Endercreeper01 Posted November 25, 2013 Share 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 Link to comment Share on other sites More sharing options...
Unity+ Posted November 25, 2013 Share 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? Link to comment Share on other sites More sharing options...
timo Posted November 25, 2013 Share 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. Link to comment Share on other sites More sharing options...
Endercreeper01 Posted November 25, 2013 Author Share 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 Link to comment Share on other sites More sharing options...
imatfaal Posted November 25, 2013 Share 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 Link to comment Share on other sites More sharing options...
Endercreeper01 Posted November 25, 2013 Author Share 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 Link to comment Share on other sites More sharing options...
John Cuthber Posted November 25, 2013 Share Posted November 25, 2013 I think you also made that mistake here http://www.scienceforums.net/topic/79261-proof-abi-abi/?p=771975 Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted November 25, 2013 Share Posted November 25, 2013 ! Moderator Note Do not reintroduce closed topics. Link to comment Share on other sites More sharing options...
Recommended Posts