anonymous
  • anonymous
a=b 1=2 where is the fallacy?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[a^{2}=b^{2}\]
terenzreignz
  • terenzreignz
What are we to do with this given equation?
anonymous
  • anonymous
Prove it?

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More answers

anonymous
  • anonymous
let me change the equation
UnkleRhaukus
  • UnkleRhaukus
so many solutions
UnkleRhaukus
  • UnkleRhaukus
second line is false
terenzreignz
  • terenzreignz
LOL... I think you're holding out on us XD
anonymous
  • anonymous
hmm I'll make it longer
UnkleRhaukus
  • UnkleRhaukus
get a real question,
anonymous
  • anonymous
a=b. a^2=ab a^2+a^2=a^2+ab 2a^2 wait I'm being called by my mom I'll continue this later http://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html
anonymous
  • anonymous
more simple: 1º: a=b 2º: ab=b^2 (multiply bouth sides by b) 3º: ab-a^2=b^2-a^2 (rest a^2 from bouth sides) 4º: a(b-a)=(b-a)(b+a) 5º: a=b+a 6º: a= a+a (since a=b) 7º: a=2a or 1=2 the falacy is in step 4º to 5º, because (b-a)=0. Remmember a=b. So we actualy devided by 0, which is not alowed
anonymous
  • anonymous
it is, if a\(\neq =0\)
anonymous
  • anonymous
what is your concern?
anonymous
  • anonymous
i am curious, :)
anonymous
  • anonymous
@Loser66 ?
anonymous
  • anonymous
a=2a a/a=2a/a 1=2
anonymous
  • anonymous
the problem is already with a=2a. But we got here because of step 4,5
anonymous
  • anonymous
it is imposible to happen that a=2a for \(a \neq 0\)

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