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1 + 1 = 1 ? ? ? relaxing ...

cK_hoangan

Posted 28 November 2007 - 10:29 PM

This is a "trick stuff" in Maths which quite interesting. Hope that anyone will enjoy it.

Assume : A = B
=> A^2 = AB (multiply both by A)
=> A^2 - B^2 = AB - B^2 (take away both to B^2)
=> (A-B)(A+B)= B (A-B)
=> A+B = B
=> 2B = B (hence A = B)
=> 2 = 1
=> 1 + 1 = 1 blink.gif BINGO !!!




I think we shouldn't study Maths anymore cause It's all wrong now ... tongue.gif

Jammie1000000

Posted 29 November 2007 - 01:31 PM

QUOTE(cK_hoangan @ Nov 28 2007, 10:29 PM) View Post
This is a "trick stuff" in Maths which quite interesting. Hope that anyone will enjoy it.

Assume : A = B
=> A^2 = AB (multiply both by A)
=> A^2 - B^2 = AB - B^2 (take away both to B^2)
=> (A-B)(A+B)= B (A-B)
=> A+B = B
=> 2B = B (hence A = B)
=> 2 = 1
=> 1 + 1 = 1 blink.gif BINGO !!!




I think we shouldn't study Maths anymore cause It's all wrong now ... tongue.gif


(A-B)=0 so you cannot divide through by (A-B) as you cannot divide by 0

cool.gif


cK_hoangan

Posted 29 November 2007 - 01:45 PM

Well-done !!!! rolleyes.gif

as I said, it's all for relaxing : rolleyes.gifD

You wanna try once more with 1 > 4 ? I will prove it soon tongue.gif

Marcus

Posted 06 January 2008 - 04:01 PM

just to go along the same lines of posting bizarre mathematics, here is a proof that \frac{0}{0}= 2 or in fact any number you want blink.gif

eqn 1) x+1 = 0
eqn 2) 2x+2 = 0

divide eqn 2 by eqn 1

\frac{2x+2}{x+1}= \frac{0}{0}

cancel the x+1 on top and bottom to get:

2=\frac{0}{0}

but this \frac{0}{0} = anything helps prove the above problem...

=> (A-B)(A+B)= B (A-B) , (A-B) = 0
=> 0(A+B)= 0B
=>2B = \frac{0}{0} B
so if \frac{0}{0} =2
then this works XD XD

ad absurdum

Posted 07 January 2008 - 07:44 PM

Here's a tough one (you'll need to have done complex numbers, can't remember when it comes up in advanced higher):

2 = e^{log(2)} = e^{i 2 \pi \frac{log(2)}{i 2 \pi}} = (e^{i 2 \pi})^\frac{log(2)}{i 2 \pi}}= 1^{\frac{log(2)}{i 2 \pi}} = 1

Where's the error?


Marcus

Posted 10 January 2008 - 11:43 PM

Right I don't think this is correct but:

e ^{ 2 i \pi } = 1 = e^0

and also i2 \pi is usually written as i(0) as angles usually lie between - \pi and  \pi

so when you multiplied the ln(2) by  \frac{i2 \pi}{i2 \pi} you were effectively multiplying by  \frac{0}{i2 \pi} which you can't do.

Am i even remotely on the right lines here? unsure.gif

englisher

Posted 21 February 2008 - 02:55 PM

QUOTE(ad absurdum @ Jan 7 2008, 08:44 PM) View Post
Here's a tough one (you'll need to have done complex numbers, can't remember when it comes up in advanced higher):

2 = e^{log(2)} = e^{i 2 \pi \frac{log(2)}{i 2 \pi}} = (e^{i 2 \pi})^\frac{log(2)}{i 2 \pi}}= 1^{\frac{log(2)}{i 2 \pi}} = 1

Where's the error?

Can't put brackets round e^2ipi ?

Marcus

Posted 22 February 2008 - 09:16 PM

QUOTE(englisher @ Feb 21 2008, 02:55 PM) View Post
Can't put brackets round e^2ipi ?

I wish it was as simple as that, but yeah you can... one of the rules of indices
x^{ab} = (x^{a})^{b}

in the case of
e^{i2\pi\frac{\ln(2)}{i2\pi}}
a would be i2\pi
and b would be \frac{\ln(2)}{i2\pi}

Wildespleen

Posted 21 January 2009 - 08:13 PM

Can I just confirm that e^2ipi is indeed one? I presume it originates from 'e^ipi + 1 = 0'? (Which we haven't really covered in class so far, we only mentioned it).
I've puzzled over this one for a while now and the only real possible fault I, personally, can find is this. Failing that, I fear the textbooks will have to be rewritten-> 1+1=1 and 2+2=2! There'll be anarchy in the streets! XD

Marcus

Posted 27 January 2009 - 02:12 PM

QUOTE (Wildespleen @ Jan 21 2009, 08:13 PM) <{POST_SNAPBACK}>
Can I just confirm that e^2ipi is indeed one? I presume it originates from 'e^ipi + 1 = 0'? (Which we haven't really covered in class so far, we only mentioned it).
I've puzzled over this one for a while now and the only real possible fault I, personally, can find is this. Failing that, I fear the textbooks will have to be rewritten-> 1+1=1 and 2+2=2! There'll be anarchy in the streets! XD


Yeah e^{2i\pi}=1, and it comes from Euler's formula: (only briefly mentioned in AH) re^{i\theta}=r(\cos{\theta}+i\sin{\theta})

I had to ask my uni profs to explain what is wrong with this proof...

Monkey Man

Posted 02 January 2010 - 06:50 PM

QUOTE Jammie1000000
QUOTE(cK_hoangan @ Nov 28 2007, 10:29 PM) View Post
This is a "trick stuff" in Maths which quite interesting. Hope that anyone will enjoy it.

Assume : A = B
=> A^2 = AB (multiply both by A)
=> A^2 - B^2 = AB - B^2 (take away both to B^2)
=> (A-B)(A+B)= B (A-B)
=> A+B = B
=> 2B = B (hence A = B)
=> 2 = 1
=> 1 + 1 = 1 blink.gif BINGO !!!



I think we shouldn't study Maths anymore cause It's all wrong now ... tongue.gif


(A-B)=0 so you cannot divide through by (A-B) as you cannot divide by 0

cool.gif

can i just add that this is not dividing through by (A-B), they cancel each other out :P :D

Marcus

Posted 27 January 2010 - 05:37 PM

but they cancel out, because you are dividing through by (a-b) << that the operation you are actually doing when you cancel out terms

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