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December 14

Validity of (ab)^m = (a^m)(b^m)

(ab)^m = a^mb^m
I have questions regarding the validity of the above law of indices depending on what a,b,m are.

I'm trying to come up with the valid values that a,b,m can take so that the above law is always true for a given set of restrictions.
I'm trying to be as inclusive as possible, meaning that if something isn't wrong or results in something being wrong, then I won't exclude it. For example, if a complex number (not just a real number) also works in some situations of the above law, then I'll also try to include and describe those situations.

Note: For all cases, any situation that results in division by 0 (or 0 to a negative real number) and 0⁰ are exceptions that have been noted. I'm not writing expressions that show I'm disallowing them so that I can reduce clutter in describing the different cases below.

Case 1:
If m ∈ {ℝ \ ℤ}, then a,b ∈ ℝ⁺
This is so we avoid the wrong results of 1 = -1 and i = -i (and other similar wrong results).

Case 2:
If m ∈ ℤ, then a,b ∈ ℂ
Apart from that note I made above, I don't see any problems with this case.

Case 3:
If m ∈ j⁄2k+1 where j,k ∈ ℤ, then a,b ∈ ℂ
The power is rational, and its denominator cannot be even because taking even roots of negative real numbers results in the same problems mentioned in Case 1.

So, my questions are:
1. I paraphrased Case 1 and 2 from this wikipedia article (http://wiki.riteme.site/wiki/Exponentiation#Failure_of_power_and_logarithm_identities). Is it correct for me to paraphrase it this way?
2. I'm not sure if Case 3 is correct though, and that is where I require help in checking its correctness. The reason I wanted another case (other than the 2 mentioned in that wikipedia article) is because (just from working a few examples out) there are some complex numbered bases that can be raised to a rational power (with an odd denominator) that satisfy this particular law of indices, so it feels too restrictive/un-inclusive if we don't cater to these cases.
Thanks. 175.156.52.140 (talk) 19:25, 14 December 2014 (UTC)[reply]

Cases 1 and 2 should be regarded as distinct functions. You cannot combine them sensibly so that the identity always holds, even though they will agree on every point for which they are both defined. If you regard the functions as distinct for case 1 (a^m ≝ exp(m⋅log a), where log is the real logarithm) and case 2 (a^m ≝ ∏_m a, extended to all integer m through division where possible), these cases work perfectly. I would phrase case 1 without the exclusion of integers, i.e.
- Case 1: $m \in ℂ$ and $a, b \in ℝ +$ is the domain of the first exponentiation function, defined in terms of $log$ .
- Case 2: $m \in ℤ$ and $a, b \in ℂ$ is the domain of the second exponentiation function, defined in terms of repeated multiplication, with $a \neq 0$ and $b \neq 0$ for $m < 0$ ; interestingly we can define $00$ for this function without problems if we wish to do so.
Case 3 does not work. There are problems like with even denominators whenever $m \in ℚ ∖ ℤ$ for most complex bases. —Quondum 21:26, 14 December 2014 (UTC)[reply]

I've extended the case 1 function domain, since you're trying to to be maximally inclusive. The $ℂ$ in the domain can be replaced with far more general objects in both cases (e.g. matrices, quaternions, etc.) if you wish to be maximally inclusive. —Quondum 01:49, 15 December 2014 (UTC)[reply]

zero knowledge comparison of two positive integers

Hi,

Can A and B run a Zero Knowledge Proof algorithm to learn whether A's or B's positive integer is larger (or if they are the same) without either of them learning (or giving away) any further information beyond this fact? They should not have to trust a third party, this is part of the field of mathematics called zero knowledge proofs. Thank you. 212.96.61.236 (talk) 20:28, 14 December 2014 (UTC)[reply]

Yes, see Secure multi-party computation for the general, highly inefficient methods, and Yao's Millionaires' Problem for a (presumably more efficient) method for this particular application. It is assumed that there is some known bound on the lengths of the numbers (which could be very loose), perhaps it can be generalized for a case that such a bound is not known a priori. -- Meni Rosenfeld (talk) 22:58, 14 December 2014 (UTC)[reply]

Meni or another: could you summarize the principle Yao's solution works on? (what is the general principle.) The math in that article is a bit hard for me to follow. 212.96.61.236 (talk) 23:14, 14 December 2014 (UTC)[reply]

As with your previous question, if A is allowed as many queries as they want, they can brute force B's integer. ("is it greater than 1?" yes. "is it greater than 2?" yes. "is it greater than 3?" no. "it's 3!"). If the range of the integers is limited, this can be done relatively quickly (see binary search) MChesterMC (talk) 09:34, 17 December 2014 (UTC)[reply]

Presumably, they will only run the process once, and they are both genuinely interested in the outcome, and thus will behave honestly. -- Meni Rosenfeld (talk) 22:17, 17 December 2014 (UTC)[reply]