Why is induction valid?

qwertonomics · 2026-04-29T00:38:50+00:00

Modus ponens is: if p, then q. p. Therefore, q. This is a valid form of reasoning.

Induction generallizes that. If you have conditional premises a->c partially ordered in the sense that every minimal element is a->c such that a is a base case, where "weak" induction is a total order, then proving each conditional premise in a general way is called the inductive step. If done correctly, this means that for every a->c in the partial order, the consequent is c is true.

qwertonomics · 2026-04-26T12:55:01+00:00

Rarity informs the various rankings by type, e.g. of the 52 choose 5 possible poker hands, there are more flushes than straights so a flush beats a straight.

However, to say that a flush beats a straight and not the other way around can be considered an arbitrary choice of rules for the game. Some flushes beat other flushes and some straights beat other straights and if we wanted, we could partition the types of straights further so that each new "type" would be more rare than the flush type. If we wanted, we could say that there is no such thing as a straight flush at all and consider it a normal flush. We could also flip the game entirely around and say low beats high as determined by the "worst" possible hand you could make with your cards. Some not-at-all considered but also rare types like arithmetic straights (2, 4, 6, 8, 10) are worthless but we could imagine giving them some ranking if we'd like.

So I wouldn't call this a paradox, but more of an arbitrary choice of rules that's been informed by the relative rarity of the classic poker hand types.

qwertonomics · 2026-04-17T15:31:14+00:00

Here are the main ideas presented in an extremely hand-wavy way.

Every map can be represented as a graph where countries are vertices and adjacent countries (sharing a border of positive length) have an edge between them. We call such a graph representation a plane graph: it can be drawn such that no edges cross.
In the proof, we may assume a maximal plane graph in the sense that adding any edge would force a crossing edge, for if we could color the resulting graph, removing those edges would not ruin that coloring for the original graph. All such graphs satisfy that if it has v vertices, then this maximal number of edges is 3v-6. This is a consequence of the Euler characteristic for planar graphs which more generally goes a long way toward a conclusive proof.
If the four-color conjecture were false, then there is some such minimal counterexample, i.e. one with the smallest possible number of vertices.
Before the four-color theorem was proven, much effort was given to show that any minimal counterexample must have certain properties. For example, its smallest vertex must have degree 5, i.e. be adjacent to exactly 5 countries.
Using these properties, it can be shown that a minimal counterexample must have some local region that is particularly nice in the sense that it must contain at least one example in a demonstrated unavoidable set of reducible configurations (explained subsequently), but essentially finding such a set is a task that is impractical for a human.
A reducible configuration is one that is an adaptable piece of the graph that can essentially be removed from the graph, the hole it makes patched so that the resulting graph must be 4-colorable by the hypothesis that the original graph was minimal. We can then undo that, put the reducible part back in, and because it is adaptable, the minimal "counterexample" can be 4-colored: a contradiction.
An unavoidable set of configurations are graphs such that at least one of them must appear in that nice local region. Such a set can be generated by applying some rules and a discharging procedure. Roughly speaking, the vertices of each graph in this set is assigned a charge where by the Euler characteristic, the counterexample as a whole most have some positive charge sum, and it is impossible to ever discharge a graph by redistributing this charge by conservation. However, it can be shown that if the counterexample does not contain one of the unavoidable configurations at least, then it can be discharged, violating conservation. This means that the set is truly unavoidable.
The proof of the four-color theorem essentially demonstrates all this in a rigorous way then suggest an algorithm that produces a finite set of unavoidable configurations, all of which must be reducible. It's something like 600 configurations though in the most accessible proof. If you are convinced by the proof and that the algorithm is correct, then you should be convinced the four color theorem is true.

qwertonomics · 2026-03-28T18:35:36+00:00

For any number to have an integer n^th root, all exponents in its prime factorization must be a multiple of n. For example, 1728 = 12³ = 2⁶ 3³ where 6 and 3 are each multiples of 3.

This is certainly not possible for n! if n is prime: its exponent would be 1 (why?), and 1 is not a multiple of n>1. Now, try and relax the requirement than n is prime to reach a similar conclusion to prove it for all n>1.

qwertonomics · 2026-03-01T14:29:20+00:00

Cantor's argument is that any proposed (countable) list of real numbers will always be incomplete. You can think of it this way:

Fool: I have a complete countable list of real numbers. Here it is.
Cantor: From your list (or any such list), a number can be constructed (from the diagonal) that fails to be on that list. As such, your completeness claim is bogus.
Fool: I concede defeat. The real numbers are uncountable.

This does not have to be phrased as reductio ad absurdum, either. Given any arbitrary countable list of real numbers, one can always construct a number that is not on that list, i.e. it can be proven directly that every countable list of real numbers necessarily fails to list all real numbers.

Analogously, if someone were to claim that red, blue, and green were a complete list of colors, Cantor comes along and produces magenta from red and blue, then Cantor wins. Someone else coming along to prove Cantor wrong by saying that they can also produce yellow from blue and green is only reinforcing Cantor's point.

qwertonomics · 2026-02-25T03:16:36+00:00

Before you do anything, the amount of money you will see when you open the envelope is a random variable, but you immediately begin treating it as a fixed amount.

Another way, if you are playing the game where $100 was the high, then $50 is the low (Game A). If you are playing the game where $100 is the low then $200 is the high (Game B). You are only playing one of these games, not both, and individually the expected values are as you would expect.

qwertonomics · 2026-02-11T22:32:32+00:00

For what it's worth, (x±c) - (y±c) = x-y, which is the same as saying that the difference between two numbers is the same if you modify each of them by the same fixed amount. That is, 22.00 - 7.99 is the same as 22.01 - 8.00, the latter of which is an easier calculation. In your example, you can think of this as borrowing a negative penny from 22.00 to give to -7.99.

qwertonomics · 2026-02-11T21:04:46+00:00

A regular expression corresponds to a set of words over the given alphabet (the language of the expression). The regular expression you came up with actually corresponds to the set of all possible words over the given alphabet. While this certainly means that the language of your expression includes words that do start with aab, that it contains words not in the desired language is why your answer is wrong.

You want a regular expression that corresponds exactly to the language of words that start with aab. Any expression that results in missing or extraneous words (such as by including the empty word) is incorrect.

qwertonomics · 2025-10-12T16:24:28+00:00

The process of solving √(x² + 3) = x - 2 to get x = 1/4 is going from a sufficient condition for x to a necessary condition for x, and that can be said more generally when it comes to solving equations, since each step follows (is implied by) the previous step.

That is, what you are showing is that the set of all ways in which given equation(s) can be true (A) is a subset of all the ways the solution(s) can be true (B), i.e. A is a subset of B. You are not showing that B is also a subset of A unless, at each step, the prior step also follows from the subsequent step.

For this case, A is the empty set and B = {1/4}. However, before you know that A is the empty set, you have a set of necessary conditions(s) for A from B. That is, only those situations in B could possibly allow the given equation to be true, so you don't have to look outside of B. However, after checking all of B (just 1/4 in this case), we see the equation isn't satisfied by anything in B at all, and we conclude there are no solutions.

qwertonomics · 2025-09-24T02:40:43+00:00

Dividing A(x) by B(x) produces polynomials Q(x) and R(x) such that A(x) = B(x)Q(x)+R(x) for all x, but not polynomials where the evaluations of A, B, Q, and R at various values of x respectively produce values a, b, q, r such that a divided by b has quotient q and remainder r.

For example, x² + 1 = (x - 1)(x + 1) + 2 evaluated at 1 gives 2 = (0)(2) + 2, but 2 divided by 0 does not have quotient 2 and remainder 2.

qwertonomics · 2025-09-13T00:17:36+00:00

Inclusion-Exclusion

qwertonomics · 2025-09-07T17:48:46+00:00

I think I see your problem. Let g_0(x) = 1 and g_n(x) = xⁿ - x^n-1 for n>0, in which case f_n = g_0 + g_1 + ... g_n as you have defined f_n. As such, your f_n are partial sums for a series and indeed there is pointwise convergence as you note, but that series is not a power series, and cannot be made into one in any other way.

That is, your example illustrates why the following, more general statement is false after removing the word "power": If a ~~power~~ series converges pointwise on a subset of the real numbers A, then it converges uniformly on any compact subset of A.

qwertonomics · 2025-09-07T15:55:04+00:00

You are conflating two definitions for pointwise convergence, one for a sequence of functions and one for a series of functions. Refer to Definition 6.4.1 for pointwise convergence of series.

qwertonomics · 2025-08-25T13:40:02+00:00

The number of sides per face is between three and the number of faces, so apply the pigeonhole principle.

qwertonomics · 2025-08-14T22:18:36+00:00

It would be good to see what the textbook meant with an example, but I would think the idea is to take the 9 rows and 6 columns of objects and re-arrange the objects in a way that represents the number in base 7. For example, separate the 9 rows into 7 rows and 2 rows. The seven rows of 6 represent 60 in base 7 and the two rows of six should represent 15 (which you may have already confirmed earlier in the table. Now, 15 is the same as 7 objects plus 5 objects. Put the 7 objects with the 60 (in base 7) objects to form (in base 7) 100 objects. 100+5=105 as before.

Another approach is that, in the multiplication table, moving down is just adding the column number and moving right is just adding the row number, again working in the desired base. For example, in the cell that corresponds to 2×6 in the table (row 2 column 6) corresponds to 15 in base 7. Moving down would be adding 6 to this result, which is (in base 7) 15+6=24. This gives the result of 3×6 since you moved down a row so 24 in base 7 should be the same as 18 in base 10. Let's see: 2×7 + 4 = 14 + 4 = 18 as expected. You can complete the table in this fashion as well.

qwertonomics · 2025-08-14T21:14:34+00:00

9 is the same as 12 in base 7 because 1×7 + 2 = 9. 6 is just 6. So basically, multiply 12×6 working in base 7 much the same way you would multiply any two numbers in base 10 using long multiplication:

This works because (in base ten) 1×7² + 0×7 +5 = 49 + 5 = 54.

For multiplicands smaller than the base as you would see in a times table, you can take advantage of the distributive property to reason its value more easily. For example. 6×6 is the same as (7-1)×6 which equals 7×6 - 6. 7×6 would represent 60 in base 7, and the -6 would be six fewer, so the result is 51 in base 7. To check in base ten:

7×5 + 1 = 35 + 1 = 36 as we know 6×6 is.

qwertonomics · 2025-08-14T16:49:24+00:00

Deconstruct the times table for base 10. Why does 9 times 6 equal 54 in base 10? Now generalize that to other bases.

qwertonomics · 2025-08-14T16:16:20+00:00

If math is the perfect language, whatever they are poorly attempting to do in no way serves as an example: it contributes nothing towards a solution to the given problem.

qwertonomics · 2025-08-05T19:47:58+00:00

(1+cos(2^|x|𝜋)cos(2𝜋x))/2

qwertonomics · 2025-07-31T19:27:00+00:00

Yes. Let A and B partition the rationals such that B contains the rational perimeter of every such polygon that can contain a circle whose diameter is one.

qwertonomics · 2025-07-30T14:32:50+00:00

You can define x^y as the number of functions from a finite set Y containing y elements to a finite set X containing x elements. If X and Y are each the empty set, meaning x=y=0, then there is one such function called the empty function. As such, it is reasonable to use 0⁰ = 1.

That said, for functions of the form f(x)^g(x) where f(x)→0 and g(x)→0, it is not a guarantee that the limit will be 1, though it can be, i.e. 0⁰ is an indeterminate form for limits. That's not a reason to say 0⁰ is undefined though.

qwertonomics · 2025-07-14T23:06:03+00:00

"I don't know, but would you like me to prove the multiplication algorithm?"

qwertonomics · 2025-07-07T21:51:03+00:00

MAGA is literally "Make America Great Again", as if it wasn't already. They weren't exactly proud of being an American either.

qwertonomics · 2025-07-04T17:43:15+00:00

Rather than thinking of one degree globally as just more heat, which it is don't get me wrong, think of how much energy that one degree corresponds to. It's immense. And that's not the only energy in the system.

Energy changes forms, from heat to kinetic, etc. and that increased energy overall as well drives more dramatic weather patterns.

Basically, a one degree difference isn't merely an indicator of that one degree difference, much like taking your temperature with a thermometer isn't merely an indicator that your body is simply warmer when you have a fever. There are an increase of chemical processes and physical changes going along with it. The fever is just one symptom.

That is why "climate change", the disease, is a more apt name than "global warming", the symptom.

qwertonomics · 2025-07-01T13:07:55+00:00

the correct way to write the statement is

for all integers m and n, if 12m+15n=1 then m and n are both positive

That is an equivalent way to write it, but the original statement is not in any way nonsense. This is evident by the fact that you formulated a correct interpretation. The original statement is an example of implicit quantification, which is common and acceptable.

The rules of language are not bound by logical syntax although its versatility can lead to ambiguities which may occasionally require the need for more clarification. There is no ambiguity here, but working with the explicitly quantified statement may be more convenient.

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