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# Algebra/Problem of advance functional analysis

Question

Question
Helo, sir I try to prove it through some ways but I can't do I feel your help if you can sort out this problem of division algebra

Any number in the complex number system can be expressed as a+bi.
If we take the inverse of a+bi we get 1/(a+bi).

If this is multiplied in the numerator and denominator by a-bi, the result is
(a-bi)/[(a+bi)(a-bi)].  Now since (a+bi)(a-bi) is aČ-(ibČ), and that is aČ+bČ,
This can be rewritten as a/(aČ+bČ) - [b/(aČ+bČ)]i.

Since either a or b is not 0 { for if a=b=0, the number is 0, and that is never invertible }
we never need to worry about the fraction being undefined.

Now since a and (aČ+bČ) are both real, a/(aČ+bČ) is real.
Also, since b is real and it was just stated that (aČ+bČ) is real,
b/(aČ+bČ) is real.

This means that a/(aČ+bČ) - [b/(aČ+bČ)]i is still in the imaginary numbers.
Since a nd b were taken to be anything, the whole set is divisible.

Algebra

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#### Scott A Wilson

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