it...

-1 = -1

-1/1 = 1/-1

root ( -1/1) = root (1/-1)

root -1 / root 1 = root 1 / root -1

i / 1 = 1 / i

cross multiply...

i^2 = 1

-1 = 1

:jaw: :wave:

 

[Dumbarse]

amazing, my grandma could come up with a better proof!, jokes

my new proof

-1 = -1

-1/1 = 1/-1

root ( -1/1) = root (1/-1)

root -1 / root 1 = root 1 / root -1

i / 1 = 1 / i

cross multiply...

i^2 = 1

i^2 - 1 = 0

-1-1 = 0

.'. -2 = 0 !!

 

[Lugia]

cool!

:wave:

 

[Dumbarse]

yeh soo whats the error, is there one?

 

[Lugia]

I don't know what's wrong with it... But I mean there HAS to be something wrong... Maybe I'll show my teacher tomolo.. hehe

oh I just remembered I've got this other one:

i = root -1
i^2 = root -1 * root -1
= root (-1 * -1)
= root 1
= 1

hehehe

but u see for the other one, if i^2 is 1, it would explain that one, but not for this one. :rofl:

 

[freaking_out]

take a look at this one:

three people go to a restaurant and they pitch in 10 dollars each for a meal and they give the 30 dollars to the waiter.

The waiter takes 30 dollars from the people to the cashier but he finds out that the meal only costed $25. so he took the change ($5) and puts $2 from the change into his pocket and he then gives the rest of the change ($3) to the 3 people.

so when you add the total money you find:

the money with the cashier ($25) + ($2) in the waiters pocket+($3) given back to the 3 people = $30 (which is what we started off with.

But if you want find the amount of money involved by multiplication, then you will find the following:

each person gave $9 (since they gave $10 and got $1 back) and there was 3 people + the 2 dollars in the waiters pocket=29

mathematically:
(9*3)+2 = 29

the second way of looking at the total amount of money gives us $29 but we were dealing with $30 in the first place.......my question is.........where is the missing dollar when you work out the money involved using the second method?




Goodluck

 

[DA]

Originally posted by Lugia

...
root ( -1/1) = root (1/-1)

root -1 / root 1 = root 1 / root -1
...

AFAIK, the error is when you say root (-1/1) is the same as (root -1) / (root 1). This law of indices does not hold for negative numbers, so you can't split up the square root like that.

Similar thing for the second one, (root -1)*(root -1) is not the same as root (-1 * -1).

 

[DA]

Originally posted by freaking_out
take a look at this one:

each person gave $9 (since they gave $10 and got $1 back) and there was 3 people + the 2 dollars in the waiters pocket=29

mathematically:
(9*3)+2 = 29

Okay, tell me if this is right - when you consider $27, that already includes the $2 the waiter pocketed, therefore by adding $27 and $2 you are not finding out the total money, so there is no contradiction. What can be done with the $27 and $2 is subtract them to get $25, the cost of the meal.

 

[RIZAL]

Here's a cool one

does 0.9999999999999999999999999 repeater = 1? Yes.

x = 0.9999999999999999repeater (1)

10x = 9.99999999999999999999repeater (2)

(2) - (1)

9x = 9
x=1

 

[Lazarus]

There's nothing wrong with that proof, either - it's true.

 

[feng]

Originally posted by Lugia
I don't know what's wrong with it... But I mean there HAS to be something wrong... Maybe I'll show my teacher tomolo.. hehe

oh I just remembered I've got this other one:

i = root -1
i^2 = root -1 * root -1
= root (-1 * -1)
= root 1
= 1

hey...this was from the Sydney boys paper....and yeah...thingo is right...the law of roots thing does not hold for complex numbers...

 

[wogboy]

quote:
--------------------------------------------------------------------------------
Originally posted by Lugia
I don't know what's wrong with it... But I mean there HAS to be something wrong... Maybe I'll show my teacher tomolo.. hehe

oh I just remembered I've got this other one:

i = root -1
i^2 = root -1 * root -1
= root (-1 * -1)
= root 1
= 1


--------------------------------------------------------------------------------


Actually there IS a mistake here, the laws of surds, indices etc also hold for complex numbers, but there's a slight catch to it. Whenever you take the nth root of any complex number (i.e. any number over the complex field) you must expect n different answers. Therefore it is insufficient to say that the square root of 1 is simply one (in this context). You must give BOTH square roots of 1 (which are 1 or -1), just as if you were finding the 4th roots of 1(in which case the answers would be 1, -1, i, -i).

Therefore from,

i^2 = root 1

therefore

i^2 = +1 OR i^2 = -1

Because of the OR, only one of these equations is neccessary to be correct, so that is how it really should be done.

 

[Lugia]

But shouldn't the square root of 1 be only 1 ? If it says root of 1^2 then it would be +/-1. Cause root(x^2) = +/- x but root(x) means only the positive square root of x^2 which is x.