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LOL-found-this-on-the-web-1-1-1-1-1-1-i-2-1-each-step-seems-right-so-where-s-the-mistake-




Question Number 44161 by MJS last updated on 22/Sep/18
LOL! found this on the web:  1=(√1)=(√((−1)(−1)))=(√(−1))(√(−1))=i^2 =−1  each step seems right, so where′s the mistake?
$$\mathrm{LOL}!\:\mathrm{found}\:\mathrm{this}\:\mathrm{on}\:\mathrm{the}\:\mathrm{web}: \\ $$$$\mathrm{1}=\sqrt{\mathrm{1}}=\sqrt{\left(−\mathrm{1}\right)\left(−\mathrm{1}\right)}=\sqrt{−\mathrm{1}}\sqrt{−\mathrm{1}}=\mathrm{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\mathrm{each}\:\mathrm{step}\:\mathrm{seems}\:\mathrm{right},\:\mathrm{so}\:\mathrm{where}'\mathrm{s}\:\mathrm{the}\:\mathrm{mistake}? \\ $$
Answered by rahul 19 last updated on 22/Sep/18
(√((−1)(−1) ))   ≠ (√(−1)) (√(−1))     General :  (√(ab)) = (√a) (√b) only  when atleast one of  a&b is non−negative....  :)
$$\sqrt{\left(−\mathrm{1}\right)\left(−\mathrm{1}\right)\:}\:\:\:\neq\:\sqrt{−\mathrm{1}}\:\sqrt{−\mathrm{1}}\:\:\: \\ $$$${General}\:: \\ $$$$\sqrt{{ab}}\:=\:\sqrt{{a}}\:\sqrt{{b}}\:{only}\:\:{when}\:{atleast}\:{one}\:{of} \\ $$$${a\&b}\:{is}\:{non}−{negative}…. \\ $$$$\left.:\right) \\ $$
Commented by MJS last updated on 22/Sep/18
you′re right!
$$\mathrm{you}'\mathrm{re}\:\mathrm{right}! \\ $$
Answered by malwaan last updated on 22/Sep/18
(√1)=±1  so (√1)=1 or (√1)=−1 wrong
$$\sqrt{\mathrm{1}}=\pm\mathrm{1} \\ $$$$\mathrm{so}\:\sqrt{\mathrm{1}}=\mathrm{1}\:\mathrm{or}\:\sqrt{\mathrm{1}}=−\mathrm{1}\:\mathrm{wrong} \\ $$
Commented by MJS last updated on 22/Sep/18
many people believe this but it′s not true.  (√x) always means the positive root, if  x=re^(θi)  ⇒ (√x)=(√r)e^((θ/2)i)   only if you′re solving an equation like  x^2 =a you get x=±(√a)  similar for (x)^(1/n)
$$\mathrm{many}\:\mathrm{people}\:\mathrm{believe}\:\mathrm{this}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}. \\ $$$$\sqrt{{x}}\:\mathrm{always}\:\mathrm{means}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{root},\:\mathrm{if} \\ $$$${x}={r}\mathrm{e}^{\theta\mathrm{i}} \:\Rightarrow\:\sqrt{{x}}=\sqrt{{r}}\mathrm{e}^{\frac{\theta}{\mathrm{2}}\mathrm{i}} \\ $$$$\mathrm{only}\:\mathrm{if}\:\mathrm{you}'\mathrm{re}\:\mathrm{solving}\:\mathrm{an}\:\mathrm{equation}\:\mathrm{like} \\ $$$${x}^{\mathrm{2}} ={a}\:\mathrm{you}\:\mathrm{get}\:{x}=\pm\sqrt{{a}} \\ $$$$\mathrm{similar}\:\mathrm{for}\:\sqrt[{{n}}]{{x}} \\ $$
Commented by Joel578 last updated on 23/Sep/18
So (√1) = 1, but (√1^2 ) = ∣1∣, right?
$$\mathrm{So}\:\sqrt{\mathrm{1}}\:=\:\mathrm{1},\:\mathrm{but}\:\sqrt{\mathrm{1}^{\mathrm{2}} }\:=\:\mid\mathrm{1}\mid,\:\mathrm{right}? \\ $$

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