Review: An Imaginary Tale

Reading the book, An Imaginary Tale, The Story of √-1, I got to page 4 where I ran into the insistence that mathematicians should slavishly follow their formulas, even when yielding ridiculous results. The problem wasn’t that the formula wasn’t followed. The problem was that for any formula for a positive volume to be strictly valid, it must include symbols for absolute values for each calculation involving roots. The problem was merely that the formula omitted the absolute value symbols which the mathematician intuitively supplied anyway! The calculation was not wrong; the formula was strictly incorrect.

is missing a couple of absolute value symbols, not the least of which is one which directs the mathematician to take the positive square root of the value rather than the negative square root of the value, which is just as likely as the internal value being negative. The strictly correct formula is:

but that’s cumbersome, so the absolute value symbols are usually not used.

If the rest of the book is like the first four pages, I’m going to be frustrated with it. Let’s make a mountain out of a mole hill and then call it the greatest thing since baseball.

But my objections do not end here. The basic premise of the book is wrong.

If √-1 is valid, why not 4√-1, 8√-1, 16√-1, 32√-1, and so forth? These numbers would give rise to new interlocking systems of imaginary and complex numbers.

(√-1) = √-1 [alternatively 2√-1 or (-1)½]
(√-1)2 = -1,
(√-1)3 = -√-1,
(√-1)4 = 1, with this sequence repeating for higher powers.

3√-1 = -1,
(3√-1)2 = 1,
(3√-1)3 = -1, repeating for higher powers, and not even creating any imaginary numbers.

But when we get to the fourth root of a negative number, we get a whole new group of numbers which don’t fit into the traditional scheme of imaginary and complex numbers (those numbers with both real and imaginary components).

This new number, 4√-1, might be called i’ (i prime: i, the traditional notation for √-1, and ‘, the symbol for something slightly different). Now i2=-1 and i4=1. But i’2=i, i’4=-1, i’6=-i, and only i’8=1.

4√-1 = 4√-1, <==
(4√-1)2 = √-1,
(4√-1)3 = (4√-1)(√-1), <==
(4√-1)4 = -1,
(4√-1)5 = –4√-1, <==
(4√-1)6 = -√-1
(4√-1)7 = -(4√-1)(√-1),<==
(4√-1)8 = 1, repeating for higher powers, but not reducible. <== super imaginary numbers

5√-1 = -1,
(5√-1)2 = 1,
(5√-1)3 = -1,
(5√-1)4 = 1,
(5√-1)5 = -1, also repeating for higher powers, and not creating any imaginary numbers.

6√-1 = √-1,
(6√-1)2 = -1,
(6√-1)3 = -√-1,
(6√-1)4 = 1,
(6√-1)5 = √-1,
(6√-1)6 = -1, also repeating.

7√-1 = -1,
(7√-1)2 = 1,
(7√-1)3 = -1, … , also repeating, no imaginary numbers.

And at the eighth root of a negative number, we have another divergence. The eighth root of a negative number is again something new; a figment of imagination which, when multiplied by itself eight times results in a negative number.
8√-1 = 8√-1 , <==
(8√-1)2 = 4√-1, <==
(8√-1)3 = (8√-1)(4√-1), <==
(8√-1)4 = √-1,
(8√-1)5 = (8√-1)(√-1), <==
(8√-1)6 = (4√-1)(√-1), <==
(8√-1)7 = (8√-1)(4√-1)(√-1), <==
(8√-1)8 = -1,
(8√-1)9 = –8√-1, <==
(8√-1)10 = –4√-1, <==
(8√-1)11 = -(8√-1)(4√-1), <==
(8√-1)12 = -√-1,
(8√-1)13 = -(8√-1)(√-1), <==
(8√-1)14 = -(4√-1)(√-1), <==
(8√-1)15 = -(8√-1)(4√-1)(√-1), <==
(8√-1)16 = 1, creating another line of super imaginary numbers!

Now, we get into the stickiest place of all, 12√-1!

12√-1 = 4√-1, <== check (if 12√-1 to the fourth power = -1, then this is the 4th root)
(12√-1)2 = 6√-1 = √-1, <== check (see 6√-1)
(12√-1)3 = ? <== whoops!!

Since this is also (-1)3/12, it is clearly 4√-1.
But it’s also (12√-1)(12√-1)2=(4√-1)( √-1), which isn’t the same at all.

(12√-1)4 = 3√-1 = -1 <=

By simple rules of mathematics, you can get two different values for (12√-1)3! By dividing the power by the root, we get the fourth root of -1, but by using the equivalence of the original twelfth root of -1, namely the fourth root of -1, we get an entirely different answer, the fourth root of -1 times the square root of -1. After this, all bets are off on what comes next. Thus, imaginary numbers don’t necessarily lead us to logical results.

16√-1 = 16√-1, <==
(16√-1)2 = 8√-1, <==
(16√-1)3 = (16√-1)(8√-1), <==
(16√-1)4 = 4√-1, <==
(16√-1)5 = (16√-1)(4√-1), <==
(16√-1)6 = (8√-1)(4√-1), <==
(16√-1)7 = (16√-1)(8√-1)(4√-1), <==
(16√-1)8 = √-1,
(16√-1)9 = (16√-1)(√-1), <==
(16√-1)10 = (8√-1)(√-1), <==
(16√-1)11 = (16√-1)(8√-1)(√-1), <==
(16√-1)12 = (4√-1)(√-1), <==
(16√-1)13 = (16√-1)(4√-1)(√-1), <==
(16√-1)14 = (8√-1)(4√-1)(√-1), <==
(16√-1)15 = (16√-1)(8√-1)(4√-1)(√-1), <==
(16√-1)16 = -1

I think the square root of i (the fourth root of -1) is perpendicular to the real and the imaginary axes and I surmise that each subsequent square root creates the need for another dimension in our mapping.

I am fairly certain that the fourth root of -1, the square root of the square root of -1, isn’t in the same quadrant as either real or standard imaginary numbers. It is something like a quarter of a minus, taking four of them to make a negative number.

Of course, the eighth root of -1 takes the eighth power to reach -1 and another eight multiplications to reach 1. While I can splay them all out radially, I think each time we split a negative number into two equal factors and then we split these factors again and again, we get increasing degrees of complication.

And, while I, too, have learned about imaginary and complex numbers in class and seen them used in various ways, I think they represent an interesting tool without necessarily representing an aspect of reality or being a reliable indicator of what is or isn’t valid in a rational universe.

x = (-1)1/32

x32 = -1 or x32+1 = 0

x16 = (-1)1/2

y = (-1)1/16

y32 = 1 or y32+1 = 2

y16 = -1 or y16+1 = 0

z = (-1)1/8

z32 = 1 or z32+1 = 2

z16 = 1 or z16+1 = 2

z8 = -1 or z8+1 = 0

©David N. Dodson, August 2017

Categories Miscellaneous

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