[unreality]

We all know the absolute value function of x: abs(x) or more commonly |x|

|25| = 25

|-25| = 25

It takes the positive part of the number, or more formally, the distance from zero (thus right-trangle trig is used to find the absolute value of complex numbers).

But how can you do it, using only the basic functions? Addition, subtraction, multiplication, division, exponentation/rooting, logarithms and the two extraction functions:

real(a + bi) = a

imag(a + bi) = b [not bi, just b]

Note that the plus/minus value of an even square root is unknowable! (ie, √25 could be 5 or -5) We'll say that in the beginning, a random decision is made of whether all square roots are going to be positive or negative, and they all will hold to this rule, but you have NO way of knowing which it is

So, can it be done?

If so, how?

If not, PROVE that it is impossible!

Good luck

[HoustonHokie]

If the basic functions include exponents and roots, then we can get part of the way there fairly easily - just follow the general rule for calculating the absolute value of a complex number. So, |a+bi|=√(a² + b²), which holds for real numbers because b=0. The real question you're asking is how to force √n to be |n| and not -|n|. If we can figure that out, the rest is easy.

The best way I can figure to do that is to take the cube root and then square it, so |a+bi|=[³√(a² + b²)]², which is always a positive number.

Is that what you were going for?

Edited November 16, 2008 by HoustonHokie

[unreality]

well here you've hit a roadblock

the 2/3 power of 'x' will be even (as 2/3 is cube root [same sign] squared [even]). However to turn that into x, you'd need to raise it to the 3/2 power, which requires an even root to work

disclaimer: your input can be assumed to be a real number

There is a solution My solution only works when real numbers (positive & negative, no 'i' part) are inputted, but that's what I was assuming in the beginning, so it fits I'm pretty sure that a solution for imaginary numbers too is impossible, not sure though

Edited November 16, 2008 by unreality

[HoustonHokie]

yeah, I see my bust now

I should have known that I needed to take the fourth root, not the third root, to make that work - and the fourth root could be positive or negative, so that doesn't help. Back to the drawing (chalk)board and reading hints...

[unreality]

Yep

Say you cube something... that should technically be the same thing as raising it to the power of 6/2, right? Which would be the square root of it being raised to the sixth power. Yet the square root could be ± while cubing is always the same.... confusing

[Guest]

abs(x) = (abs(√x))² = (real(√x))² + (imag(√x))²

This involves square root operations but since you've stated that these would involve a random decision positive or negative, that works for our purposes, since the squaring operations that follow would give the same positive answer regardless. That would also work regardless of whether x was real or complex.

[unreality]

Nice! I'm pretty sure that works. Good job

This is my solution:

|x| = (real(√x) + imag(√x)*i) * (real(√x) - imag(√x)*i)

Which is like multiplying the square root against its conjugate

|25| = (±5 + 0i)*(±5 - 0i) = 25

|-25| = (0 ± 5i)*(0 -+ 5i) = -25*i² = 25

The first term of the |-25| solution is plus-minus while the second is minus-plus, as it will be the opposite sign of the other one (because the square roots are unknown but constant), and you add the imag(√x)*i in the first term but subtract it in the second. Thus this works for all real numbers

edit: it might actually work with imaginary numbers too, one sec

edit2: yep, I just checked, it works with 25i

Edited November 17, 2008 by unreality

[Guest]

Nice! I'm pretty sure that works. Good job

This is my solution:

|x| = (real(√x) + imag(√x)*i) * (real(√x) - imag(√x)*i)

Which is like multiplying the square root against its conjugate

|25| = (±5 + 0i)*(±5 - 0i) = 25

|-25| = (0 ± 5i)*(0 -+ 5i) = -25*i² = 25

The first term of the |-25| solution is plus-minus while the second is minus-plus, as it will be the opposite sign of the other one (because the square roots are unknown but constant), and you add the imag(√x)*i in the first term but subtract it in the second. Thus this works for all real numbers

edit: it might actually work with imaginary numbers too, one sec

edit2: yep, I just checked, it works with 25i

...that was the same answer?

(real(√x) + imag(√x)*i) * (real(√x) - imag(√x)*i) =

(real(√x))² - (imag(√x))²*i² =

(real(√x))² + (imag(√x))²

ta-da!

[unreality]

oh yeah... haha I forgot that conjugate rule:

(a+bi)*(a-bi) = a² + b²

hehe Good one

thanks for spotting that I did have a knack that they were probably the same after some manipulation, but not that little ;D hehe

Edited November 17, 2008 by unreality