Proof: C*=C

[Endercreeper01]

I have a proof that proves that the conjugate of a complex number is equal to the number. Proof:

1. You have the complex number a+bi
2. A+bi is also equal to (i(a+bi))/i
3. Distributing gives you (ai - b)/i
4. Distributing the division gives you a - bi

Or, in equation form:

a+bi = (i(a+bi))/i = (ai - b)/i = a - bi

[Unity+]

The mistake happened in step (ai-b)/i = ai-b because they are not equal. You removed a term without even removing i from the top portion.

 

And what do you mean by distributing the division? Do you mean simplifying?

[timo]

[math] -b/i = +bi \neq -bi [/math]. Not to mention that a simple check whether 0+1i equals 0-1i would have sufficed to sanity-check your alleged proof.

[Endercreeper01]

The mistake happened in step (ai-b)/i = ai-b because they are not equal. You removed a term without even removing i from the top portion.

 

And what do you mean by distributing the division? Do you mean simplifying?

No, it was a - bi

[math] -b/i = +bi \neq -bi [/math]. Not to mention that a simple check whether 0+1i equals 0-1i would have sufficed to sanity-check your alleged proof.

Why does -b/i=bi? Edited November 25, 2013 by Endercreeper01

[imatfaal]

[latex]-b/i=(-1.b)/i=(i.i.b)/i=i.b [/latex]

No, it was a - bi
Why does -b/i=bi?

[latex]-b/i=(-1.b)/i=(i.i.b)/i=i.b [/latex]

[Endercreeper01]

You can also make the argument that it equals both bi and -bi since -b/i=((-b)^2/i^2)^0.5=(-b^2)^(1/2), which is both bi and -bi

[John Cuthber]

I think you also made that mistake here

http://www.scienceforums.net/topic/79261-proof-abi-abi/?p=771975

[hypervalent_iodine]

!

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