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Question Number 56244 by Kunal12588 last updated on 12/Mar/19

$$Is\mathrm{\infty }acomplexnumber.$$$$IfnotsowhatisIt.$$

Commented by Joel578 last updated on 12/Mar/19

$$itisnotanumber$$

Commented by Kunal12588 last updated on 12/Mar/19

$$okiedokie.$$$$waitingforthenwhatitis$$$$andithoughtwecanrepresentitas\frac{1}{0}.$$$$so\frac{1}{0}isnotanumber.butbothnumerator$$$$anddenominatorarenumbers.$$

Commented by 121194 last updated on 12/Mar/19

$$\mathrm{one}\mathrm{may}\mathrm{think}\mathrm{that}\frac{1}{0}=\mathrm{\infty },\mathrm{but}\mathrm{that}\mathrm{not}\mathrm{entire}\mathrm{true}$$$$\mathrm{think}\mathrm{of}$$$$\underset{x\to 0}{\lim }\frac{1}{x}$$$$\mathrm{lets}\mathrm{aproach}\mathrm{it}2\mathrm{ways}$$$$1\to 1$$$$0.5\to 2$$$$0.1\to 10$$$$0.01\to 100$$$$0.001\to 1000$$$$asyoucanseeascloseweaproachx1/xgrow$$$$withoutanylimit,so$$$$\underset{x\to 0^{+}}{\lim }\frac{1}{x}=+\mathrm{\infty }$$$$butthatonly1path,takesthe2ndone$$$$-1\to -1$$$$-0.5\to -2$$$$-0.1\to -10$$$$-0.01\to -100$$$$asyouseeitgrowincompletopostdirection,as$$$$\underset{x\to 0^{-}}{\lim }\frac{1}{x}=-\mathrm{\infty }$$$$sincethe2limitsarediferentswe{can}^{\prime }tsaythat$$$$thelimitexistso$$$$\underset{x\to 0}{\lim }\frac{1}{x}=\nexists$$$$btwthereanyanalogousof\mathrm{\infty }tocomplex,the$$$$complexinfinite\stackrel{\sim }{\mathrm{\infty }}$$$$itbasicalya‘‘{number}^{″}withhighmodulosbut$$$$unknowargument$$

Commented by Joel578 last updated on 12/Mar/19

$$\frac{1}{0}\mathrm{is}\mathrm{actually}\mathrm{undefined}$$$$\mathrm{Now},\mathrm{let}f\left(x\right)=\frac{1}{x}$$$$\mathrm{If}\mathrm{we}\mathrm{take}x=0.00000001,f\left(x\right)=100000000$$$$\mathrm{Next},\mathrm{if}x\mathrm{getting}\mathrm{smaller},\mathrm{very}\mathrm{close}\mathrm{to}\mathrm{zero}$$$$\mathrm{then}\mathrm{our}f\left(x\right)\mathrm{is}\mathrm{getting}\mathrm{larger}$$$$\mathrm{Next},x\mathrm{now}\mathrm{is}\mathrm{very}\mathrm{very}\mathrm{very}\mathrm{smaller},\mathrm{then}$$$$\mathrm{our}f\left(x\right)\mathrm{is}\mathrm{very}\mathrm{very}\mathrm{very}\mathrm{large}.$$$$\mathrm{In}\mathrm{this}\mathrm{context},\mathrm{we}\mathrm{can}\mathrm{denote}\mathrm{this}\mathrm{as}$$$$\underset{x\to 0^{+}}{\lim }\frac{1}{x}=\frac{1}{0}=\mathrm{\infty }$$$$\mathrm{As}x\mathrm{is}\mathrm{getting}\mathrm{smaller},\mathrm{very}\mathrm{close}\mathrm{to}\mathrm{zero}$$$$\mathrm{then}\mathrm{its}\mathrm{value}\mathrm{will}\mathrm{reach}\mathrm{an}\mathrm{arbitrarily}\mathrm{large}\mathrm{number}.$$$$\mathrm{How}\mathrm{large}?\mathrm{Very}\mathrm{large}.\mathrm{And}\mathrm{we}\mathrm{denote}\mathrm{this}\mathrm{with}‘‘{\mathrm{\infty }}^{″}$$$$\mathrm{So},\mathrm{in}\mathrm{my}\mathrm{opinion}‘‘{\mathrm{\infty }}^{″}\mathrm{is}\mathrm{a}\mathrm{concept}\mathrm{to}\mathrm{describe}$$$$\mathrm{an}\mathrm{arbitrarily}\mathrm{large}\mathrm{number},\mathrm{but}\mathrm{it}{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{a}\mathrm{number}$$

Commented by Joel578 last updated on 12/Mar/19

$${\mathrm{That}}^{\prime }\mathrm{s}\mathrm{what}\mathrm{I}\mathrm{know}.\mathrm{I}\mathrm{am}\mathrm{open}\mathrm{for}\mathrm{any}\mathrm{critics}\mathrm{if}$$$$\mathrm{I}\mathrm{made}\mathrm{any}\mathrm{mistakes}$$

Commented by Kunal12588 last updated on 12/Mar/19

$$thankyouverymuchsirs.$$

Commented by prakash jain last updated on 12/Mar/19

please visit the below link for definition of complex infinity http://mathworld.wolfram.com/ComplexInfinity.html

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