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October 4

Are there only two types of Truth in Mathematics?

Are there only two types of Truth in Mathematics? The only two type I can think of are

Type 1: Truth in boolean algebra. Truth is denote by 1 or T and it is arbitrary constant (ie by definition). Conversely False is also an arbitrary constant.

Type 2: Truth in Proof Theory. A Mathematical statement is true if the statement that are provable in a formal axiomatic system. A statement is false if the proof leads to a contradiction. A statement is unproveable if it cannot be proven true or false.

It seems to me that the type 1 is pretty arbitrary while type 2 isn't about truth at all but merely about proveability of statement(s). Can I safely say that there are no "truths" in mathematics, where the word "truths" refers to real world certainties like "It is TRUE that the IRS expects me to pay my taxes."

110.22.20.252 (talk) 01:48, 4 October 2017 (UTC)[reply]

Well, you've completely left out truth as understood by mathematical realists (aka Platonists). For realists, mathematical truth is about the behavior of real objects, and is quite distinct from provability. --Trovatore (talk) 02:18, 4 October 2017 (UTC)[reply]

I think 2 is usually called provable rather than true, but there are schools of thought, e.g. intuitionism, which takes 2 to mean true. So statements can be true but unprovable for most people, but some would say these are statements which are not false, which is not the same as being true. Falsity has shades of meaning as well; while most say it's the other Boolean value, the intuitionists would say that a statement is only false if by assuming it you can derive a contradiction. There many flavors of non-standard logic and their study is a specialty in itself; any of them could be said to have its own interpretation of truth and falsity. Also, my understanding is that a mathematical statement must be about mathematical objects. Since the IRS is not a mathematical object, statements about it aren't mathematical statements. There are types of truth outside mathematics, e.g. scientific truth, and mathematics doesn't claim to cover everything in the universe. --RDBury (talk) 03:45, 4 October 2017 (UTC)[reply]

It's true that intuitionists take truth to equal provability, but it's important to keep in mind that they disidentify provability from provability in any fixed axiomatic system. That's the "intuition" part — the human mathematician is supposed to be able to tell what is a proof and what's not, and doesn't delegate that role to any stinking formal deductive system. --Trovatore (talk) 04:11, 4 October 2017 (UTC)[reply]

I should probably say a little more about that. It's not as arbitrary as it sounds. Here's the idea:

A proof of A∧B is a proof of A together with a proof of B.
A proof of A∨B is either a proof of A or a proof of B.
A proof of ∃xP(x) is a value n together with a proof of P(n).
A proof of ¬A is a proof of A→false.

So far this is completely judgment free. But we haven't said what to do about implication (note that A→B is not the same thing as ¬A∨B for intuitionists; if it were, the clause defining a proof of ¬A would be circular, since presumably you can't prove false). So here are the final clauses:

A proof of A→B is a method for turning any proof of A into a proof of B.
A proof of ∀xP(x) is a method for obtaining, for any value n, a proof of P(n).

This is where it ceases to be clearly well-defined. Generally, intuitionists do not nail themselves down as to what is meant by a "method". As I understand it, the human mathematician is supposed to understand the argument underlying the method, and agree that it does indeed turn any proof of A into a proof of B.

I was put off by that when I first encountered it, but in the end had to agree that it is not that different from what mathematicians who claim to be using the axiomatic method do in practice. They almost never actually prove things directly from the axioms. Rather, they give an argument in prose which, in principle, is supposed to be translatable into an argument from the axioms, using routine techniques, given a competent mathematician with unlimited time and patience. --Trovatore (talk) 04:38, 4 October 2017 (UTC)[reply]

Thanks for the clarification. I'm not with you on never being able to prove false though since that's basically how proof by contraction works, but hopefully you can't prove false without adding additional assumptions to the standard axioms. What you're saying about the usual practice for proof is true, but it seems to me that that will change in the next few decades as automated proof checkers gain acceptance. Btw, [1] is a rather entertaining (for an hour long lecture on mathematical logic) video covering some of these ideas. --RDBury (talk) 14:42, 4 October 2017 (UTC)[reply]

So for purposes of this discussion I'm using words like "prove" in their intuitionistic acceptions (wikt:acception; a useful word maybe not that common in English). Axioms have not been mentioned.

For intuitionists, ¬A is shorthand for A→false. A proof of ¬A is a method that, given any proof of A, would allow you to transform it into a proof of false. But presumably no proof of false exists, because then false would be true. This is the intuitionistic version of proof by contradiction.

(Occasionally people make the mistake of thinking that intuitionists don't accept proof by contradiction. That's not true — actually it's the only way you can intuitionistically prove a negative statement. It's true that they don't accept it for proving positive statements.) --Trovatore (talk) 17:39, 4 October 2017 (UTC)[reply]

Well that is rather an open question. Maybe it depends on one's point of view, and maybe it doesn't. The article Philosophy of mathematics is a start. Perhaps I could get invites on a dinner circuit by adopting the Mathematical universe hypothesis and saying we're in a superposition of worlds where undecidable results are true or false :) By the way Truth has a small section about mathematics which is like what you said, but as always philosophical questions become murkier the more you look at them. 12:41, 4 October 2017 (UTC)

MU (Markup?) key on a calculator

I recently got a "Comix CS-3136" calculator for $11.99. [2]

Mostly I wanted something with big keys, but the 16 digit display might be useful for calculating the exact size of very large sparse arrays. A 14 digit calculator can handle up to 2^46 (70,368,744,177,664 Bytes, 70TB). A 16 digit calculator can handle up to 2^53 (9,007,199,254,740,992 Bytes, 9007TB).

Mostly it acts like a normal calculator, or perhaps a 10-key adding machine, but it has a "MU" key that is poorly documented. Being curious, I tried to figure out what this key does.

It appears to be related to percent, so I started by doing an experiment with the % key

Example #1: 200 + 20% = ????

Enter: 200 Display: 200
Enter + Display: 200+
Enter 20 Display: 20+
Enter % Display: GRAND TOTAL 240=
Enter = Display: GRAND TOTAL 440=

Yes, there is a small + or = to the right of some results, and sometimes it says GRAND TOTAL above the number.

OK, fairly normal, other than the + key working 10-key style instead of calculator style.

So I tried the same thing with MU:

Example #2: 200 + 20MU = ????

Enter: 200 Display: 200
Enter + Display: 200+
Enter 20 Display: 20+
Enter MU Display: 1100=
Enter = Display: GRAND TOTAL 1100=

So what did the MU key just do?

If anyone has any other test they would like me to run, just let me know. --Guy Macon (talk) 17:40, 4 October 2017 (UTC)[reply]

Yes, that is hard to understand. One thought is that it determined that a markup of 20 (dollars, cents, Euros, or whatever) on a 200 unit item (same units, of course), is 10% markup. It then applied that markup to a base of 1000 units to get a price of 1100, at that markup. Not sure where you define the base, but it apparently defaults to 1000 units. StuRat (talk) 18:57, 4 October 2017 (UTC)[reply]

Just Google calculator mu. The first result is https://www.quora.com/What-does-MU-on-a-calculator-mean. PrimeHunter (talk) 20:37, 4 October 2017 (UTC)[reply]

That page say "Suppose a shopkeeper wants to sell the product at 100 after 20% discount. Then he can use mark up button. He will input 100 MU 20 % and the answer he gets is the price he will have to write on the product", So I tried it:

Example #3:

Enter: 100 Display: 100
Enter: MU Display: (no change)
Enter: 20 Display: 10010
Enter: % Display: (no change)
Enter: = Display: GRAND TOTAL 10010=

I also am having trouble making that concept work with example 2. Mark 1100 on the item so that it sells for 200 after a 20% discount?

Hmm. Maybe I am getting the final price and the discount percentage backwards. Let's try that:

Example #4:

Enter: 20 Display: 20
Enter + Display: 20+
Enter 200 Display: 200+
Enter MU Display: 110=
Enter = Display: GRAND TOTAL 110=

Is the above saying that if I buy an item for 200 and sell it for 220 I have a 110% markup? Would that be a useful calculation for a merchant? --Guy Macon (talk) 08:56, 5 October 2017 (UTC)[reply]

Possibly this manual http://files.sharpusa.com/Downloads/ForHome/HomeOffice/Calculators/Manuals/cal_man_CS1194H.pdf will help. It's about a calculator model(s) different from yours, but I suppose the MU key is the same "Multiple Use" function. --CiaPan (talk) 09:56, 5 October 2017 (UTC)[reply]
Forgot to ping: @Guy Macon: --CiaPan (talk) 09:59, 5 October 2017 (UTC)[reply]

Maximizing Inner Product

I have an M x 3 matrix A and a row vector <v|, I want to find |u> so that <v|A|u> is maximal. The specifics of <v| vary around, but it can be assumed that every element is positive. The elements of A and |u> are all in the interval [0, 1]. If it matters, M will never be huge, but in the vicinity of 20 to 30. While I would prefer to be able to find the true maximal |u>, and in a computationally quick way, an approximation would work, as long as it is relatively close to the actual value. I have a large data set of A's for which I will need to find |u>, for fixed <v|, for all points (numbering in the millions), so any method that takes a while (like discretely stepping through all values of |u> with elements of the form an integer / 1000, or some such, are not going to be viable due to time constraints (in this case on the order of 1000000 * 1000 * 1000 * 1000, or a quadrillion operations for the entire data set - but, if I could narrow the elements of |u> down to a much smaller interval...)). Thank you for any, and all, help:-)Phoenixia1177 (talk) 19:59, 4 October 2017 (UTC)[reply]

Since all elements of

<v|

and

A

are non-negative then all elements of raw vector

<v|A

are also non-negative. Since all three elements of the column vector

|u>

are required to be non-negative and less or equal to the unity then the column vector that maximizes the product

<v|A|u>

is

|u>=(1,1,1)^{T}

. Ruslik_Zero 20:44, 4 October 2017 (UTC)[reply]

Oh my! I forgot a constraint, sorry about that - there is <L| = (a, b, c) so a + b + c = 1 and a, b, c are all in the unit interval. Let the transpose of the first row in A be |d>, then I need the maximal |u> from the set {|u> : <L|u> = <L|d>}. I've been mulling this over in my head for a while and this is the first time I've put it on paper, I still can't believe I left that out. Since this is an approximation problem to start with; I'd be just as happy with help on a variation (if the above is a pain), instead of this added constraint, assume that there is fixed x in (0, 1) so that given a transpose of a row in A, call it |d>, <L|d> = x, then the |u> can come from {|u> : <L|u> = x}. Either constraint does the trick, the latter is probably a better solution, but I haven't been able to test either, so I'm not sure about that. Sorry again for the confusion. (Or, for another variation,you can assume that the |u> are all a fixed distance from the transpose of A's first row, that that first row is in the interior of the unit cube, and that the fixed distance is enough that the ball around the row is within the cube as well.)Phoenixia1177 (talk) 20:57, 4 October 2017 (UTC)[reply]