proof-that-1-2-

Question Number 149485 by abdurehime last updated on 05/Aug/21

$$\mathrm{proof}\:\mathrm{that}\:\mathrm{1}=\mathrm{2}?? \\ $$

Answered by Olaf_Thorendsen last updated on 05/Aug/21

if A = B A×A = A×B A^2 = AB = BA A^2 −B^2 = BA−B^2 (A+B)(A−B) = B(A−B) A+B = B 2B = B 2 = 1

$$\mathrm{if}\:\mathrm{A}\:=\:\mathrm{B} \\ $$$$\mathrm{A}×\mathrm{A}\:=\:\mathrm{A}×\mathrm{B} \\ $$$$\mathrm{A}^{\mathrm{2}} \:=\:\mathrm{AB}\:=\:\mathrm{BA} \\ $$$$\mathrm{A}^{\mathrm{2}} −\mathrm{B}^{\mathrm{2}} \:=\:\mathrm{BA}−\mathrm{B}^{\mathrm{2}} \\ $$$$\left(\mathrm{A}+\mathrm{B}\right)\left(\mathrm{A}−\mathrm{B}\right)\:=\:\mathrm{B}\left(\mathrm{A}−\mathrm{B}\right) \\ $$$$\mathrm{A}+\mathrm{B}\:=\:\mathrm{B} \\ $$$$\mathrm{2B}\:=\:\mathrm{B} \\ $$$$\mathrm{2}\:=\:\mathrm{1} \\ $$

Commented by cherokeesay last updated on 05/Aug/21

you can only simplify A−B on both sides of the equality (5th line) when A≠B ! but A = B ⇒ A − B = 0 !!! so you will have 0 of the sides of this equality !

$$ \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{only}\:\mathrm{simplify}\:\mathrm{A}−\mathrm{B}\:\mathrm{on}\:{both}\:{sides}\:{of}\:{the}\:{equality}\:\left(\mathrm{5}{th}\:{line}\right)\:{when}\:{A}\neq{B}\:! \\ $$$${but}\:{A}\:=\:{B}\:\Rightarrow\:{A}\:−\:{B}\:=\:\mathrm{0}\:!!! \\ $$$${so}\:{you}\:{will}\:{have}\:\mathrm{0}\:{of}\:{the}\:{sides}\:{of}\:{this}\:{equality}\:! \\ $$

Commented by peter frank last updated on 05/Aug/21

$${A}+{B}={B} \\ $$$${A}=\mathrm{0}\: \\ $$

Commented by Olaf_Thorendsen last updated on 05/Aug/21

Of course sir ! But in the real life 1 ≠ 2 too.

$$\mathrm{Of}\:\mathrm{course}\:\mathrm{sir}\:! \\ $$$$\mathrm{But}\:\mathrm{in}\:\mathrm{the}\:\mathrm{real}\:\mathrm{life}\:\mathrm{1}\:\neq\:\mathrm{2}\:\mathrm{too}. \\ $$

Commented by MJS_new last updated on 06/Aug/21

I remember someone here proved i=(√(−1))=±1 but I don′t remember how... that guy was seriously convinced... as I stated before: those who know nothing must believe everything

$$\mathrm{I}\:\mathrm{remember}\:\mathrm{someone}\:\mathrm{here}\:\mathrm{proved} \\ $$$$\mathrm{i}=\sqrt{−\mathrm{1}}=\pm\mathrm{1} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{remember}\:\mathrm{how}…\:\mathrm{that}\:\mathrm{guy}\:\mathrm{was} \\ $$$$\mathrm{seriously}\:\mathrm{convinced}…\:\mathrm{as}\:\mathrm{I}\:\mathrm{stated}\:\mathrm{before}: \\ $$$${those}\:{who}\:{know}\:{nothing}\:{must}\:{believe}\:{everything} \\ $$

Commented by MJS_new last updated on 06/Aug/21

...on the other hand f(x)=(x/x) lim_(x→0) f(x) =1 ⇒ (0/0)=1 g(x)=((2x)/x) lim_(x→0) g(x) =2 ⇒ (0/0)=2 (0/0)=1∧(0/0)=2 ⇒ 1=2 q.e.d. now what′s you argument Mr. Peter Frank?

$$…\mathrm{on}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hand} \\ $$$${f}\left({x}\right)=\frac{{x}}{{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({x}\right)\:=\mathrm{1}\:\Rightarrow\:\frac{\mathrm{0}}{\mathrm{0}}=\mathrm{1} \\ $$$${g}\left({x}\right)=\frac{\mathrm{2}{x}}{{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{g}\left({x}\right)\:=\mathrm{2}\:\Rightarrow\:\frac{\mathrm{0}}{\mathrm{0}}=\mathrm{2} \\ $$$$ \\ $$$$\frac{\mathrm{0}}{\mathrm{0}}=\mathrm{1}\wedge\frac{\mathrm{0}}{\mathrm{0}}=\mathrm{2}\:\Rightarrow\:\mathrm{1}=\mathrm{2}\:{q}.{e}.{d}. \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{what}'\mathrm{s}\:\mathrm{you}\:\mathrm{argument}\:\mathrm{Mr}.\:\mathrm{Peter}\:\mathrm{Frank}? \\ $$

Commented by MJS_new last updated on 06/Aug/21

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Commented by iloveisrael last updated on 06/Aug/21

the other way lim_(x→0) ((sin x)/x)=1=(0/0) lim_(x→0) ((tan (4x))/(2x))=2=(0/0) then 1=2

$$\mathrm{the}\:\mathrm{other}\:\mathrm{way}\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}=\mathrm{1}=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\left(\mathrm{4x}\right)}{\mathrm{2x}}=\mathrm{2}=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\mathrm{then}\:\mathrm{1}=\mathrm{2}\: \\ $$

Commented by Olaf_Thorendsen last updated on 06/Aug/21

1 = (√1) 1 = (√((−1)×(−1))) 1 = (√(−1))×(√(−1)) 1 = i×i = i^2 = −1 1 = −1

$$\mathrm{1}\:=\:\sqrt{\mathrm{1}} \\ $$$$\mathrm{1}\:=\:\sqrt{\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right)} \\ $$$$\mathrm{1}\:=\:\sqrt{−\mathrm{1}}×\sqrt{−\mathrm{1}} \\ $$$$\mathrm{1}\:=\:{i}×{i}\:=\:{i}^{\mathrm{2}} =\:−\mathrm{1} \\ $$$$\mathrm{1}\:=\:−\mathrm{1} \\ $$

Commented by mathmax by abdo last updated on 06/Aug/21

$$\mathrm{but}\:\mathrm{its}\:\mathrm{correct}\:\mathrm{dans}\:\mathrm{Z}/\mathrm{2Z} \\ $$

Commented by MJS_new last updated on 06/Aug/21

many people (maybe most people) forget that the rules in R are different from those in C. (we usually don′t have to think about the differences between the rules of N, Z and Q as we just “walk through” them while we start to study)

$$\mathrm{many}\:\mathrm{people}\:\left(\mathrm{maybe}\:\mathrm{most}\:\mathrm{people}\right)\:\mathrm{forget}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{rules}\:\mathrm{in}\:\mathbb{R}\:\mathrm{are}\:\mathrm{different}\:\mathrm{from}\:\mathrm{those}\:\mathrm{in}\:\mathbb{C}. \\ $$$$\left(\mathrm{we}\:\mathrm{usually}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{to}\:\mathrm{think}\:\mathrm{about}\:\mathrm{the}\right. \\ $$$$\mathrm{differences}\:\mathrm{between}\:\mathrm{the}\:\mathrm{rules}\:\mathrm{of}\:\mathbb{N},\:\mathbb{Z}\:\mathrm{and}\:\mathbb{Q} \\ $$$$\mathrm{as}\:\mathrm{we}\:\mathrm{just}\:“\mathrm{walk}\:\mathrm{through}''\:\mathrm{them}\:\mathrm{while}\:\mathrm{we} \\ $$$$\left.\mathrm{start}\:\mathrm{to}\:\mathrm{study}\right) \\ $$

Commented by ajfour last updated on 06/Aug/21

2^(nd) step to 3^(rd) step is not allowed, there are certain rules here that need be followed..

$$\mathrm{2}^{{nd}} \:{step}\:{to}\:\mathrm{3}^{{rd}} \:\:{step}\:{is}\:{not} \\ $$$${allowed},\:{there}\:{are}\:{certain} \\ $$$${rules}\:{here}\:{that}\:{need}\:{be} \\ $$$${followed}.. \\ $$

Commented by peter frank last updated on 06/Aug/21

$${no}\:{argument}\:{sir}.{understood} \\ $$

Answered by puissant last updated on 06/Aug/21

a=b ⇒ a^2 =ab ⇒a^2 +b^2 =b^2 +ab ⇒a^2 +b^2 −2ab=b^2 −ab ⇒ 2a^2 −2ab = b^2 −ab car a=b ⇒ 2a(a−b)=a(a−b) ⇒ 2a=a soit 2=1...

$$\mathrm{a}=\mathrm{b} \\ $$$$\Rightarrow\:\mathrm{a}^{\mathrm{2}} =\mathrm{ab} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{b}^{\mathrm{2}} +\mathrm{ab} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2ab}=\mathrm{b}^{\mathrm{2}} −\mathrm{ab} \\ $$$$\Rightarrow\:\mathrm{2a}^{\mathrm{2}} −\mathrm{2ab}\:=\:\mathrm{b}^{\mathrm{2}} −\mathrm{ab}\:\mathrm{car}\:\mathrm{a}=\mathrm{b} \\ $$$$\Rightarrow\:\mathrm{2a}\left(\mathrm{a}−\mathrm{b}\right)=\mathrm{a}\left(\mathrm{a}−\mathrm{b}\right) \\ $$$$\Rightarrow\:\mathrm{2a}=\mathrm{a}\:\mathrm{soit}\:\mathrm{2}=\mathrm{1}… \\ $$

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