Oh my bad, the condition for the numbers to work is them having a difference of 1. Calling numbers with difference of 1 consecutive was by a force of habit. They're not actually consecutive . Anyways thank you for pointing that out ! I would think the problem works on Z. For positive and negative integers when using consecutive.
@@samylol99 Yes, it works also for negative integers because the product of two negatives (the smallest and the largest) give a positive number and the square of the middle negative is always positive. So that proof works for any three numbers that are consecutive and are in the Z set.
@filippocontiberas Exactly. And the only possibility that involves a negative times a positive also happens to have 0 in the middle so when we substarct by 1, it also becomes negative. Take the consecutive integers : -1 , 0 , 1 1 × - 1 = 0² - 1 = -1
@@samylol99 yes, because that -1 is in the point (0, -1) which is the minimum of the function f(x) = x^2 -1 for both discrete and continuous version of f(x).
I'd solve level 2 like this: If the set {1,2,...,2n-1,2n} contains a subset with n+1 elements and we must choose the largest sum of two elements of that subset, it is forced that we add the largest two elements of the subset. We ask: Is it possible to construct a subset of {1,2,...,2n-1,n} such that the largest two elements don't sum to at least 2n+1? Logically, the subset with the smallest largest elements is the set {1,2,...,n, n+1}. Adding the two largest elements gives 2n+1. This is our solution. QED.
Neat solve, It just occurred to me but your proof is not rigorous enough, you also need to take into account the possibility that the smallest terms of the greatest subset possible don't sum up to 2n+1, so you have to add this step to your proof : We ask is it possible that we pick a subset with elements so big that their pairs always sum up to more than 2n+1 ? To answer that we'll take the greatest subset which is logically {n,n+1,.............,2n-1,2n} and we sum the smallest elements in it: n + n + 1 = 2n+1. Therefore when picking any subset from {1,2,.......2n} there is no way that all the sums in these subsets are greater than or smaller than 2n+1 which means that there exists a pair of numbers that sum up to 2n+1 in any of these subsets. Just in case you don't get why your proof is not rigorous enough it's because it applies perfectly with a set such as {1,2.....2n,2n+1} when picking n+1 elements to sum up to 2n+1 and it shouldn't since every pair of elements in the greatest subset of this set for instance sum up to more than 2n+1.
very interesting video! you should definitely have more subscribers for the quality content you make. Just try next time to make "an" more clear with subscripts, but i believe someone already pointed this out. Anyways keep up the good work!!
Good video! Several inaccuracies: 1. natural numbers are positive integers, no need to add "non-zero" (L5). 2. "positive non-zero integer" (L2), again no need for non-zero. 3. "3 consecutive real numbers" (L1) should be "3 consecutive integers", plus there is consecutive real numbers because they are "dense" (infinitely more reals between any two reals).
Alright thanks for letting me know! I was already told about 3. That was by force of habbit to call numbers with difference of 1 consecutive, sorry. As for nonzero I felt I needed to mention it because I already stumbled upon domains like ℕ and ℕ⁺ when making math studies.
I'd argue that strong induction is easier than normal induction. The reason being is by increasing the restriction on our requirements, we are also narrowing the focus of our solutions. It's a bit counterintuitive at first, but you have "stronger" evidence with more restriction. I suppose it is preference and takes experience.
I see, that's an interesting take and I agree with you. But I can very easily see a beginner failing to note down all restrictions and making mistakes. But the upside of being able to use all the terms before n while narrowing requirements is really good. (Also the video wouldn't make sense if I started with strong induction then induction)
@@jorianweststrate2580 I think it does, the 5th level is seemingly more challenging to prove with normal induction. But it's true that these occurrences are rare but they can't be ignored.
@@samylol99 you're right in that it might be easier to prove some things using strong induction. However, because strong induction implies weak induction, a proof written using strong induction can always be transformed into a proof using weak induction.
[level 3] For the average person I meeting, I (would) explain/ing, that if sqrt(2) = p/q, therewhere 2 = p^2 / q^2 . We do can find for q being 1 or -1, but there is no p as an interger. Q.E.D.
One thing i got a bit confused with in level 1 is when you said consecutive "real" numbers rather than integars or whole numbers. I honestly thought it was like a trick question since consecutive real numbers technically aren't possible
Yeah I've wrote "from 3blue1brown" The problem caught my interest yet there would have been no point for me to make a video on it, since whatever I end up making will be redundant and objectively less good than 3Blue 1Brown's video as they would treat the same theme.
Computing went with the leading ! ("bang") character as the logical negation. ~ (tilde) is the symbol for approximation. As a 40+ year computing guy (and a chem engineer) I gotta say I prefer "bang".
Yeah. There exists many ways to denote not p Such as ¬p, !p,p̄, and I would like to mention that ~p definitely means not p. (You can easily find it when researching about logical negation) But I suppose it could be confused with approximately p. ≈ is what I use for it.
Construction is for existence quantors Pigeonhole is for existence quantors (actually it's a specific case of contradiction, because you prove that non-existence isn't possible) Contradiction/Contraposition are for negations Induction is for all quantors I think you forgot case distinction for disjunctions (or-statements)
I didn't include proof by cases because (in my opinion) it falls into direct proofs and it's too intuitive. Basically you split cases and you construct according to conditions you've split from the original statement.
@@samylol99 I personally would reserve the term "direct proof" for just glueing implications together to bigger implications. Cases have different logical structure. Imo that's overlooked too much. Proof by cases is very important in discrete mathematics.
I see. I don't doubt at all the importance of proof by cases. But could you tell me how different their structures are from construction proofs? I was always under the impression that proof by cases was a repeated construction proof with different subsets or possibilites of the entire statement you're asked to prove, I would like to be corrected if I'm wrong.
@@samylol99 yea latex is the most powerful tool for this but also you can just use an autohotkey script that uses unicode math symbols called PlainTeX made by someone on the r/math subreddit, you can type a_n^2 and it will appear as aₙ² although it's limited and currently doesnt have +, -, (, ) symbols in the subscripts part but you could just add these lines to the script yourself so you could write stuff like a₃₍ₙ₋₂₎₊ₖ as a_3_(_n_+_2_)_+_k or ℤ⁺ as \mathbb{Z}^+ :*?C:^-::⁻ :*?C:^+::⁺ :*?C:^(::⁽ :*?C:^)::⁾ :*?C:_-::₋ :*?C:_+::₊ :*?C:_(::₍ :*?C:_)::₎ it tries to emulate latex syntax even tho it's limited but maybe it's enough for you, or maybe not but either way latex is good to learn regardless for typing math
I found a simpler way to prove the first one. **Proof:** Let ( n-1 ), ( n ), and ( n+1 ) be three consecutive real numbers. Let ( n^2 - 1 ) be the number preceding the square of the second number. We can factor ( n^2 - 1 ) as follows: (difference of squares) n^2 - 1 = (n+1)(n-1) Therefore, ( n-1 ) and ( n+1 ) are factors of ( n^2 - 1 ).
Well, I understand your frustration but no. I can't voice reveal nor do I want to. And as of now, I don't have any other alternatives, I'm sorry for that.
1:19 - not 3 consecutive real numbers, but natural number. Is impossible to get 3 consecutive real numbers because that set is dense.
Oh my bad, the condition for the numbers to work is them having a difference of 1. Calling numbers with difference of 1 consecutive was by a force of habit. They're not actually consecutive . Anyways thank you for pointing that out ! I would think the problem works on Z. For positive and negative integers when using consecutive.
@@samylol99 Yes, it works also for negative integers because the product of two negatives (the smallest and the largest) give a positive number and the square of the middle negative is always positive. So that proof works for any three numbers that are consecutive and are in the Z set.
@filippocontiberas Exactly. And the only possibility that involves a negative times a positive also happens to have 0 in the middle so when we substarct by 1, it also becomes negative. Take the consecutive integers : -1 , 0 , 1
1 × - 1 = 0² - 1 = -1
@@samylol99 yes, because that -1 is in the point (0, -1) which is the minimum of the function f(x) = x^2 -1 for both discrete and continuous version of f(x).
amazing video, would love to watch all mathematical proofs techniques or principles with an example.
I'd solve level 2 like this:
If the set {1,2,...,2n-1,2n} contains a subset with n+1 elements and we must choose the largest sum of two elements of that subset, it is forced that we add the largest two elements of the subset.
We ask: Is it possible to construct a subset of {1,2,...,2n-1,n} such that the largest two elements don't sum to at least 2n+1?
Logically, the subset with the smallest largest elements is the set {1,2,...,n, n+1}. Adding the two largest elements gives 2n+1. This is our solution.
QED.
Neat solve,
It just occurred to me but your proof is not rigorous enough, you also need to take into account the possibility that the smallest terms of the greatest subset possible don't sum up to 2n+1, so you have to add this step to your proof :
We ask is it possible that we pick a subset with elements so big that their pairs always sum up to more than 2n+1 ? To answer that we'll take the greatest subset which is logically {n,n+1,.............,2n-1,2n} and we sum the smallest elements in it: n + n + 1 = 2n+1.
Therefore when picking any subset from {1,2,.......2n} there is no way that all the sums in these subsets are greater than or smaller than 2n+1 which means that there exists a pair of numbers that sum up to 2n+1 in any of these subsets.
Just in case you don't get why your proof is not rigorous enough it's because it applies perfectly with a set such as {1,2.....2n,2n+1} when picking n+1 elements to sum up to 2n+1 and it shouldn't since every pair of elements in the greatest subset of this set for instance sum up to more than 2n+1.
very interesting video! you should definitely have more subscribers for the quality content you make. Just try next time to make "an" more clear with subscripts, but i believe someone already pointed this out. Anyways keep up the good work!!
Thank you for the encouragement ^^
I will definitely work on that.
Good video! Several inaccuracies: 1. natural numbers are positive integers, no need to add "non-zero" (L5). 2. "positive non-zero integer" (L2), again no need for non-zero. 3. "3 consecutive real numbers" (L1) should be "3 consecutive integers", plus there is consecutive real numbers because they are "dense" (infinitely more reals between any two reals).
Alright thanks for letting me know!
I was already told about 3. That was by force of habbit to call numbers with difference of 1 consecutive, sorry. As for nonzero I felt I needed to mention it because I already stumbled upon domains like ℕ and ℕ⁺ when making math studies.
I'd argue that strong induction is easier than normal induction. The reason being is by increasing the restriction on our requirements, we are also narrowing the focus of our solutions. It's a bit counterintuitive at first, but you have "stronger" evidence with more restriction. I suppose it is preference and takes experience.
I see, that's an interesting take and I agree with you. But I can very easily see a beginner failing to note down all restrictions and making mistakes.
But the upside of being able to use all the terms before n while narrowing requirements is really good.
(Also the video wouldn't make sense if I started with strong induction then induction)
Strong induction and normal induction are equivalent anyway, it doesn't really matter that much which one you use
@@jorianweststrate2580 I think it does, the 5th level is seemingly more challenging to prove with normal induction. But it's true that these occurrences are rare but they can't be ignored.
@@samylol99 you're right in that it might be easier to prove some things using strong induction. However, because strong induction implies weak induction, a proof written using strong induction can always be transformed into a proof using weak induction.
[level 3]
For the average person I meeting, I (would) explain/ing, that if sqrt(2) = p/q,
therewhere 2 = p^2 / q^2 .
We do can find for q being 1 or -1, but there is no p as an interger.
Q.E.D.
Nice, but not rigours enough. Why is this integer p nonexistent ?
I do have an answer to make it more rigours, let me know if you're interested.
@@samylol99 Between 1 and 2 , there are no other natural numbers.
So what?
@@samylol99 (that's also an axiom)
One thing i got a bit confused with in level 1 is when you said consecutive "real" numbers rather than integars or whole numbers. I honestly thought it was like a trick question since consecutive real numbers technically aren't possible
Silly habit sorry for confusing you
@@samylol99 no it's alright, just wanted to check about it. Loved the rest of the video
(0:43) Recommendation to _Moser's circle problem_ is the video from _3Blue1Brown_
Yeah I've wrote "from 3blue1brown"
The problem caught my interest yet there would have been no point for me to make a video on it, since whatever I end up making will be redundant and objectively less good than 3Blue 1Brown's video as they would treat the same theme.
Computing went with the leading ! ("bang") character as the logical negation. ~ (tilde) is the symbol for approximation. As a 40+ year computing guy (and a chem engineer) I gotta say I prefer "bang".
Yeah. There exists many ways to denote not p
Such as ¬p, !p,p̄, and I would like to mention that ~p definitely means not p. (You can easily find it when researching about logical negation)
But I suppose it could be confused with approximately p. ≈ is what I use for it.
fantastic video!
Construction is for existence quantors
Pigeonhole is for existence quantors (actually it's a specific case of contradiction, because you prove that non-existence isn't possible)
Contradiction/Contraposition are for negations
Induction is for all quantors
I think you forgot case distinction for disjunctions (or-statements)
I didn't include proof by cases because (in my opinion) it falls into direct proofs and it's too intuitive.
Basically you split cases and you construct according to conditions you've split from the original statement.
@@samylol99 I personally would reserve the term "direct proof" for just glueing implications together to bigger implications. Cases have different logical structure. Imo that's overlooked too much. Proof by cases is very important in discrete mathematics.
I see. I don't doubt at all the importance of proof by cases. But could you tell me how different their structures are from construction proofs? I was always under the impression that proof by cases was a repeated construction proof with different subsets or possibilites of the entire statement you're asked to prove, I would like to be corrected if I'm wrong.
cool video but maybe some stuff were hard to make out at a glance due to lack of typesetting for subscripts
I see, thank you for the feedback. I will improve on that for future videos. I'm already starting to learn how to use LaTeX.
@@samylol99 yea latex is the most powerful tool for this but also you can just use an autohotkey script that uses unicode math symbols called PlainTeX made by someone on the r/math subreddit, you can type a_n^2 and it will appear as aₙ² although it's limited and currently doesnt have +, -, (, ) symbols in the subscripts part but you could just add these lines to the script yourself so you could write stuff like a₃₍ₙ₋₂₎₊ₖ as a_3_(_n_+_2_)_+_k or ℤ⁺ as \mathbb{Z}^+
:*?C:^-::⁻
:*?C:^+::⁺
:*?C:^(::⁽
:*?C:^)::⁾
:*?C:_-::₋
:*?C:_+::₊
:*?C:_(::₍
:*?C:_)::₎
it tries to emulate latex syntax even tho it's limited but maybe it's enough for you, or maybe not but either way latex is good to learn regardless for typing math
Thank you so much for the suggestion I really appreciate this. I didn't know of PlainTeX I'm gonna make some research on it and see if I can use it
I like how straightforward you about the proofs
I found a simpler way to prove the first one.
**Proof:**
Let ( n-1 ), ( n ), and ( n+1 ) be three consecutive real numbers.
Let ( n^2 - 1 ) be the number preceding the square of the second number.
We can factor ( n^2 - 1 ) as follows: (difference of squares)
n^2 - 1 = (n+1)(n-1)
Therefore, ( n-1 ) and ( n+1 ) are factors of ( n^2 - 1 ).
Yes ! This is basically what I did just in a much slower manner for beginners to grasp it. Well done ^^
Use a real voice. I can't stand the AI voice. Good content, bad voice.
Well, I understand your frustration but no. I can't voice reveal nor do I want to. And as of now, I don't have any other alternatives, I'm sorry for that.