Stupid, stupid. I read x∊ℤ as a complex number. z is a common choice of a complex number, z=a+bi, z=1+i. Unconsciously until now I have wanted the capital letter further down the alphabet to be more complicated classes of numbers, as in ℕ, ℚ, ℝ. So Using ℂ for complex goes against the lettered trend. I can remove the problem by learning the word zahlen, the German for number.
Original problem was to solve 2|x| + (-1)ˣ = 17
ℤ is the set of all integers, ℕ is the natural numbers (including 0 sometimes, ℕ⁺ defintely excludes 0), ℚ is rational numbers, ℝ is the real numbers, ℂ is complex numbers. Of those used less often, I found irrational numbers expressed as ℙ, 𝔽 and ℚ' (not-ℚ, ℚ prime, ℚ⋃ℚ'=ℝ). I think the last of these makes the most sense. There are quarternions, ℍ; and octonions, 𝕆. I'm a little surprised that there is confusion and that there is no convention for a transcendental number (I think all transcendentals are irrational). Adding two irrationals may (!!) not be irrational. (homework)
Here's a fine diagram offering a glimmer of understanding, c/o wikipedia.
Quarternions: 3-d complex a+bi+cj+dk and think of i,j,k as being unit vectors along mutually perpendicular axes. ℍ for Hamilton. Fundamentally, i²=j²=k²= ijk=-1.
Octonions: Obscure, but maybe so fundamental that we ought to study them a lot more. There are eight fundamental unit octonions, e₀ to e₇; to this one must add a multiplication table, though there are many (480) possible tables. For example one might make e₃e₄=e₇ and e₄e₇=e₃ while e₇e₄=-e₃ (because 4+7=3 mod 8). This is an 8-d space, one real dimension and seven imaginary, just as the quaternion is one real and three imaginary. Octonions generate descriptions of algebras not previously explored and provide further understanding of the well-understood (we thought) other algebras. It seemed immediately clear to me to wonder whether octonions (or some similar multidimensional algebra) is what we need to explain quantum theory. Provided we use sufficient dimensions, of course. If octonions are adequate, then finding the right multiplication matrix is going to require an awful lot of expensive research, though my understanding is that the work should be repeatable, which means we don't have to do it many times to be sure we have a piece of the puzzle.
Solve 2|x| + (-1)ˣ = 17
Solution: for x∈ℤ, (-1)ˣ is ±1, so 2|x| is 16 or 18, so x is ±8 and ±9. Not at all hard, though it does look it.
Unless you allow x to stop being merely an integer. Suppose x is complex, x=a+bi; 2|a+bi| + (-1)^(a+bi) = 17. Replace -1 with i² and we have 2|a+bi| + (i²)^(a+bi) = 17, so 2|a+bi| + (i)^(2a+2bi) = 17 and then we have to find things to do with i^i, which I can write as iᴵ.
Let's consider iᴵ. Any eᴵˣ = cosx + isinx so when eᴵˣ = i, sinx is one and cosx is zero, so x is π/2 + 2kπ, k∈ℤ. S So iᴵ = e^(-π/2 + 2kπ) which is entirely real. For k=0, iᴵ= e^(-π/2) which is roughly 0.208. The next value, k=1, is e^(3π/2) = 111.3 and for k=-1, e^(-5π/2)=3.9x10⁻⁴
So in 2|a+bi| + (i)^(2a+2bi) = 17 we should look at the second term, (i)^(2a+2bi) = (i²)^a + iᴵ^2b = (-1)^a + iᴵ^2b. This too is only real, which means that the only imaginary term is in the first term, so that 2b = 0. The real part, 2|a| +(i^i)^2b + (i^2)^a = 17 reduces immediately (b=0 and (anything non-zero)^0 =1) to 2|a| + 1 + (-1)^a = 17, which suggests that a = 15/2 or 17/2. Which is great until you try to evaluate that, because( -1)^a = (-1)^15/2 = i^15 = -i, which means we have an unresolved imaginary part coupled with a value of x=a=±8. I think this means we reject both complex and non-integer real values of x, leaving only the solutions we have found already.
Which is a pity, because I'd guessed there to be solutions near to 8±15i and 15±8i, since these with have length of 17. Please agree or correct me.
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Related topic, in matrices, that (3 4 )(7 2) = (37 42)
(8 7 )(4 9) (84 79). I am looking for others......
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Found quite by accident, aˣbʸ = axby, a four digit number. Set all variables non-zero, since 0⁰.0⁰ might be zero (discuss, a possibility is unity). I have just one solution though I'd love hear otherwise. Maybe there are other answers in other bases? Octal or hex, anyone? How would you do that?
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