Welcome under Another Roof, where there is always another proof. Videos about mathematics, mathematical logic, and the history of mathematics. Did I mention I like mathematics?
So is it ironic that one of Sherlocks most quoted lines "Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth." is deductive reasoning.
Actually Abduction is very important un Science. As note by Peirce, Abduction plays a crucial rol in the formulation of Hypotesis. Cause de "most probable explanation" for an observation does not come derive logically from the premese, nor comes from multiple observation. It comes from the researcher prior experience, knowldege, creativity, intuition etc. Abduction in a nutshell is what prevents scientist the need to (imposible) test the infinte set of hypotesis thar could explain observations. Great vid!!
03:30 No. The problem is that you assume that propositional logic actually exists. Propositional logic is a metaphysical concept that cannot be proven and is paradoxical in itself. I am talking about the Münchhausen-Trilemma. The Münchhausen-Trilemma says that there is no real justification. There are only these three possibilities: - Circular reasoning - Infinite regress - Dogmatization Besides, Sherlock Holmes is a fictional character described by the author as using deduction in his head, and since we can't see inside his head and Holmes is entirely fictional, the question of whether Holmes uses deduction is moot anyway.
Just because Holmes does not present all the arguments to you to make it seem like deduction, does not mean he is not deducing. He can very easily be omitting what to him are obvious facts that led him to his conclusion.
I think a more accurate title would be, "Deduction doesn't mean what you think," but you needed something to bait people into clicking, so you used a somewhat disingenuous title. I think that's clickbait.
@@Selrisitai With that title, someone would complain "um, actually, I knew what deduction was so this title is clickbait." You can't please everybody and it's often subjective. I truly believe Holmes never makes a genuine deduction!
I'll never forget when I became a serial adulterer as an inexperienced teen because I have a habit of regularly removing before showers the ring gifted to me by my parents 😔
This is not what I expected from a video called "How to count". I'm so glad I watched it though, because I've been studying lambda calculus for awhile now and this filled in a few gaps on the pure math side of things, e.g., why is zero indexing in a for loop correct. I had an argument on the CS side but no equivalent argument on the pure math side, but you've provided it! It's because to count to five, one has to enumerate the elements starting from zero, not one.
Here is the proof of every statement found toward the end of the video: 1. Statement: x ∈ Z ⇒ ∃y ∈ Z [x = 2y ∨ x = 2y ± 1] Proof: Assume, on the contrary, x ∈ Z ∧ ∀y ∈ Z [x ≠ 2y ∧ x ≠ 2y ± 1] By Euclid’s Division Lemma, ∃! q, r ∈ Z [x = 2q + r ∧ |r| < 2] ⇒ −2 < r < 2 ∀y ∈ Z [x ≠ 2y] ⇒ r ≠ 0. ⇒ r ∈ {−1, 1} If r = 1, then ∃q ∈ Z [x = 2q + 1] But ∀y ∈ Z [x ≠ 2y ± 1] A contradiction! If r = −1, then ∃q ∈ Z [x = 2q − 1] But ∀y ∈ Z [x ≠ 2y ± 1] A contradiction! ¬ [x ∈ Z ∧ ∀y ∈ Z [x ≠ 2y ∧ x ≠ 2y ± 1]] ⇒ x ∈ Z ⇒ ∃y ∈ Z [x = 2y ∨ x = 2y ± 1] Q.E.D. 2. Statement: ∀x ∈ R [x > 0 ∨ x < 0] Counter example: Since 0 ∈ R, when x = 0, ¬ (x > 0 ∨ x < 0) ⇒ ∃x ∈ R [¬ (x > 0 ∨ x < 0)] ⇒ ¬∀x ∈ R [x > 0 ∨ x < 0] Q.E.D. 3. Statement: ∀x ∈ R [ ¬ (x > 0) ⇒ x < 0] Counterexample: Since 0 ∈ R, when x = 0, ¬ (x > 0) ∧ ¬ (x < 0) ⇒ ∃x ∈ R [¬ (x > 0) ∧ ¬ (x < 0)] ⇒ ¬∀x ∈ R [ ¬ (x > 0) ⇒ x < 0] Q.E.D. 4. Statement: ∃(x ≠ 1) ∈ N [∀y, z ∈ N [ x = yz ⇒ (y = 1 ∨ z = 1)] ∧ ∃a ∈ Z [x = 2a]] Proof: Let x = 2. Assume, on the contrary, that ∃y, z ∈ N [2 = yz ∧ ¬ (y = 1 ∨ z = 1)] By De Morgan's Laws, ∃y, z ∈ N [2 = yz ∧ (y ≠ 1 ∧ z ≠ 1)] ⇒ y ≥ 2 ∧ z ≥ 2 ⇒ yz ≥ 2z > 2, by the multiplication axiom of inequality. ⇒ yz > 2 But 2 = yz. A contradiction! ⇒ ¬ ∃y, z ∈ N [2 = yz ∧ (y ≠ 1 ∧ z ≠ 1)] ⇒ ∀y, z ∈ N [ 2 = yz ⇒ (y = 1 ∨ z = 1)] By the multiplicative identity axiom, 2 = 2 × 1. Let a = 1. ⇒ 2 = 2a ⇒ ∃a ∈ Z [x = 2a] Therefore, ∃(x ≠ 1) ∈ N [∀y, z ∈ N [ x = yz ⇒ (y = 1 ∨ z = 1)] ∧ ∃a ∈ Z [x = 2a]] Q.E.D. 5. Statement: ∃x ∈ R, ∀y ∈ R [x + y = 0] Counterexample: Consider an arbitrary x ∈ R. Let y = |x| + 1. Then, x + y = x + |x| + 1. x < 0 ⇒ x + |x| + 1 = 1 > 0 x = 0 ⇒ x + |x| + 1 = 1 > 0 x > 0 ⇒ x + |x| + 1 = 2x + 1 > 0 ⇒ ∀x ∈ R, ∃y ∈ R [x + y ≠ 0] ⇒ ¬ ∃x ∈ R, ∀y ∈ R [x + y = 0] Q.E.D. 6. Statement: ∀x ∈ R, ∃y ∈ R [x + y = 0] This is the additive inverse axiom, one of the field axioms, and is therefore, true. 7. Statement: ∀x, y ∈ R [(x ≠ y ∧ x^2 = y^2) ⇒ x = −y] Proof: Assume ∃x, y ∈ R [x ≠ y ∧ x^2 = y^2 ∧ x ≠ −y] Case 1: |x| < |y| |x||x| < |x||y| ⇒ |x||x| = |x.x| = |x^2| = x^2 < |x||y|, for x^2 is non-negative. Also, |x||y| < |y||y| ⇒ |x||y| < |y||y| = |y.y| = |y^2| = y^2, for y^2 is non-negative. Thus, x^2 < y^2. But x^2 = y^2 (Contradiction!) Case 2: |x| > |y| |x||x| > |x||y| ⇒ x^2 > |x||y| Also, |x||y| > |y||y| ⇒ |x||y| > y^2 Thus, x^2 > y^2. But x^2 = y^2. (Contradiction!) Therefore, ¬ ∃x, y ∈ R [x ≠ y ∧ x^2 = y^2 ∧ x ≠ −y] Thus, ∀x, y ∈ R [(x ≠ y ∧ x^2 = y^2) ⇒ x = −y] Q.E.D. 8. Statement: ∀x, y ∈ R [(x ∈ Q ∧ y ∈ Q) ⟺ (x + y) ∈ Q] Counterexample: Let x = √3, and y = -√3 ⇒ x + y = 0 ∈ Q ⇒ ∃x, y ∈ R [(x + y) ∈ Q ∧ (x ∉ Q ∨ y ∉ Q)] ⇒ ¬ ∀x, y ∈ R [(x + y) ∈ Q ⇒ (x ∈ Q ∧ y ∈ Q)] ⇒ ¬ ∀x, y ∈ R [(x ∈ Q ∧ y ∈ Q) ⟺ (x + y) ∈ Q] Q.E.D. 9. Statement: ∀x, y ∈ R [x < y ⇒ ∃z ∈ R [x < z < y]] Proof: Consider arbitrary x, y ∈ R for which x < y. ⇒ x + y < 2y, by the addition property of inequality. Also, 2x < x + y. ⇒ 2x < x + y < 2y ⇒ x < (x + y)/2 < y, by the division property of inequality. Let (x + y)/2 = z ∈ R, by the closure axiom of addition and division within real numbers, where 2 ≠ 0. We have, ⇒ x < z < y Therefore, ∀x, y ∈ R [x < y ⇒ ∃z ∈ R [x < z < y]] Q.E.D. 10. Statement: ∃!x ∈ R, ∀y ∈ R [y > 0 ⇒ x^2 < y] Proof: Let x = 0. Consider an arbitrary y ∈ R for which y > 0. ⟺ 0^2 = 0 < y. ⇒ ∃x ∈ R, ∀y ∈ R [y > 0 ⇒ x^2 < y] Proposition: ∀(z ≠ 0) ∈ R, ∃y ∈ R [y > 0 ∧ z^2 ≥ y] Proof: Consider an arbitrary (z ≠ 0) ∈ R. ⇒ z^2 > 0, by the multiplication property of inequality. Let y = z^2. z^2 = z^2, by the reflexive axiom of equality. ⇒ z^2 ≥ y This concludes the proof of the proposition. ⇒ ¬ ∃(z ≠ 0) ∈ R, ∀y ∈ R [y > 0 ⇒ z^2 < y] ⇒ ∃!x ∈ R, ∀y ∈ R [y > 0 ⇒ x^2 < y] Q.E.D. Looking forward to your thoughts and questions. We would be happy to discuss! :)
LOL 22:24 I'm trans so I obvi used my current name and it matched imho pretty correctly so i wondered what if i use my old names, and guess what, it doesn't match at all
24:52 Yes! A fellow scholar. House > Guy Ritchie (also very enjoyable), and the Cumberbatch one (while I like him) was not very good, and then became very bad
Kant said that syllogisms don’t increase knowledge, but math increases knowledge and works on deductive reasoning. What made Kant (and many others like Aristotle ) say that?
I have a story of how someone got totally wrong conclusions because they were a Sherlock fan. Someone came up to me on a bus, said they liked Sherlock, and asked if they could check their powers of observation by drawing conclusions about me based on my appearance. I liked Sherlock too so I agreed. He said I have a cat and a dog because I had two types of pet hair on my skirt. I had two cats. He made a correct observation but drew the wrong conclusion. He then said my boots were new because they were clean. They weren’t new (or particularly clean) but I do take care of my clothes. He again drew the wrong conclusion. Unfortunately I can’t remember if he said anything else but I do remember he didn’t get anything about me right.
I detest (hyperbole) how much this reminded me of the proofs unit in my calculus class. Edit: you used to be a math teacher, that makes a lot of sense now
I think it's a little weird to present unique existence as a separate form of quantification when it it merely shorthand for ∃x. (P(x) ∧ ∀y. (P(y) => y = x)). For the alternative propositions, 1 and 2 are not standalone propositions, but rather predicates (since they contain free variables), and I disagree with the implicitly presented convention that free variables should be treated as universally quantified. If you want universal quantification, write universal quantification!
about people who take their rings off: my friend is a baker, who's married and until I see evidence to the contrary, faithful to his wife. He never wears his ring, he's not even allowed to. There are more professions like this
You know, I accidentally insulted my philosophy professor in my first class by challenging an argument as contrary to “basic math” - he agreed that it wasn’t valid because the logic didn’t work, but insisted it wasn’t math 😏
Talk about logical reasoning. In my country there's a test before enroling in university, we will be given syllogism and what-not. I used to be good at the "pattern" of the test and it made me feel big brain. Now I forgot it entirely and it makes me feel like lvl 1 crook.
Oh, wow. Such a source of frustration to me. Thanks for explaining that the reason I never really understood what “deductive” means is because people almost always misuse it. I think the true meanings do matter because (in the US)… well, if people use these words at all, it’s usually some sort of power play. Like, they don’t remember anything except that deduction is supposedly the “best” type of reasoning, so they say something is deductive to imply that it’s unassailable, not that it’s sound. Whether or not I say it out loud, it’s going to be a huge help for me to be able to say, “Aha, but what you’re talking about isn’t actually deduction!”
21:05 my solution is simpler to me, just aiming to hit 450 with the 25 means I need an 18. So 5+4+9 is 18, multiply 25, 450. Then I have a 7 and 6 left to make the final 1, which is easy. I'm assuming your code stops as soon as it hits a valid solution, and isn't looking for the most human-simple one!
17:19 that may be intuitive but that not how a real Mexican wave works, see all the people doing the wave meet at jon's house every last fan doing the wave, they order them self then they match up their clocks work out the timing to wave where the first person waves on seconds ending in 1 then the next person waves on 2 so on... and it only looks like they are looking to the right when they are really acting all independently by mod of the clock you could do it with a finite amount of people such as the amount of digits in the base doing the wave but everyone was going to jon's house anyway. my point being their might be different behavior with different bases because it's true for the mod of a 12 hour clock it might not be true for a 10 hour clock. take 1/3 can you count on it repeating 0.3 forever in all bases?
You said that this is also how negative numbers are defined but I can't actually wrap my head around the negative numbers defined as sets: haw can you remove something from an empty set???
“At the foundation of mathematics we have something strong and secure “ but the way I see it (as a non mathematician) is that this foundation is built of nothingness encompassed and systemized by language. Pretty cool, nothing more sturdy than nothingness! 😂
This reminds me a little bit about the sixes challenge. The ideia is "can you make 6 by using only 3 copies of an integer?", like 2+2+2 = 6 or 3*3-3 = 6, can you do that for numbers 0 to 9? Like: 0 0 0 = 6 1 1 1 = 6 ... 9 9 9 = 6 In this case in particular you do need to add factorial to solve the first two, but I believe the rest can be solved by using only the 4 basic operations, parentheses and square root. Of course, then you start to ask questions like, how far can you go, what numbers other than 6 can you do, what if you use more or fewer numbers, what if you use different numbers, what if you allow for different operations, etc. The show basically presents the same concept, but it changes how many numbers you have, what those numbers can be, what number you need to make, how many numbers you need to use and what operations you can use.
A very rudimentary theorem prover is an accessible project for a computer science undergraduate without any LLM or neural net tech. Any axiom or theorem can be translated to disjunctive normal form, and any set of such statements can be trivially brute-force churned until a proof-by-contradiction is reached. The hard part is making it _good_ and pruning the junk out of the proofs.
As a lawyer, I just realized that the whole justice system is based on abduction. When someone is convicted "beyond reasonable doubt", it's just an abductive argument.
Book Sherlock was a lot smarter when compared to his other adaptations because Doyle TOOK THE EFFORT TO SHOW THE READER HE IS; something most writers don't bother doing because to them the book was just something to reference, and so they ignore the nuances of Sherlock's investigative work. Also he was kind and polite, if not a bit eccentric to people he works with, the victims, and even regular people unrelated to te crime, unlike the incorrigible, pretentious assfucker that was BBC's (and even Ritchie's) Sherlock.
I watch a shitton of videos on "debunking" many different things: creationism / christianis apologetics, flat earth and other conspiracy theories. And especially the apologetics type people can't seem to grasp the concept, that one might disagree with their premises. But even privately (when casually arguing with friends or family myself) I find it extremely frustrating that people often don't understand that their premises are VERY flawed. No matter what you tell them, they keep pushing that conclusion and expect you to disprove their conclusion and ONLY the conclusion and they don't understand that the conclusion is worth NOTHING without true premises. I want to say: I agree that it's a real problem in the real world that people think they are using deductive reasoning when they really aren't.
