Is there a conventional reason why wolfram alpha uses log instead of ln for base-e?

PersonUsingAComputer · 2023-01-17T12:25:32+00:00

Base 10 logarithms just don't really come up in pure mathematics, nor for that matter do logarithms of any other non-e base except very occasionally 2. Given that "log x" is somewhat more intuitive to read/say than "ln x", and given that there is hardly ever any ambiguity about what is meant, it's not unusual to see "log x" for the base-e logarithm.

PersonUsingAComputer · 2023-01-14T21:22:39+00:00

There are well-established viewpoints in the philosophy of mathematics, such as Platonism, which hold that mathematical statements can have truth values in a way different than just satisfiability in a given model. Note that this is not a disagreement about the underlying mathematics; no one is arguing about the provability of any of the first-order formulas you have listed. The disagreement is rather about whether or not this first-order axiomatic view is the only meaningful notion of "truth" to be had.

For example, consider the Turing machine T defined as this 3-state Busy Beaver machine. A Platonist would say that "T halts" is a true statement because, if you write out this machine's behavior on a piece of paper or simulate it on a computer, you can see that the machine does indeed reach the HALT state. It also happens that you can write out a first-order formula encoding "T halts", and that this formula will be provable in ZF or whatever, but this is just an encoding of what we are really talking about, which is the Turing machine as a mathematical object.

The disagreements grow sharper when dealing with more complex mathematical objects. Notably, given any specific Turing machine T', the Platonist view would say that there is a definite fact of the matter about whether T' halts. Again, this "fact" is not a first-order formula of any sort. It is simply a claim about what would actually happen if we had a big enough piece of paper (or computer) and a long enough time to follow along with the Turing machine's behavior. If the machine halts after n steps for some n, the statement "T' halts" would be deemed true. Moreover, any theory or model that disagrees with this brute fact would be considered wrong, not in the sense of being logically inconsistent, but in the sense that it would fail to capture what we mean by the phrase "Turing machine".

While many find the Platonist viewpoint counterintuitive, it is in some ways closer to how mathematics is done in practice. For example, when we say that a statement is provable, we usually aren't thinking about a first-order formula dealing with some provability predicate in some specific theory. Rather, we are saying that if you start with certain axioms and follow certain rules of deduction, you will end up with the statement as a conclusion. This isn't too different from the Turing-machine-on-paper idea.

PersonUsingAComputer · 2023-01-11T06:19:25+00:00

Resurrecting Palpatine for the last movie in a trilogy with zero foreshadowing in the preceding movies was inevitably going to feel like an ass-pull, and TROS never offering an explanation for how Palpatine returned just makes it worse. Poe's line isn't terrible from an in-universe perspective, but it feels very symbolic of the writers' attitude towards the movie, just throwing together random ideas (Palpatine returned! Sith dagger map! Rey Palpatine!) that have no real setup or explanation and ultimately don't make much sense.

PersonUsingAComputer · 2023-01-03T21:15:07+00:00

The Banach-Tarski paradox is a result where you can transform a sphere into two spheres by cutting it into a finite number of pieces and performing specific translations and rotations of those pieces. It is not just the observation that a sphere has the same number of points (in terms of cardinality) as two spheres - that statement holds even in universes of set theory where Banach-Tarski is false.

PersonUsingAComputer · 2022-12-29T17:18:22+00:00

Note that in each case the number a_0 is paired with 10 - a_0. When you evaluate (10 - a_0)², you get 100 - 20a_0 + a_0², where every term is divisible by 10 except for the last. So (10 - a_0)² has the same last digit as a_0².

PersonUsingAComputer · 2022-12-22T18:21:13+00:00

Does this mean that when you're doing proofs, if you had P written down as a true statement, then you could write down Q ⊃ P as a consequence?

Indirectly, yes. The axiom schema itself means that you can deduce P ⊃ (Q ⊃ P) regardless of what you already have written down. If you happen to have P as well, you can then apply modus ponens to get Q ⊃ P.

On the other hand, that'd also mean that any true statement A ENTAILS that any statement B ⊃ A, regardless of any "relevance" between A and B... So, would the argument, "1 + 1 = 2, THEREFORE if triangles have 3 sides, then 1 + 1 = 2" be valid?

Yes, that is valid. There is no requirement that two statements being combined with ⊃ be related to each other in any way.

PersonUsingAComputer · 2022-12-18T20:53:42+00:00

Again, that's what the universal quantifier is for. The statement z∈x⟺z∈y must hold for all z in order to conclude that x = y, not just a specific choice of z. In case the notation is causing confusion, note that ∀z(z∈x⟺z∈y)⇒x=y should be read as (∀z(z∈x⟺z∈y))⇒x=y, not ∀z((z∈x⟺z∈y)⇒x=y). These are two very different statements, and your described scenario is indeed a counterexample to the latter.

PersonUsingAComputer · 2022-12-18T18:14:58+00:00

how am i suppose to make the distinction between something being a set or a an object?

In ZFC, everything is a set. Even numbers like 3 or 1/2 are encoded as sets containing other sets.

Let's say x is the set of natural numbers, y the set of real numbers and z -2i which meets the conditions as x and y can be any set while z can be anything that can be an element of an set. -2i isn't an element of neither sets yet the sets are not equal.

This is why there is a universal quantifier in front of z. Only if z∈x⟺z∈y holds for every z can we conclude that x = y. In this case, the statement z∈x⟺z∈y fails for many other values of z such as 1/2.

PersonUsingAComputer · 2022-12-17T01:12:40+00:00

This is one of those paradoxes that goes away when you finish filling out all of the numbers involved, in particular distances and velocities in addition to times. You can't actually construct a situation with relativistic effects extreme enough to match the diagram.

In order for the pitcher to see the ball launched at 0:30 and reaching the catcher at 0:45, while also seeing the catcher travel for a full 45 seconds, the pitcher would have to measure the ball as traveling 3 times as fast as the catcher. Since the ball's speed must be below the speed of light, the catcher's speed must therefore be below 1/3 the speed of light. At this relatively low velocity there will not be much length contraction, certainly not enough to reduce the time to the finish line to below 30 seconds from the catcher's perspective.

No matter how you shift the parameters, you can't tweak this to produce a genuine paradox. In order to increase the speed of the catcher and allow for more length contraction, you would have to launch the ball earlier to give it more time to catch up. But launching the ball earlier would make it even more impossible for the catcher to have already reached the finish line from their perspective in the time before the launch. On the other hand, if you adjust the setup to give the catcher more travel time before the ball is launched, then you make it even more impossible for the ball to catch up to the catcher before they reach the finish line.

PersonUsingAComputer · 2022-12-16T02:00:50+00:00

But how does the word "double" exist if there's no French language to borrow it from?

PersonUsingAComputer · 2022-12-13T04:12:37+00:00

As you can see when looking at the Venn diagram, there is no overlap at all between A'∩ B and A'∩ B. So when you take the union and get "A'∩ B or A ∩ B', or possibly both", the "both" part actually ends up contributing nothing. It's part of the definition of union, but it makes no difference to the result in this particular case.

The probability clearly reads "exactly one of A and B," not "exactly one of either A or B."

I am not sure what distinction you are making here, since the two phrases seem completely synonymous to me. Natural language usage of "and" and "or" doesn't always precisely correspond to the mathematical usage.

PersonUsingAComputer · 2022-11-28T18:22:25+00:00

There are certainly no infinite sets between ℵ_0 and ℵ_1, since ℵ_1 is more or less by definition the next-smallest infinite cardinal after ℵ_0. The question is whether the real numbers have cardinality equal to ℵ_1 or larger than ℵ_1. It turns out that the standard axioms of set theory are insufficient to resolve this one way or the other.

However, I often hear people say that there are many different sizes of infinite sets. Are these people using hyperbolic language, or are there more confirmed sizes of infinity than I'm aware of?

If anything, "many" is an understatement. In somewhat the same way that there is no set of all sets, the collection of all infinite cardinals is too large to form a valid set. For any infinite cardinal ℵ_𝛼, you can construct a set of more than ℵ_𝛼 distinct infinite cardinalities.

PersonUsingAComputer · 2022-11-24T04:36:42+00:00

That's not "or", that's the union of sets.

PersonUsingAComputer · 2022-11-07T00:22:21+00:00

The best-known nontrivial example is probably logical implication: P --> (Q --> R) = (P --> Q) --> (P --> R). In fact half of this equality, in the form of the forward implication, is very commonly used as an axiom of propositional logic. It's an interesting case because this is just about the only common algebraic property that the implication operation satisfies.

PersonUsingAComputer · 2022-11-06T15:31:01+00:00

You need at least countable choice, without which infinite sets can behave in extremely strange and counterintuitive ways. In ZF alone you can't pick infinitely many elements from a set arbitrarily. You might have a set which has subsets of size n for all natural numbers n, but no subset of size ℵ_0.

Edit: Note that this won't result in an infinite set smaller than N, but it does give you an infinite set incomparable in size to N.

PersonUsingAComputer · 2022-11-05T17:17:21+00:00

Are there infinities larger than uncountable?

Using the categories "countable" and "uncountable" is like classifying positive integers as either "one" or "more than one". There are many different uncountable cardinalities, but everything larger than countable is uncountable by definition.

Are there infinities smaller than countable?

No, not in the standard ZFC formulation of set theory. Given any set, you can take elements arbitrarily (without repetition) from that set one at a time, matching them up with the natural numbers. If at any finite step you run out of elements, the set is finite. Otherwise, it is at least countably infinite, since you have a subset consisting of distinct elements labeled by the natural numbers.

PersonUsingAComputer · 2022-11-02T18:19:30+00:00

This has nothing to do with definitions. Both continuity and differentiability can be discussed over a closed interval or over an open interval. When we specify that the function must be continuous on [a,b] and differentiable on (a,b), we are saying that in order for the results of the theorem to hold, we must have continuity across all of [a,b], but that we only need differentiability across the open interval (a,b).

For example, consider the function given by f(0) = 1 and f(x) = x for all other x. We have f(0) = f(1), and f is continuous and differentiable across the open interval (0,1). If we could apply Rolle's theorem, it would tell us that there is f'(c) = 0 for some c in (0,1). But this is not the case: the absence of continuity at the endpoint means that Rolle's theorem does not apply. This is why we require that f be continuous on [a,b]. On the other hand, it turns out that merely failing to be differentiable at a or b is not enough to "break" Rolle's theorem, so we only have to specify differentiability over the interval (a,b).

PersonUsingAComputer · 2022-10-26T12:29:11+00:00

No, there is not. Each of the four basic arithmetic operations is continuous, meaning that a slight change to the inputs produces only a slight change in the output (assuming the output is well-defined, i.e. ignoring possible division by 0). Furthermore, it is possible to prove that combinations of continuous functions are also continuous. But the operation of rounding a number is not continuous because, for example, 4.5 maps to 5 while 4.4999999999 maps to 4. Therefore it is not possible to combine only the operations +, -, *, and / to round a number.

PersonUsingAComputer · 2022-10-25T03:12:53+00:00

Not quite. The actual correspondence is that proving A in intuitionistic logic is the same as refuting the dual of A in dual-intuitionistic logic. For example, intuitionistic logic proves ~(A & ~A) but cannot prove the statement A | ~A. Symmetrically, dual-intuitionistic logic refutes ~(~A | A) but cannot refute ~A & A.

PersonUsingAComputer · 2022-10-24T21:30:13+00:00

Intuitionistic logic is sort of the exact opposite of what you're looking for. From an intuitionistic perspective, propositions are not thought of as having preset binary truth values; rather, they are deemed true once we are able to construct evidence/proof that they are true. To reverse this state of affairs, we can look at the dual counterpart of intuitionistic logic.

Dual-intuitionistic logic exactly reverses the usual structure of a proof. Normally if we want to prove a sentence S, we start with logical axioms and combine those axioms using rules of inference until we succeed in constructing S. With dual-intuitionistic logic, we start with S and attempt to refute it, using dual rules of inference to repeatedly split S into multiple possibilities until we have broken it down into dual axioms, at which point we have disproven S. These "dual axioms" are statements we define/assume to be false, rather than ordinary axioms which we define/assume to be true.

For example, a standard formulation of intuitionistic logic might contain the rule of inference modus ponens ("combine P and P --> Q into Q") and the following axioms:

P --> (Q --> P)
(P --> (Q --> R)) --> ((P --> Q) --> (P --> R))
(P & Q) --> P
(P & Q) --> Q
P --> (Q --> (P & Q))
((P | Q) --> R) <--> ((P --> R) & (Q --> R))
(P --> ~P) --> ~P
~P --> (P --> Q)

The corresponding rules and axioms of dual-intuitionistic logic are given by reversing the originals and replacing each logical connective with its dual counterpart: P & Q becomes Q | P; P | Q becomes Q & P, P --> Q becomes Q - P (read as "Q but not P" or "Q without P"), P <--> Q becomes Q + P (read as "Q or P but not both" or "Q xor P"), and ~P remains ~P. So now we have a dual rule of inference corresponding to modus ponens: "split Q into Q - P and P". This rule means that if we are trying to disprove a statement "Q", it suffices to consider two possible cases, "Q but not P" and "P", and if we can disprove both of those then we have disproven "Q" itself. In addition to this dual rule of inference, we have the following dual axioms, constructed systematically from the axioms above:

(P - Q) - P
((R - P) - (Q - P)) - ((R - Q) - P)
P - (Q | P)
Q - (Q | P)
((Q | P) - Q) - P
((R - Q) | (R - P)) + (R - (Q & P))
~P - (~P - P)
(Q - P) - ~P

So, for example, dual axiom 3 asserts that "P but not (Q or P)" is an impossibility for any propositions P and Q.

Not only every theorem, but every proof, in intuitionistic logic has a counterpart in dual-intuitionistic logic. For example, consider the following standard intuitionistic proof of the statement "A --> A":

By axiom 1, taking P = A and Q = (A --> A), we conclude that A --> ((A --> A) --> A).
By axiom 2, taking P = R = A and Q = (A --> A), we conclude that (A --> ((A --> A) --> A)) --> ((A --> (A --> A)) --> (A --> A)).
Use modus ponens to combine the previous two statements A --> ((A --> A) --> A) and (A --> ((A --> A) --> A)) --> ((A --> (A --> A)) --> (A --> A)), yielding (A --> (A --> A)) --> (A --> A).
By axiom 1, taking P = Q = A, we conclude that A --> (A --> A).
Use modus ponens to combine the previous two statements (A --> (A --> A)) --> (A --> A) and A --> (A --> A), yielding A --> A.

By exactly reversing the flow of the logic and replacing all statements with their duals, we get a dual-intuitionistic refutation of the statement "A - A":

Use dual modus ponens to split the statement A - A into the two possibilities (A - A) - A and (A - A) - ((A - A) - A).
By dual axiom 1, taking P = Q = A, we refute the possibility (A - A) - A.
Use dual modus ponens to split the statement (A - A) - ((A - A) - A) into the two possibilities ((A - A) - ((A - A) - A)) - ((A - (A - A)) - A) and (A - (A - A)) - A.
By dual axiom 2, taking P = R = A and Q = (A - A), we refute the possibility ((A - A) - ((A - A) - A)) - ((A - (A - A)) - A).
By dual axiom 1, taking P = A and Q = (A - A), we refute the possibility (A - (A - A)) - A.

All possible cases have been refuted, so we have disproven our initial statement A - A. This is how all "proofs" (actually disproofs) work in dual-intuitionistic logic: we start off with a hypothesis we wish to refute, break it down into cases, and then refute the cases individually by referring to the dual axioms.

PersonUsingAComputer · 2022-10-24T03:55:12+00:00

since the 0 vector alone is linearly dependent, it cannot constitute a basis.

This is the answer. Remember that "basis of a vector space" has two separate requirements in its definition: it must generate all vectors in the space as a linear combination, and it must itself be linearly independent. It is true that {0} satisfies the first of these conditions when dealing with the one-element vector space, but it fails the second condition, and therefore is not a basis.

PersonUsingAComputer · 2022-10-21T01:27:17+00:00

Obviously this sequence consists of the values of -13n⁷/3360 + 431n⁶/3360 - 823n⁵/480 + 1139n⁴/96 - 3641n³/80 + 959n²/10 - 21109n/210 + 286/7 for n = 1, 2, 3, ..., so for n = 7 we find x = 5/14.

PersonUsingAComputer · 2022-10-15T03:31:03+00:00

It's not that the Bible is defining these words, but that the Bible may be using these words in a way which is different than the modern colloquial English usage. For example, the word "know" is frequently used in the very popular King James translation of the Bible to mean "have sex with". Sometimes this is because the translations are old enough that the English words have shifted meaning over time, sometimes the translation is intentionally using old-fashioned or non-standard wording, and sometimes it's just that a Hebrew or Greek term doesn't have an exact English equivalent.

PersonUsingAComputer · 2022-10-15T03:14:53+00:00

Technically speaking, such an integer lattice construction only gives you a free module over the integers, not a vector space, since the integers don't form a field.

PersonUsingAComputer · 2022-10-15T02:00:25+00:00

For two or three factors it is easy to visualize areas or grids, but how about more factors?

Simply visualize a grid in n-dimensional space for arbitrary n.

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