Действительно ли мнимая единица равна 1?
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notice that the fraction I over theare
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root of I can be rationalized by
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multiplying the numerator and the
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denominator by the square root of I now
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we can simply cancel out these two
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imaginary units on the top and on the
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bottom and this gives us the square root
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of I so we found our first equation I
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over the < TK of I equal the < TK of
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I notice also
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that 1 over the square root of I
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equal the < TK of 1 over the < TK of I
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which equal the < TK of 1 over I which
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is the same as the square < TK of 1 over
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theare < TK of -1 which is the same as
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the square < TK of theare < TK of 1 over
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the square < TK of
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minus1 and this is the same as the
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square < TK of theare < TK of 1 / -1
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which equals the square < TK of the squ
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root of min-1 in other words the square
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root of
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I thus we found our second
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equation it turns out that these two
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equations are the same so we can say
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that I over the < TK I = 1/ the < TK of
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I but wait a
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second this means that if we cancel out
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these two terms in the denominator
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we get I =
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1 and that’s it thanks for
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watching no this is wrong but why is it
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wrong I mean what was our mistake here
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well let’s replay the
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scene 1/ the < TK of I = the < TK of 1
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over the < TK of I which equal the squ
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root of 1 / I which is the same as the
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Square < t of 1 over the square < TK of
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-1 which is the same as the square < TK
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of theare < TK of 1 over theare < TK of
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minus1 and this is the same as the
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square root of the square < TK of 1 / -1
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which
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equals this is not
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true so actually our second equation is
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not
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true but
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why well the answer is that the property
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that the square root of a over the
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square otk of b equal the square < TK of
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A over B is valid if and only if a is
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greater than or equal to zero and B is
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greater than zero so the problem is that
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we cannot apply this property for the
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square root of minus1 and that’s where
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the contradiction lies if you enjoyed
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