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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
2 years, 8 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 · 2 years, 8 months ago

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@Mursalin Habib Thanks for replying you are awsome in proof problems Aman Sharma · 2 years, 8 months ago

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@Aman Sharma Thanks! But I wouldn't say I'm awesome. Mursalin Habib · 2 years, 8 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 · 2 years, 8 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 · 2 years, 8 months ago

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@Zi Song Yeoh I am editing it thanks for pointing out the mistake Aman Sharma · 2 years, 8 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 · 2 years, 8 months ago

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@Agnishom Chattopadhyay Edited Aman Sharma · 2 years, 8 months ago

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