You can, and some devices do, use Newton's algorithm. For square roots, it consists of taking an initial guess z ; it can be pretty bad, say you guess sqrt(a) to just be a.

Then you calculate (z + a/z) / 2 *. This is a much better estimate. Now set that result to be z and plug it in again; and again; and again. It becomes a very good approximation to the square root pretty fast.

Example: calculating sqrt(5), starting with the estimate z=2 to get nice simple fractions instead of ugly decimals (a calculator obviously would not care, but I do).

• Step 0: z=2 ; z² = 4.
• Step 1: z = (2 + 5/2)/2 = 9/4 ; z² = 5.0625
• Step 2: z = (9/4 + 5·4/9)/2 = 161/72 ; z² ~ 5.00019
• Step 3: z = (161/72 + 5·72/161)/2 = 51841/23184 ; z² ~ 5.00000000186

It took us only 3 iterations to get 8 digits down. It can actually be shown that the number of correct digits roughly doubles each step. So with one more, we easily get the 10 or 12 digits most calculators use.

*: Note how (z + a/z) / 2 is in the middle between our previous guess z and a/z, which is another plausible guess (indeed, if z²=a, then a/z is just z; and if z is too small, then a/z is too large, and vice versa). The true reason behind the algorithm is based in calculus, one approximates using tangents at a curve to find places that should be closer to solutions.

Let me quote Wikipedia, but this is also in agreement with what I learned and sources are given there as well: '''The strength of chemical bonds varies considerably; there are "strong bonds" or "primary bonds" such as covalent, ionic and metallic bonds, and "weak bonds" or "secondary bonds" such as dipole–dipole interactions, the London dispersion force and hydrogen bonding.'''

So bonds are just that, bonds. Some are just stronger or more commonly encountered.

Also, dissolving salts involves their ionic bonds, so if we are pedantic, this part involves even more bonds. But I get some feeling that even ionic bonds are not "bonds" in your mind, only covalent ones?

> Something being dissolved can be endo or exothermic, there's no hard and fast rule about whether dissolving something will heat or cool a liquid. Saying it's one way or another just isn't accurate, it depends entirely on what you're dissolving.

Look, you get hinged on this again, yet the original post never said it is just one or another. It differentiated between the exothermic "reaction" of the formation of hydrogen bonds (or van-der-Waals forces or others, unmentioned ones) and the endothermic dissolution of the (often ionic, but also not explicitely mentioned) bonds in the solid. Those are two things, their sign is (almost always) pretty clear and opposite, and the heat/cold of dissolution comes from their (signed!) sum. Hence it can be either way, and this was mentioned in that post.

Yes, they left out some detail and might( have wrongly implied that the total is always negative, i.e. dissolution is endothermic, but this was not what you called out and this is ELI5, not a journal paper.

Well, I will leave my expertise of chemistry to the other readers (and my post history), but you clearly can only do insults (and down-votes, it seems; surely a sign of a healthy adult scientific discussion!).

Are you seriously saying that "hydrogen bonding" is not "bonding""? Because that is clearly silly.

And yes, they said the new bonds (the hydrogen ones, duh!) release energy; after more(!) energy was used to break ionic bonds.

You are just trying to be pedantic and fail at it.

Maybe some examples for lesser solubility at higher temperatures that are salts:

• many calcium salts such as its carbonate (gypsum), sulfate (chalk), phosphate, etc.;
• sodium sulfate has its highest solubility at 33°C, it falls off in both directions;
• unlike what many believe, table salt's solubility increases almost not with temperature; only the speed of dissolving does.

It is also interesting to note that some salts produce a lot of heat (e.g. NaOH a.k.a. caustic soda) or cold (e.g. ammonium nitrate a.k.a. the stuff that blew up Beirut, or urea). This and solubility are related, but all combinations can happen.

> But new bonds are NOT created. There is hydrogen bonding, but there's no chemist in the world who would say that hydrogen bonds are a form of chemical bonding.

They didn't say "chemical bond", but just "bond". Just read what you wrote yourself again, even you used "hydrogen bonding", and indeed, there are many types of bonds and all of them involve some energetic changes. It doesn't matter if hydrogen bonds, ionic bonds, chemical bonds, or magical bonds.

> Secondly, this process doesn't release energy.

They didn't claim it does. Even in what you quoted they say "This refunds some of the heat used to break up the solid in the first place", implying that in total, energy is still spent. And it is correct that the break-up requires energy to overcome the ionic bonds, yet some other salts release quite a bit of energy when dissolved in water, exactly due to the reason they gave.

First, there is not really a "and so on". It depends on what you allow, but it ends as soon as you add some common requirements. We often encounter:

• the reals ℝ, an (actually, the archimedean complete) ordered field, i.e. we have ≤ and it is compatible in various way with + and ·,
• the complex numbers ℂ, an (algebraically closed, complete) commutative division algebra, i.e. a·b = b·a holds,
• the quaternions ℍ, an associative division algebra, i.e. a·(b·c) = (a·b)·c holds,
• the octonions 𝕆, an alternative division algebra, i.e. a·(a·b) = (a·a)·b and a·(b·b) = (a·b)·b hold
• the sedenions 𝕊, which are not even a division algebra, i.e. for some x there is no y with x·y = 1; they at least still satisfy a·(a·a) = (a·a)·a and (a·a)·(a·a) = ((a·a)·a)·a.

The cursive words are the most important aspect here: each of them is also true for the ones above them, but not below. At each step, something is lost.

With 𝕊, almost none of the typically used properties are left, we barely(!) can define powers x^n without having to write down in which order we multiply them. Most people hence stop the list there, only very few people seriously work with them.

Generally, classification theorems exist for algebras over the reals, which say that the dimensions 1, 2, 4, 8 are the only ones where algebras with the above properties exist. However, depending on what exactly you require, it is not always true that the above list contains all such algebras; only the dimensions are certain powers of 2!

For convenience, I will now write "ℝ-algebra" for what is typically called a "unitary finite dimensional algebra over ℝ": a set A with addition + and multiplication · containing (and compatible with) ℝ, satisfying:

• distributivity (that is: a·(b+c) = a·b+a·c and (a+b)·c = a·c+b·c) for any a,b,c in A,
• commutativity of addition (a+b = b+a),
• centrality of ℝ (s·a = a·s for all a in A and s in ℝ, in other words, multiplication of two numbers is commutative if at least one factor is in ℝ)
• negativity (a+(-1)·a = 0, noting that -1 as a real number is in A),
• addition&multiplication turn A into a real vector space of finite dimension dim(A) (there is a finite list a1, a2, ..., an of elements of A such that every element of A can be written in the form s1·a1 + s2·a2 + ... + sn·an for suitable s1, s2, ..., sn in ℝ).

If you then require that A satisfies some sane other properties, e.g. allowing division, or associativity or commutativity, then one can show that only very few examples exist. The exact version depends on what you want, there are many different results.

Lets maybe dive a bit into one such result by Hurwitz on "composition algebras", where the standard absolute value |·| on ℝ extends linearly to the algebra and furthermore satisfies |x·y| = |x|·|y|. Then such algebras exist only in dimensions 1, 2, 4 and 8. Some examples are

• ℝ,
• ℂ, with |a+bi|² = a²+b²,
• ℍ, with |a+bi+cj+dk|² = a²+b²+c²+d²,
• 𝕆 with the same, but 8 terms.
• ℝ², with |(a,b)| = a²-b²,
• and many more.

Note that the last one is not on the list above, and there are similar examples of dimensions 4 and 8.

But the really interesting part about this theorem is that |x·y| = |x|·|y| on ℂ corresponds, with x = a+bi and y = c+di, to the formula (a²+b²)·(c²+d²) = (ac-bd)²+(ad+bc)², hence a product of two sums of two squares is again a sum of two squares! And we can verify this formula directly, never even talking about complex numbers, or even composition algebras; it actually holds in any ring! Similarly, we get such formulas for sums of 4 or 8 squares by invoking ℍ or 𝕆, they are just a bit longer.

Yet, almost magically, there cannot be such a formula for other numbers of variables by the theorem!

Lastly, a quick statement on the proof behind such results:

• Proving that ℝ is the only archimedean ordered complete field is typically done in the first year of a calculus course and not difficult, just a bit technical.
• The only ℝ-algebra (except ℝ itself) that also happens to be a field is ℂ, as follows from the fundamental theorem of algebra (i.e. that every non-constant polynomial with complex coefficients has a root within ℂ); this famous result is shown similarly early when studying mathematics.
• There is a very vast theory of "Brauer groups" that classifies, for any field K, all the K-algebras that also happen to be division rings (similar to fields, but without requiring commutativity). For K=ℝ, the only examples are ℝ, ℂ and ℍ. This is already pretty advanced, typically a Master's course, and often done when also dealing with "Group/Galois cohomology".
• Almost all proofs where the answer is either 1, 2, 4, 8 or exactly the list ℝ, ℂ, ℍ, 𝕆 are seriously involved and usually done by invoking advanced methods from algebraic topology such as K-theory or higher homotopy groups. Even sketching anything here would go way beyond this already long post.

Edit: forgot to mention:

Those results on composition / division algebras also have ramifications on very different looking things that are quite interesting in their own rights, such as:

• "Sphere eversions" (turning a sphere inside-out without creasing or tearing it) exists for spheres of dimensions 0, 2 and 6 within 1, 3 and 7 dimensional space, respectively. There are some nice videos of the 2-sphere being everted in 3-space such as https://www.youtube.com/watch?v=OI-To1eUtuU&t=1131s, and I would recommend the entire video as worth your time.
• A "cross product" exists only in dimensions 0, 1, 3 and 7. The 3-dimensional one is widely known among multiple fields and is related to quaternions, but the 7-dimensional cross product coming from octonions is something even many mathematics professors don't know.

> Consider a set S that is a subset of the reals,

That's where you went wrong: the set is simply not required to be contained in the reals. You can give any set X a total ordering such that the induced topology is discrete; similarly, we can give X a metric that makes it discrete, the discrete metric.

> One can't "count" all of the decimals because you can always pick another decimals between A and B.

This is unrelated to "counting". The rationals satisfy the very same property, yet can be counted. But all decimals, i.e., real numbers, cannot be counted, they are "uncountable".

Conversely, there are uncountable "discrete" ordered sets where nothing is between a number and its two neighbours. Hence the property you speak of and being (un)countable are independent, neither implies the other.

This. "Infinity" has different meanings or interpretations depending on the field (mathematics, physics, philosophy, ...), context (sets, numbers, size, mind, ...) and much more. In other words, there is not the infinity, and listing&explaining all available options would be almost impossible, and definitely too long.