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May 9

Can all integers >1 coprime to 6 be written as sum of a 3-smooth number and a prime?

Prove or disprove: All integers >1 coprime to 6 can be written as sum of a 3-smooth number and a prime. 218.187.64.147 (talk) 13:38, 9 May 2023 (UTC)

Try getting some estimate for the number of 3-smooth numbers less than N and the number of primes less than N.The estimates don't have to be good. The product will give the number of sums of one and the other. Compare that to the number of numbers less than N which are coprime to 6 NadVolum (talk) 16:53, 9 May 2023 (UTC)

Does this argument take into account that a given number may be expressed as such a sum in multiple ways? For example, 115885 = 2 + 115883 = 6 + 115879 = 8 + 115877 = 12 + 115873 = ... = 104976 + 10909, altogether 50 different ways.  --Lambiam 20:27, 9 May 2023 (UTC)

No sorry it doesn't - it just gives an upper bound. But that's quite good enough for this. It's fine and easier to use a generous overestimate for the number of 3-smooth numbers below a limit too. NadVolum (talk) 21:55, 9 May 2023 (UTC)

Can every number greater than 1 be written as the sum of a square and a prime? There are 5 squares less than 35. There are 11 primes less than 35. 5 × 11 = 55, which is more than 35. Yet the number 34 cannot be written as the sum of a square and a prime.  --Lambiam 05:37, 10 May 2023 (UTC)

Sorry. Error on my part. I was just wrong when I calculated it. I thought the chances of a number being able to be expressed that way went down for sufficiently large numbers but they don't, they go up. NadVolum (talk) 09:34, 10 May 2023 (UTC)

Just did a Google search on power of two plus a prime as the same argument about the chances going up would apply there, got an interesting article

Cook, John (4 May 2020). "Prime plus power of 2 | Generalized Polignac conjecture". www.johndcook.com.

NadVolum (talk) 10:34, 10 May 2023 (UTC)

So I looked around online for some resource that might have an answer to this, and as it turns out someone asked a similar Math StackExchange question; in particular, they postulate that all primes greater than 2 (a subset of integers > 1 coprime to 6, ignoring 3) can be written as the sum of a 3-smooth number and a prime. No luck in that post either. GalacticShoe (talk) 23:40, 15 May 2023 (UTC)

May 12

Number sets and pros and cons

Do successively larger number sets have pros and cons that are improved or worsened by the next number set?

For example, the natural numbers, integers, rational numbers, real numbers, complex numbers, quaternions, octonions, and sedenions have pros and cons as follows:

Pros and cons of various number sets

Number set	Pros	Cons
Natural numbers ($\mathbb {N}$)	Natural numbers can always be added or multiplied. Mathematical induction can easily be done with natural numbers.	Natural numbers cannot always be subtracted. Differences m - n of natural numbers m and n with m ≥ n cannot be expressed as sums of natural numbers.
Integers ($\mathbb {Z}$)	Integers can always be subtracted. Differences of integers can always be expressed as sums.	Integers cannot always be divided. Quotients m / n of integers m and n with n | m cannot be expressed as products of integers. Induction with integers would require a reversed arrow of time, by not only going forward to the future (from P(n) to P(n + 1)), but also backward to the past (from P(n) to P(n - 1)).
Rational numbers ($\mathbb {Q}$)	Rational numbers can always be divided (unless one is dividing by zero, of course). Quotients of rational numbers can always be expressed as products.	The rational numbers are not complete as either a poset (Dedekind completeness) or a metric space (metric completeness). Nonnegative rational numbers do not always have rational square roots. The rational numbers do not include important constants like π and e.
Real numbers ($\mathbb {R}$)	The real numbers are complete both as a poset (Dedekind completeness) and as a metric space (metric completeness). Nonnegative real numbers always have real square roots. The real numbers include important constants like π and e. The exponential function on the real numbers is injective.	The real numbers are not an algebraically closed field (in particular, square roots of negative numbers do not exist in the real numbers).
Complex numbers ($\mathbb {C}$)	Multiplication of complex numbers is still commutative. The complex numbers are an algebraically closed field. Every complex number has exactly two complex square roots.	The complex numbers cannot be turned into an ordered field. The exponential function on the complex numbers is not injective.
Quaternions ($\mathbb {H}$)	Multiplication of quaternions is still associative.	Multiplication of quaternions is no longer commutative. Negative real numbers have infinitely many square roots in the quaternions.
Octonions ($\mathbb {O}$)	Multiplication of octonions is still alternative. The octonions still do not contain any zero divisors.	Multiplication of octonions is no longer associative.
Sedenions ($\mathbb {S}$)	Multiplication of sedenions is still power-associative.	Multiplication of sedenions is no longer alternative. The sedenions contain zero divisors.

GeoffreyT2000 (talk) 06:03, 12 May 2023 (UTC)

• To me it's a weird question. Pros and cons??? These things are what they are. They are necessarily what they are. They could not, in any possible world, be different (though first-order logic is too weak to capture this).
  So what sense does it make to speak of pros and cons? Relative to what? --Trovatore (talk) 06:12, 12 May 2023 (UTC)

  Just as a point of order, the notion that the "could not...be different", isn't quite right, especially for the more esoteric number systems. For example, is there any world where quaternions don't exist? Sure; quaternions are useful, but they are a construct that solves a problem, not an inevitable conclusion one arrives at by following a line of thinking that is inevitable from first principles. It's been said that the only parts of math that exist outside of human invention are counting numbers, and that everything else is an invention for convenience, or as Leopold Kronecker put it, "God made the integers, all else is the work of man". Every other number system was invented to make mathematics more efficient, or to solve a problem; back to the quaternion example, complex numbers allow for vastly simplified two-dimensional mechanics. Working with the square root of negative 1, and Euler's formula takes a very messy trigonometry problem and makes it much friendlier to work with; strictly speaking neither cartesian coordinates nor complex numbers are necessary, but they sure are freaking useful. However, it only works in two dimensions. Hamilton saw this problem, and worked out a solution for three dimensions, and that's the quaternions, but that's also not strictly necessary either. It's all "made up", none of it is "inevitable" or "necessary". It's just "what is useful". There certainly exists universes which are perfectly consistent in which no one invented quaternions. Similarly, there are any arbitrary number of potential number systems which haven't yet been invented, but could. --Jayron32 14:49, 12 May 2023 (UTC)

  No, there is no world where quaternions don't exist. Hope this helps.

  That's a bit glib, of course. I'm a Platonist. For what it's worth, I didn't start out that way. It took a long time into my Ph.D. studies, being dragged along little by little, seeing how empty the alternatives are and how things fit together in ways that are not easily explained by the "invention" narrative.

  But I can't expect you to have followed the same path, and we aren't going to settle it here. Interested readers can start with our philosophy of mathematics article. --Trovatore (talk) 15:54, 12 May 2023 (UTC)

• One could say that several extensions were "invented" to overcome a limitation (listed as a "con" above): the integers allow solving $x+7=3,$ the rationals allow solving $7x=3,$ the reals allow solving $x^{7}=3,$ the complex numbers allow solving $x^{2}+7=3.$ It is a bit strange (IMO) to label these limitations as "cons". Is it a con of screwdrivers that they cannot saw wood? Or of saws that they don't turn screws? I think of these sets and their attendant theories as tools in the mathematician's toolbox; each tool has its uses and some tools are better suited for certain jobs than others. Extending a system making it suitable for one type of job may entail losing some capability: the extension to the uncountable real numbers means use of mathematical induction becomes impossible. (Also, in general, the theory of a more powerful system is more complicated, so being more powerful is at the same time a pro and a con.) There are many other extensions of the naive number concept: p-adic numbers, transfinite numbers, ordinal numbers, cardinal numbers, infinitesimals, hyperreal numbers, surreal numbers. These cannot be ordered in a chain in which each is viewed as an extension of the previous one. Each number system has its uses and each has its limitations.  --Lambiam 06:55, 12 May 2023 (UTC)  I forgot to mention the Gaussian integers. Denoting these as $\mathbb {Z[i]} ,$ we have the canonical injections $\mathbb {Z} \hookrightarrow \mathbb {Z[i]} \hookrightarrow \mathbb {C} .$ Yet they are incomparable to $\mathbb {Q}$ and $\mathbb {R} ,$ which are also situated between $\mathbb {Z}$ and $\mathbb {C} .$  --Lambiam 13:36, 12 May 2023 (UTC)
• The pros and cons thing is nonsense; each number set was invented (either explicitly by a specific mathematician at a specific time, or implicitly as humanity sort of settled on it as a "thing") to serve a specific purpose, which is to say there was some deficiency in mathematics (some operation the old set couldn't do, or something like that) that the prior sets didn't handle, so the concept of what counted as a number was expanded to include more things. It represents an overall growth in the understanding of how to think about what a number is, but there's no "pros and cons". --Jayron32 11:48, 12 May 2023 (UTC)

"Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk." -- Leopold Kronecker. Does that count as a Pro? --RDBury (talk) 17:12, 12 May 2023 (UTC)

It counts as a piece of nonsense. --Trovatore (talk) 17:39, 12 May 2023 (UTC)

I thought Giuseppe Peano created the natural numbers. Perhaps the SF created the negative integers.  --Lambiam 18:35, 12 May 2023 (UTC)

Sum of the reciprocals of numbers in a set

For given positive integer n, let the sequence A_n be the sequence n*(generalized pentagonal numbers)+1, i.e. n+1, 2*n+1, 5*n+1, 7*n+1, 12*n+1, 15*n+1, 22*n+1, 26*n+1, ...

For all positive integer n, find the sum of the reciprocals of all numbers in the sequence A_n 2001:B042:4005:4133:18C0:328F:4CBA:E181 (talk) 20:05, 12 May 2023 (UTC)

Umm ... why? --Trovatore (talk) 00:00, 13 May 2023 (UTC)

If you just want a formula, it's $-{\frac {8\pi \sin({\frac {\pi }{3}}{\sqrt {\frac {n-24}{n}}})}{{\sqrt {n(n-24)}}(1-2\cos({\frac {\pi }{3}}{\sqrt {\frac {n-24}{n}}}))}}$, or any equivalent formulation thereof. I can try to draft up how I got to this, but it make take a little while. GalacticShoe (talk) 01:59, 13 May 2023 (UTC)

For $n<24$ this formula sends us on a little trip through the complex domain. For $n=24$, the formula gives us $0/0.$ For $n=25$ the result appears to be negative, which is curious for the sum of an everywhere positive series.  --Lambiam 06:13, 13 May 2023 (UTC)

At $n=25$ the expression yields $-{\frac {8\pi \sin({\frac {\pi }{15}})}{5-10\cos({\frac {\pi }{15}})}}\approx 1.0928$. For $n<24$, though the complex values yielded in the numerator and denominator are unfortunate, the expression is still real valued. For example, $n=23$ yields $-{\frac {8\pi \sinh({\frac {\pi }{3{\sqrt {23}}}})}{{\sqrt {23}}(1-2\cosh({\frac {\pi }{3{\sqrt {23}}}}))}}\approx 1.1007$. I think the only really problematic value here is indeed the indeterminate value given at $n=24$, though the limit does exist there and is equal to ${\frac {\pi ^{2}}{9}}\approx 1.0966$. GalacticShoe (talk) 11:47, 13 May 2023 (UTC)

Sorry, I overlooked the unary minus sign.  --Lambiam 15:25, 13 May 2023 (UTC)

OK, this is pretty impressive if it's real. I half thought it might be a joke. You want to give us a 30000-foot view of the ideas involved? I'm not super-interested in the plug-and-chug part, just the technique. --Trovatore (talk) 19:40, 13 May 2023 (UTC)

Yeah, for sure! I confess that the most involved part, which is converting the sum into a closed-form expression, was done through WolframAlpha. However, after spending some time looking at WolframAlpha's answer (and learning more about digamma functions than I think I will ever use in my entire life), I will supply what I think to be how it was done in-engine. Also, I know you don't want to go super into the nitty-gritty, but I will include some specific expressions just to highlight what's happening.

So first, note that the generalized pentagonal numbers are given by:

$p_{k}={\frac {3k^{2}-k}{2}}$

This means that the expression sought is:

$\sum _{k=-\infty }^{\infty }{\frac {1}{{\frac {3k^{2}-k}{2}}n+1}}$

or:

$\sum _{k=-\infty }^{\infty }{\frac {2}{(3k^{2}-k)n+2}}$.

Using WolframAlpha, specifically, splitting this sum into:

$(\sum _{k=-\infty }^{0}{\frac {2}{(3k^{2}-k)n+2}})+(\sum _{k=0}^{\infty }{\frac {2}{(3k^{2}-k)n+2}})-1$

it's possible to convert this sum into an expression using the digamma function $\psi (n)$. Specifically:

${\frac {2(\psi ({\frac {1}{6}}{\sqrt {\frac {n-24}{n}}}-{\frac {1}{6}})-\psi (-{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}}+{\frac {1}{6}})+\psi ({\frac {1}{6}}{\sqrt {\frac {n-24}{n}}}+{\frac {1}{6}}))-\psi (-{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}}-{\frac {1}{6}}))}{\sqrt {n(n-24)}}}-1$

Using the digamma reflection formula and recurrence formulas to find:

$\psi (f(n))-\psi (-f(n))=-{\frac {1}{f(n)}}-\pi \cot(\pi f(n))$

this can be converted into a completely elementary expression. The last set of steps which resulted in the formula supplied above was just a bunch of expression manipulation and some trig identities, nothing particularly complicated or worth mentioning.

Now, as to how I think WolframAlpha got that expression (and I must note here, I would have never thought of doing this had I not used WolframAlpha):

${\frac {2}{(3k^{2}-k)n+2}}={\frac {2}{3n}}{\frac {1}{k^{2}-{\frac {1}{3}}k+{\frac {2}{3n}}}}={\frac {2}{3n}}{\frac {1}{(k-({\frac {1}{6}}-{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}}))(k-({\frac {1}{6}}+{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}}))}}$

Splitting this up with partial fraction decomposition yields:

${\frac {2}{\sqrt {n(n-24)}}}({\frac {1}{k-({\frac {1}{6}}+{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}})}}-{\frac {1}{k-({\frac {1}{6}}-{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}})}})$

or:

${\frac {2}{\sqrt {n(n-24)}}}({\frac {1}{k+(-{\frac {1}{6}}-{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}})}}-{\frac {1}{k+(-{\frac {1}{6}}+{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}})}})$

Now, when taking the sum of this over $k$ from $0$ to $\infty$, we can use the digamma series formula:

$\psi (z)=-\gamma +\sum _{k=0}^{\infty }({\frac {1}{k+1}}-{\frac {1}{k+z}})$

to obtain:

$\psi (z_{1})-\psi (z_{2})=\sum _{k=0}^{\infty }{\frac {1}{k+z_{2}}}-\sum _{k=0}^{\infty }{\frac {1}{k+z_{1}}}$

yielding the closed-form expression for that half of the infinite series. Meanwhile, for $k$ from $-\infty$ to $0$, we can flip it to a sum from $0$ to $\infty$ again by just taking terms with $-k$ instead to yield:

${\frac {2}{\sqrt {n(n-24)}}}({\frac {1}{k+({\frac {1}{6}}-{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}})}}-{\frac {1}{k+({\frac {1}{6}}+{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}})}})$

The full closed-form expression for the result follows.

GalacticShoe (talk) 22:36, 13 May 2023 (UTC)

Generalizing the earlier comment about $n=23$, using the fact that $\cosh(ix)=\cos(x)$, $\sinh(ix)=i\sin(x)$, $\cosh(-x)=\cosh(x)$, and $\sinh(-x)=-\sinh(x)$, we can make this formula work for $n<24$ without having to work with complex values. In particular, it becomes:

$-{\frac {8\pi \sinh({\frac {\pi }{3}}{\sqrt {\frac {24-n}{n}}})}{{\sqrt {n(24-n)}}(1-2\cosh({\frac {\pi }{3}}{\sqrt {\frac {24-n}{n}}}))}}$

Although this still only exists at $n=24$ in terms of limits. Not sure if there's an expression that combines everything together without having to deal with complex values, and that evaluates to a non-$0/0$ expression at $24$. GalacticShoe (talk) 14:56, 15 May 2023 (UTC)

The $\operatorname {sinc}$ function solves the $0/0$ issue, but not explicitly the complex excursion. One could use the function

$\operatorname {singamajig} (x)=\sum _{i=0}^{\infty }{\tfrac {(-x)^{i}}{(2i+1)!}},$

and a similar replacement for the $\cos$ bit, but these functions have no standard closed-form notation.  --Lambiam 19:54, 15 May 2023 (UTC)

On the off chance I had a seach for singamajig on Google - and it exists!, albeit as a species of various cute fluffy toys that sing. There's also cosymajig which are sustainable fabrics.😀 NadVolum (talk) 21:01, 15 May 2023 (UTC)

May 14

Maximum packing density of *edge connected* Regular Pentagons

The maximum packing density of Regular Pentagons according to the article is 92.131. What is the maximum packing density of Regular Pentagons if *all* Pentagons in the packing must connect from one to another in a sequence of where an edge is shared between the pentagons in each step? Naraht (talk) 03:24, 14 May 2023 (UTC)

Here is a non-trivial lower bound. Warning: I have not checked everything carefully. All edge lengths will be $1$. Take a regular dodecagon. On its ten sides we can glue ten regular pentagons, on the outside. I think there is also room on the inside to glue on three more. The combined area of these pentagons equals $13A$, where

$A={\tfrac {1}{4}}{\sqrt {5(5+2{\sqrt {5}})}}.$

The whole combination fits in a circle of radius

$R={\sqrt {5+2{\sqrt {5}}}}.$

These combinations of 13 pentagons can be packed edge to edge in the plane in a square lattice pattern with a lattice spacing of slightly less than $2R.$ (Note: none of the edges will be parallel to a lattice axis.) If I'm not mistaken, this packing leaves space to squeeze in an extra pentagon for each group of 13. All combined, this gives a packing density slightly greater than

${\frac {14A}{\,4R^{2}}}\approx 0.6357.$

This can be improved by calculating the exact lattice spacing, but since there may be a simple way of establishing a far better lower bound, the slight improvement may not be worth the effort.  --Lambiam 10:03, 15 May 2023 (UTC)

I don't think Lambian has it right here. regular dodecagon says the interior angle is 150 deg. In pentagon it says the interior angle is 108 deg. Therefore each corner has 150+108+108 deg = 366 deg which can't be done flat without overlap. -- SGBailey (talk) 11:32, 15 May 2023 (UTC)

Sorry, you're right, but I meant to write decagon, as can be seen from fitting just 10 pentagons along the outside.  --Lambiam 12:50, 15 May 2023 (UTC)

The lattice will not be rectangular but somewhat skewed. Without detailed calculation, I'm not so sure a 14th pentagon will fit in the in-between space.  --Lambiam 14:38, 15 May 2023 (UTC)

May 15

Can 2*12^n+1 be square for n>2?

2*12^n+1 is square for n=1 and n=2, but can 2*12^n+1 be square again for n>2? 210.244.74.74 (talk) 21:04, 15 May 2023 (UTC)

It can't. Assume otherwise. Expand out the expression as $2^{2n+1}3^{n}+1=k^{2}$, so that $2^{2n+1}3^{n}=(k-1)(k+1)$. Since $(k+1)-(k-1)=2$, and $n>2$, $3^{n}$ divides one term entirely.

If $3^{n}|k-1=2^{a}3^{n}$ for $0\leq a\leq 2n+1$, then $2^{a}3^{n}+2=2^{2n-a+1}$. $a=0$ would imply that $3^{n}+2=2^{2n-1}$, which for $n>2$ can't be true by parity. So $a>0$ (and also by evenness, $a<2n+1$.) Divide both sides by $2$ to get $2^{a-1}3^{n}+1=2^{2n-a}$. $a=2n$ would once again lead to a problem, as $2^{2n-1}3^{n}=0$ would not make sense, so $2^{2n-a}$ is even, implying that $2^{a-1}3^{n}$ is odd, and thus $a=1$. So $3^{n}+1=2^{2n-1}$. There is no integer $n>2$ for which this is the case, so we discard this case.

Now if $3^{n}|k+1=2^{a}3^{n}$ for $0\leq a\leq 2n+1$, then $2^{a}3^{n}=2^{2n-a+1}+2$. $a=0$ would lead to $3^{n}=2^{2n+1}+2$, which would be impossible by parity, so $a>0$. If $a=2n+1$, then $2^{2n+1}3^{n}=3$, which cannot be for $n>2$, so $a<2n+1$. Thus we can divide both sides by $2$ to get $2^{a-1}3^{n}=2^{2n-a}+1$. $a=2n$ once again would again lead to the impossible $2^{2n-1}3^{n}=2$, so the right side is odd. Thus, $a=1$ and $3^{n}=2^{2n-1}+1$. This time, there are integer solutions to this, but they are $n=1,2$. So we discard this case.

But since those are the only possible cases, there's a contradiction; so no $n>2$ has the property that $2*12^{n}+1$ is square. GalacticShoe (talk) 23:28, 15 May 2023 (UTC)

A problem of the Aliquot sum

Let s(n) = sigma(n)-n = OEIS: A001065(n), s^k is the iterated function, we list the largest k such that a given natural number n is in the range of s^k.

k	such natural numbers n	OEIS sequence
0	2, 5, 52, 88, 96, 120, 124, 146, 162, 188, 206, 210, 216, 238, 246, 248, 262, 268, 276, 288, 290, 292, 304, 306, 322, 324, 326, 336, 342, 372, …	OEIS: A005114
1	208, 250, 362, 396, 412, 428, 438, 452, 478, 486, 494, 508, 672, 712, 716, 772, 844, 900, 906, 950, 1042, 1048, 1086, 1090, 1112, 1132, 1140, 1252, 1262, 1310, 1338, 1372, …	OEIS: A283152
2	388, 606, 696, 790, 918, 1264, 1330, 1344, 1350, 1468, 1480, 1496, 1634, 1688, 1800, 1938, 1966, 2006, 2026, 2202, 2220, 2318, 2402, 2456, 2538, 2780, 2830, 2916, 2962, 2966, 2998, …	OEIS: A284147

Can you find the sequence of k=infinity (i.e. n is in the range of s^k for all natural numbers k)? Assuming the strong version of Goldbach conjecture is true, i.e. all even number >6 are the sum of two distinct primes. 210.244.74.74 (talk) 21:15, 15 May 2023 (UTC)

May 16

Adding matrices with different dimensions

Why is it necessary to leave adding matrices with different dimensions undefined?? Georgia guy (talk) 16:40, 16 May 2023 (UTC)

Have you found some use for the idea? If you have a use for the idea then you must have some meaning to assign to the business, I'd be interested in what that could be! NadVolum (talk) 16:46, 16 May 2023 (UTC)

Matrix mathematics is only sensible if you have the same dimensions between the matrices. For example, if you're adding:

${\begin{bmatrix}1&3\\1&0\\4&2\end{bmatrix}}+{\begin{bmatrix}0&0\\7&5\end{bmatrix}}$

To what are you adding the 4 and 2 on the bottom row? --Jayron32 16:52, 16 May 2023 (UTC)

Some matrix math packages use "broadcasting" to define operations on inputs that have different dimensions. For example, in NumPy: [1]. This is just a convenience that makes sense in some contexts, and to make use of it, you have to know the broadcasting rules used by a particular tool. --Amble (talk) 17:32, 16 May 2023 (UTC)

I mean, one could define what it means to add matrices with different numbers of dimensions in any number of ways. That's one way to do it. Anyone could do anything. However, under the 'normally understood' method of matrix addition, one adds the numbers termwise to create a new matrix of the same size with the summed totals in the same locations. When the matrices are different sizes, it raises all kinds of issues with how to perform said operation, not the least of which is "to what do we add the numbers in the larger matrix". Workarounds such as the NumPy one are possible, but they are not part of the standard algorithms. --Jayron32 17:56, 16 May 2023 (UTC)

NumPy is not some obscure footnote along the lines of "anyone could do anything." Like I said, it makes sense in some contexts, and you have to know the rules that are being applied. --Amble (talk) 18:22, 16 May 2023 (UTC)

Oh, no, I wasn't trying to imply that NumPy was obscure. The point was that NumPy has built in some functionality to deal with asymmetric matrix addition, that functionality should not be meant to imply that, under the standard algorithms of matrices, that matrix addition is normally possibly between non-matching matrices. Someone obviously has a use for such a functionality (which is why it was built in to NumPy). The standard method of matrix addition, however, is not one of those uses. --Jayron32 18:34, 16 May 2023 (UTC)

May 17

How do we know 1 + 1 = 2?

Who told us this was the case and why did we so readily believe them? ___ 216.168.139.240 (talk) 14:36, 17 May 2023 (UTC)

I can demonstrate that to you right now.

Here is one dot: ⋅

Here are two dots: ⋅⋅

If I take one dot: ⋅ and i put another dot: ⋅ next to it, I get two dots: ⋅⋅

Therefore, one and one is the same thing as two. Q.E.D.. Thus endeth the lesson. --Jayron32 14:39, 17 May 2023 (UTC)

The proof in Whitehead and Russell's Principia Mathematica (second figure in the article) is only slightly more complicated... --Wrongfilter (talk) 14:44, 17 May 2023 (UTC)

Explain how those two dots are in any way related to each other. I see two seperate instances of dots connected by nothing, explain how those two dots which still maintain their individual forms end up creating 2? Mathematics seems like subjective bologne is what I'm trying to say 216.168.139.240 (talk) 14:50, 17 May 2023 (UTC)

This isn't the place to play silly games. Please stop if you aren't interested in being useful or asking serious questions. --Jayron32 14:53, 17 May 2023 (UTC)

Oh I'm sorry, "Jayron32" I seem to have missed the section of the Reference Desk Guide that says that questions have to benefit anyone other than the asker. 216.168.139.240 (talk) 14:56, 17 May 2023 (UTC)

I've yet to be convinced that numbers aren't just social constructs... AndyTheGrump (talk) 14:51, 17 May 2023 (UTC)