proof for 1=2

chethiyakb

Member
Apr 18, 2007
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Let a=b

therefore a*a=ab

a*a +a*a=a*a+ab

2a*a=a*a+ab

2a*a-2ab=a*a+ab-2ab

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

2(a*a-2ab)=1(a*a-ab)

when both sides are divided by (a*a-ab) :D




ps: we can't divide a number by zero. If we divide, the result is undefined (usually said the infinity)
 

tdevinda

Member
Nov 16, 2006
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Mattegoda
:S

this is total nonsense. When you substitute a for b at even the first instance you end up with 0=0....

Looks like you're new to maths. Did you just start AL? Finish the pure maths and you'll turn purple every time you remember you posted something like this :D

LoL...
 

yasasrd

Member
Mar 9, 2007
5,962
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In elakiri
tdevinda said:
:S

this is total nonsense. When you substitute a for b at even the first instance you end up with 0=0....

Looks like you're new to maths. Did you just start AL? Finish the pure maths and you'll turn purple every time you remember you posted something like this :D

LoL...


:yes: :yes: :yes: :yes: :yes: :yes:
 

chethiyakb

Member
Apr 18, 2007
314
4
0
tdevinda said:
:S

this is total nonsense. When you substitute a for b at even the first instance you end up with 0=0....

Looks like you're new to maths. Did you just start AL? Finish the pure maths and you'll turn purple every time you remember you posted something like this :D

LoL...

I posted this for newcomers to the mathematics field. If u were at my age you gonna take this serious for sure.

Anyway this is a joke ( not when my math teacher gave this!!)
 

henderson

Active member
  • Nov 24, 2007
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    Senior Executive, Goldman Sachs
    Let a=b

    therefore a*a=ab

    a*a +a*a=a*a+ab

    2a*a=a*a+ab

    2a*a-2ab=a*a+ab-2ab

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

    2(a*a-2ab)=1(a*a-ab)

    when both sides are divided by (a*a-ab)

    Now the issue is your math is perfectly correct until last line,
    how can the (a * a - 2ab)/(a*a-ab) = 1

    so that is not a proof or nowhere near
     

    madurax86

    Member
    Jun 29, 2006
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    mokadda me? :0
    a=b
    therefore a*a=ab

    a*a +a*a=a*a+ab

    2a*a=a*a+ab

    2a*a-2ab=a*a+ab-2ab

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

    2(a*a-ab)=1(a*a-ab) ...ur proof should be lyk dis i think but as u assigned a=b; then a*a-ab=0 how da hell are u gonna divide by 0? its limit goes to infinty! u better learn math b4 posting dude
     

    alagakkonara

    Member
    Sep 25, 2008
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    Guys this is a common problem, where we can prove any number is equal to any number.

    This is something we normally learn while doing A/L maths. And of course we use division by zero, so yeah no real meaning in the proof.

    Everybody who did maths knows that. So this proof is for those who dont know it to have some funtime figuring it out.

    Give him a break,
     

    magicman

    Member
    Feb 8, 2008
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    In the magic world
    tdevinda said:
    :S

    this is total nonsense. When you substitute a for b at even the first instance you end up with 0=0....

    Looks like you're new to maths. Did you just start AL? Finish the pure maths and you'll turn purple every time you remember you posted something like this :D

    LoL...



    :yes::yes::yes::yes::yes::yes::yes::yes::yes::yes::nerd::nerd::D:D:D
     

    aragon

    Well-known member
  • Oct 16, 2008
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    Very poor attempt to prove it, that's completely nonsense. U better to start learning maths from it's beginning.