Wikipedia:Reference desk/Archives/Mathematics/2018 April 21

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April 21

Squares with odd digits

For which bases b is it true that the odd squares less than b are the only perfect squares with only odd digits in base b? Is this true for b = 10? GeoffreyT2000 (talk) 04:37, 21 April 2018 (UTC)

It is true for b = 10. (10x + y)² = 100x² + 20xy + y²: the first term cannot affect the last two digits, and the second term conserves the parity of the tens digit. So we only need to look at the single-digit squares, which all have at least one even digit (counting a leading zero in the tens place, since that will show up for positive x). Double sharp (talk) 04:47, 21 April 2018 (UTC)

Actually, allowing for negative values of y cuts down the numbers you have to check to the odd digits up to b/2. As for other bases: this is always false for odd b, as then (b − 1)² has two odd digits, so we can focus on even bases alone. An easy way to get a counterexample is to find an odd square between b and 2b. So for 14 ≤ b ≤ 24, 5² is a counterexample; for 26 ≤ b ≤ 48, 7² is a counterexample; for 50 ≤ b ≤ 80, 9² is a counterexample; and because (1 + 2/k)² ≤ 2 for k ≥ 5 this approach yields counterexamples for all bases above 12. So we only need to try it for b = 2, 4, 6, 8, 10, 12. Of these the statement is only true for b = 2, 4, 10, and 12; counterexamples for the other two are 3² = 13₆ = 11₈. Double sharp (talk) 06:20, 21 April 2018 (UTC)

(P.S. The argument in my first post is the same as that in this old Math Forum answer to the same question for b = 10 specifically; I extended it for my second post.) Double sharp (talk) 06:41, 21 April 2018 (UTC)

Decidability of Arithmetic

Let's denote by G1 the proposition, almost identical - to the proposition G - i.e. to the arithmetical proposition undecidable in PA as discovered by Goedel, except that for every quantifier Q and for every variable x - if the term "Q(x)" appears in G - then it's replaced in G1 by the term "Q(x∈N)", N being the set of natural numbers as defined in ZF.

I guess that G1 is decidable in ZF (after having translated the symbols "+" and "⋅" into the respective analogous operations of addition and multiplication in ZF). I also guess that G1 is provable (rather than refutable) in ZF. Am I right? HOTmag (talk) 18:40, 21 April 2018 (UTC)

Yes. Put another way, $ZF\vdash [(\omega ,+,\cdot )\vDash {\text{con}}({\text{PA}})]$.--73.58.152.120 (talk) 04:29, 24 April 2018 (UTC)

Decidability of Arithmetic (a second question)

For every arithmetical proposition P, let's denote by P1 the proposition almost identical to P, except that for every quantifier Q and for every variable x - if the term "Q(x)" appears in P - then it's replaced in P1 by the term "Q(x∈N)", N being the set of natural numbers as defined in ZF.

Does every arithmetical proposition P satisfy that P1 is decidable in ZF (after having translated the symbols "+" and "⋅" into the respective analogous operations of addition and multiplication in ZF)? HOTmag (talk) 18:42, 21 April 2018 (UTC)

No. Con(ZF) is an arithmetical proposition, which Gödel tells us is undecidable in ZF.--73.58.152.120 (talk) 04:43, 24 April 2018 (UTC)

But Con(ZF) is an arithmetical proposition (as you've pointed out), while I was not looking for an arithmetical proposition P, but rather for its analogous one, P1, which is not arithmetical (because it contains also the symbol "∈", besides "+" and "⋅"). HOTmag (talk) 07:28, 24 April 2018 (UTC)

A statement about the decidability of an arithmetical proposition in ZF is actually a statement about its translate into the language of ZF, because that's the only way to formalize such a statement. So saying Con(ZF) is undecidable in ZF actually means that its translation in the manner you describe is undecidable.--2601:245:C500:754:513D:EF2C:25FD:51A (talk) 04:14, 25 April 2018 (UTC)

As an aside, I'd like to encourage saying "independent of ZF" rather than "undecidable" here. "Undecidability" is not a truth value; in context it means simply that it can neither be proved nor refuted from ZF. Also it risks confusion with the notion of an undecidable problem, which is quite different (but in a closely related enough field to be confusing). --Trovatore (talk) 04:23, 25 April 2018 (UTC)

Is there any arithmetical proposition about natural numbers that is equivalent to Continuum hypothesis?

After having translated - every variable in the original arithmetical proposition - into a variable over the set N of natural numbers in ZF, and after having translated - the symbols "+" and "⋅" in the original arithmetical proposition - into the respective analogous operations of addition and multiplication in ZF. HOTmag (talk) 18:38, 21 April 2018 (UTC)

No. Most methods of proving the undecidability of CH (e.g. Cohen forcing, Gödel's L) preserve $\omega$, and thus preserve the truth value of all arithmetical propositions while changing the truth value of CH.--73.58.152.120 (talk) 04:47, 24 April 2018 (UTC)

Well, to quibble a little, it does depend on what you mean by "equivalent". If CH is true, then CH is equivalent to 0=0 (in the sense that the proposition "CH if and only if 0=0" is true). If CH is false, then CH is equivalent (in the same sense) to 0=1. So yes, there is such an arithmetical proposition; we just don't know which one it is.

Of course that isn't what HOTmag means by "equivalent". Probably. But the quibble does illustrate why it takes a little care to make the original question precise. --Trovatore (talk) 05:13, 24 April 2018 (UTC)

Wasn't it clear, that I was looking for one single arithmetical proposition p (after it was translated as mentioned in my first post), that satisfies $ZF\vdash CH\leftrightarrow p?$ HOTmag (talk) 07:06, 24 April 2018 (UTC)

Well, that's one reasonable interpretation of what you wrote, but you didn't actually say it. You mentioned ZF a couple of times, but not in reference to the notion of equivalence you were interested in.

(As a side note, the ways that you did mention ZF don't necessarily make much sense. There is no "set of natural numbers in ZF", for example. There's a set of natural numbers, period, which ZF makes assertions about. Probably you meant something like "the set of natural numbers, as coded in the language of ZF".) --Trovatore (talk) 17:57, 24 April 2018 (UTC)

As for $ZF\vdash CH\leftrightarrow p:$ Well, in my opinion, it's more than a reasonable interpretation only (as you've put it). If it hadn't been the most likely interpretation (as it is in my opinion), you wouldn't have written "to quibble a little".

As for the "set of natural numbers in ZF": It's an abbreviation of what I have already elaborated on in my previous threads about "Decidability of Arithmetic"; See ibid. "N being the set of natural numbers as defined in ZF". In the current thread, I meant that every variable in the arithmetical proposition p satisfying $ZF\vdash CH\leftrightarrow p$, should be interpreted as running over the set of (standard) natural numbers as defined in ZF, i.e. as defined in the system ZF for which I've wanted p to satisfy $ZF\vdash CH\leftrightarrow p,$ i.e. as opposed to the system PA in whose language we formulate p - because the set of (standard) natural numbers can't be defined in PA. That's why I added "in ZF".

HOTmag (talk) 07:08, 27 April 2018 (UTC)

Flat response

By what function must i muktiply a positive half gaussian curve ro obtain a constant product upto an arbitrary value of the horizontal variable?86.8.201.80 (talk) 19:42, 21 April 2018 (UTC)

sorry that's just a stupid question. It's obviously the reciprocal of the gaussian. Op 86.8.201.80 (talk) 13:25, 22 April 2018 (UTC)

There are no stupid questions, but there are rude respondents. "Obvious" is a very vexed term, rarely so glibly applicable in mathematical discourse. -- Jack of Oz ^{[pleasantries]} 09:52, 24 April 2018 (UTC)

Why I strongly agree that rudely responding to a good-faith question is deplorable, rudely responding to yourself is surely an allowed exception. Tigraan^{Click here to contact me} 10:23, 24 April 2018 (UTC)

I didn't notice they were the same editors. Still, being rude to oneself is surely something to be discouraged. -- Jack of Oz ^{[pleasantries]} 21:53, 24 April 2018 (UTC)

I once posted something about what "some idiot called Trovatore" had done, which I thought might get a wry smile out of someone. But I've regretted it, because every now and then I happen across it, and feel my stress response flaring up before I notice it was I myself who posted it. --Trovatore (talk) 04:19, 25 April 2018 (UTC)

Just post an incisive retort to yourself. Your stress response will probably remain flat if you first notice that. Basemetal 10:43, 25 April 2018 (UTC)

"x must be positive"

In defining the sqrt function in Microsoft Excel, the rule is that x must be positive. Does positive mean greater than 0 here or does it mean greater than or equal to 0?? Georgia guy (talk) 19:58, 21 April 2018 (UTC)

What happens is you set cell A1 to 0, and then another cell to =SQRT(A1)? I don't have Excel to test this myself, but I don't see this being anything other than 0. The reason square roots are only valid for positive inputs is because the square root of a negative number is not a real number, but a complex number instead. Iffy★Chat -- 20:27, 21 April 2018 (UTC)

Negative means less than 0. Positive is intended to mean greater than 0 (not greater than or equal to 0.) But here positive is in fact being used to mean greater than or equal to 0. Can you correct me?? Georgia guy (talk) 20:29, 21 April 2018 (UTC)

Excel isn't made for people who are pedantic about whether 0 is a positive number or not. Iffy★Chat -- 21:09, 21 April 2018 (UTC)

This is ultimately just a convention, and other languages may define things differently. For example, in standard French usage zero is both positive and negative: see fr:Signe (arithmétique). Double sharp (talk) 10:36, 23 April 2018 (UTC)

Thank you for that link. I'd no idea that the French think so differently. They do point out that the "anglo-saxons" regard zero as neither positive nor negative. Are there any other countries that think like the French? Dbfirs 11:33, 23 April 2018 (UTC)

Well, from the interwiki links that I am able to understand (and assuming the articles are correct!): Italian, Spanish, German and Swedish all follow the English convention (i.e. 0 is not positive). I do not know why French differs in that regard, but from my experience we often use positif ou nul ("positive or zero", i.e. nonnegative) and strictement positif ("strictly positive", i.e. positive but not zero) rather than simply positif - there is still a little bit of ambiguity, even if the official convention is that positif = positif ou nul. Tigraan^{Click here to contact me} 19:39, 23 April 2018 (UTC)

Consider also that the two square roots of x may be expressed as +√x and −√x, where √x (which is always non-negative) is properly called the "principal square root". It is natural enough to think of +√x as the "positive square root" even though it is only guaranteed to be non-negative, not positive. And that's without considering how numbers are represented on computers, where in some cases a "negative zero" representation is possible. --69.159.62.113 (talk) 20:15, 23 April 2018 (UTC)

For more detail on the French idiosyncrasy, see Positive number on the French Wiktionary. The Académie française seems to take the Anglo-Saxon view. Do we need to put a caveat in fr:Signe (arithmétique) that the view is that of one French author? Dbfirs 20:21, 23 April 2018 (UTC)

For what I know, positive is the opposite of negative. I think that positive means "more than zero," but I am not really sure. UnbeatableFlame154 (talk) 01:23, 27 April 2018 (UTC)