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April 25

Non-integer "modular arithmetic"

What's the most common equivalent term to "modular arithmetic" when the modulus is a non-integer, such as with angles expressed in radians (which are equivalent "modulo" 2π)? Neon Merlin 01:35, 25 April 2014 (UTC)[reply]

Good question! A point of time modulo a period may be called a phase. Bo Jacoby (talk) 09:14, 25 April 2014 (UTC).[reply]

The general term (maybe too general for your question) is that you are working with cosets in the factor group

\mathbb {R} /2\pi \mathbb {Z}

. —Kusma (t·c) 17:51, 26 April 2014 (UTC)[reply]

I guess that you could refer to it as periodic equivalence class, which I think is a less wordy way of saying what Kusma has said. Note that it does not share all the properties of modular arithmetic; in particular, while you can add or subtract two equivalence classes, you cannot multiply or divide them as you can in modular arithmetic. —Quondum 18:14, 26 April 2014 (UTC)[reply]

Basic Proof About Supremum

I'm trying to understand the following basic proof:

Given two non-empty sets A,B in R, with A ⊆ B and B bounded above. Show sup(A) ≤ sup(B).

The proof establishes that sup(B) exists and is an upper bound for both A and B, which I understand. Then it says:

Imagine sup(B) < sup(A). Then sup(B) is not an upper bound for A (by definition of sup(A)), which is a contradiction.

I don't understand the contradiction: if sup(B) is an upper bound for A, then all a ∈ A are ≤ sup(B). But sup(A) is not necessarily in A, so sup(B) < sup(A) doesn't mean that there exists an element of A > sup(B). What am I missing? OldTimeNESter (talk) 14:18, 25 April 2014 (UTC)[reply]

The contrapositive makes that proof unnecessarily complicated. A better proof goes: sup B is an upper bound of B, and so also an upper bound of A, because A is a subset of B. Therefore sup A, being the least upper bound of A cannot exceed sup B. Sławomir Biały (talk) 14:34, 25 April 2014 (UTC)[reply]

I agree with Sławomir Biały that the proof can be reformulated in a neater fashion. However, the proof as written is still valid. The essential part is the line "(by definition of sup(A))". As the least upper bound, there are no upper bounds less than it. So if sup(B) < sup(A), we conclude that sup(B) isn't an upper bound. 129.234.186.11 (talk) 11:38, 26 April 2014 (UTC)[reply]

Thanks, I understand it now.OldTimeNESter (talk) 19:43, 28 April 2014 (UTC)[reply]