Endercreeper01 Posted November 25, 2013 Posted November 25, 2013 I have a proof that proves that the conjugate of a complex number is equal to the number. Proof: You have the complex number a+bi A+bi is also equal to (i(a+bi))/i Distributing gives you (ai - b)/i Distributing the division gives you a - bi Or, in equation form: a+bi = (i(a+bi))/i = (ai - b)/i = a - bi -2
Unity+ Posted November 25, 2013 Posted November 25, 2013 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 Posted November 25, 2013 Posted November 25, 2013 [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 Posted November 25, 2013 Author Posted November 25, 2013 (edited) 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 Posted November 25, 2013 Posted November 25, 2013 [latex]-b/i=(-1.b)/i=(i.i.b)/i=i.b [/latex] No, it was a - biWhy does -b/i=bi? [latex]-b/i=(-1.b)/i=(i.i.b)/i=i.b [/latex] 1
Endercreeper01 Posted November 25, 2013 Author Posted November 25, 2013 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 Posted November 25, 2013 Posted November 25, 2013 I think you also made that mistake here http://www.scienceforums.net/topic/79261-proof-abi-abi/?p=771975
hypervalent_iodine Posted November 25, 2013 Posted November 25, 2013 ! Moderator Note Do not reintroduce closed topics.
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