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Prove that a purely real number can never became equal to a purely imaginary number (Provided that the number is not 0+0i)

Note by Aman Sharma
3 years, 2 months ago

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A pure imaginary number is of the form \(bi\) where \(b\) is a [non-zero] real number and \(i^2=-1\).

If \(a\) real number were equal to \(bi\), then we would have \(a^2=-b^2\) which is impossible since the right hand side is negative while the left hand side is non-negative.

Mursalin Habib - 3 years, 2 months ago

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Thanks for replying you are awsome in proof problems

Aman Sharma - 3 years, 2 months ago

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Thanks! But I wouldn't say I'm awesome.

Mursalin Habib - 3 years, 2 months ago

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@Mursalin Habib No, you realy are awsome, also your set ''proof problem of the day'' is realy cool one

Aman Sharma - 3 years, 2 months ago

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Actually, by the definition of complex numbers, a real number is also considered a complex number. Think of it as like, we can write x as x + 0i.

Zi Song Yeoh - 3 years, 2 months ago

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I am editing it thanks for pointing out the mistake

Aman Sharma - 3 years, 2 months ago

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Sorry, all real numbers are complex numbers. If this is from NCERT, exercise and problem number please?

Agnishom Chattopadhyay - 3 years, 2 months ago

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Edited

Aman Sharma - 3 years, 2 months ago

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