[QUOTE=BudgieJane;376470]Dunno about the porcupines, but there is this porpuquine
[url]http://mathcs.wilkes.edu/~rpryor/mth231/RecurrencesandRecursion.pdf[/url][/QUOTE]I can define "mutual recursion", but only in terms of "recursion, mutual". Is that OK?

"preceded by its negation yields FALSE" preceded by its negation yields FALSE. Just quining a phrase there

The above paragraph is based on something I found in "Copper, Silver, Gold: An Indestructible Metallic Alloy" in Egbert B Gebstadter or, possibly, in a work referenced in its index.

Hey, knock it off. We're not here to have fun, we're here to do mathematics stuff and make sure that each and every word is on-topic, true and accurate without deception, mistake or falsiness. :Gordon:

[QUOTE=retina;376474]Hey, knock it off. We're not here to have fun, we're here to do mathematics stuff and make sure that each and every word is on-topic, true and accurate without deception, mistake or falsiness. :Gordon:[/QUOTE]

But, ...
[url]http://3.bp.blogspot.com/-qJ_vaX1ZejU/UJzk8VgxnjI/AAAAAAAAFc0/eDqkbIFIXPA/s1600/math+is+fun.jpg[/url]

[QUOTE=retina;376474]Hey, knock it off. We're not here to have fun, we're here to do mathematics stuff and make sure that each and every word is on-topic, true and accurate without deception, mistake or falsiness. :Gordon:[/QUOTE]

At least in the DATA forum, want some [URL="http://www.mersenneforum.org/forumdisplay.php?f=7"]fun and frolics?[/URL]

I am still interested in why this

[TEX]n=\prod_{p^e||n}p^e[/TEX]

fails to meet the requirements of this definiton

[QUOTE=R.D. Silverman;375973]Start by defining "completely factored".

My definition would be:

A number is completely factored when it is represented as the product of primes.[/QUOTE]

One of the issues is that "completely factored" is not a property of the number, it's a property of our knowledge. The unmathematical word "when" is in the definition because our knowledge can can change over time.

I think the expression does not fail the "prime" or "product" components, so it must fail the "represented" part. We need "represented" to mean something like "represented in decimal notation" or some equivalent wording that precludes a representation whose evaluation requires "first factor the number." Is there a standard way or saying that?

[QUOTE=wblipp;376498]I think the expression does not fail the "prime" or "product" components, so it must fail the "represented" part. We need "represented" to mean something like "represented in decimal notation" or some equivalent wording that precludes a representation whose evaluation requires "first factor the number." Is there a standard way or saying that?[/QUOTE]

"Explicitly represented", in the sense of the fundamental theorem of arithmetic.

[QUOTE=R.D. Silverman;375973]You are bandying words and undefined terminology.

Start by defining "completely factored".

My definition would be:

A number is completely factored when it is represented as the product
of primes.

Since the number in question has not been represented as the product of
primes, then it most definitely has NOT been completely factored.[/QUOTE]

What's wrong with this?

C = A and B where A, B, and C are the propositions

A: N is represented as a product
B: Each multiplicand in the product is a prime number
C: N is completely factored

(not C) = (not A) or (not B)

Bob has asserted (not C) but cannot prove (not A) nor (not B)

This is a good framework for understanding your difference of opinions.
Bob says: "There is no proof of (B), therefore (not B)"
William says: "There is no proof of (not B), therefore (B)"

The second argument seems far less convenient. By this logic, any sufficiently large number (for example 2[SUP]2[SUP]195[/SUP][/SUP]+1) has no proof of (not B) and is therefore completely factored. If you want to disprove that F195 is not completely factored, you have to prove that the cofactor is composite, which you can't.

[QUOTE=Batalov;377899]William says: "There is no proof of (not B), therefore (B).[/QUOTE]

I have never claimed B. I have consistently claimed that both "B" and "not B" are unproven. My only complaint is that Bob has failed to follow implications of his own definition.

The original poster briefly claimed "B." I don't think anybody has claimed "B" since early in the thread.

[QUOTE=Batalov;377899]The second argument seems far less convenient. By this logic, any sufficiently large number (for example 2[SUP]2[SUP]195[/SUP][/SUP]+1) has no proof of (not B) and is therefore completely factored. If you want to disprove that F195 is not completely factored, you have to prove that the cofactor is composite, which you can't.[/QUOTE]

This is one of the reasons I don't like Bob's definition. If the definitions don't give you the results you want, then you need to change the definitions. You cannot keep the definition and ignore the mathematical consequences of the definition.

In [URL="http://en.wikipedia.org/wiki/Three-valued_logic#Logics"]three-valued logic[/URL], most of these (A, B, C) statements will fall into Unknown value pretty fast.

What Bob is saying, I think, is (C) [TEX]\ne[/TEX] True. What we all seem to agree with, is that presently (C) evaluates as Unknown. As well as, obviously, (not C) evaluates as Unknown, too.