14:30 I guess here we should write that "there are 2 truth-tellers" contradicts the combination of R and "β is a liar". It is not contradicting R by itself. 15:28 "there are 3 thruth-tellers" is not at all contradicting ¬P ("there is at least one truth-teller"). It is contradicting the statement "β is a liar" (as well as the previously established statement R that α is a liar).
The discussion forced Q's perspective onto R & assumes that Q would not trap themselves in a lie when Q may in fact say many other distinctly deceptive things that may be falsely assumed as true around our aloof figure R
I deeply enjoy these videos & learning together, I must say that the equation cannot be solved from Q's oblique obfuscating statement. Parallel conditions presented by Q are assumption. If Q wished to not be deceptive, but forthright & truthful Q would have made a succinct reference to P & be known as true. Q may say a number of oblique statements so what Q said is conditional truth with aloof R. If Star Trek taught me anything it's that Q, oh Q must say something distinct for who knows what people say when we also cannot attest to truth
because we already have enough evidence on our hands to conclude that there's one and only one truth-teller, and that is beta or the second person. So we don't need gamma to say anything for us to decide which tribe they belong to, considering we've already made a decision on who's who. Hope this helped :).
@@tomreingold4024 In the real world we wouldn't know what γ would be. But in this hypothetical scenario, we know that people MUST be either truth-tellers or liars. It is a binary scenario, there is no in-between. Since β is the one and only truth-teller, that means that γ is a liar.
Teacher Eddie: in the end you said 3 truth tellers contradicts notP statement. Thats not true, the right conclusion is 3 truth tellers contradicts statement R
The first native was a liar, the second was a truth-teller, and the third was a liar. If the first native was telling the truth, then there is a problem with his statement, "we are all liars." So, he must be a liar. That means that his statement is false, and there must be at least one person telling the truth. If the second native is a liar, then either there are two people telling the truth or none. But we know that they can't all be liars, and there can't be two truth-tellers since the first one is known to be a liar. Hence, the second native is telling the truth, and that means he is the only truth-teller. So the third native is a liar (even though he didn't speak).
A quick tip: if you hold down when you are drawing a line, circle etc, Notability automatically converts it to a perfect line, circle etc
He's used that handy feature many times in videos where geometric precision was more relevant to the lecture.
These classes setups are going amazing.
Very nice introduction. Only thing to mention: 3 truth-tellers doesn't contradict "not P" but the fact that alpha is a liar ;-)
And that beta is a liar. Doesn't make much of a difference in this example
Eddie your lessons are fantastic.
Watching Eddie's lessons makes me think of Lockhart's Lament. I think Eddie manages it.
I am remarkably envious: this handwriting is pretty ^-^
Getting things split increases excitement thanks eddie for the idea
i am just thankful that i attempted the hsc in '71; 1st level maths did not include implications in the syllabus, mark
I love this channel, can someone suggest more channels like these.. of any country..
3 truth-tellers doesn't contradict ¬P, it contradicts R. (It also contradicts ¬Q.)
3 truth-tellers is an implication of Beta being a liar =P
@@auronwintermoon6064 Yes, I know. What I'm saying is that "at least one truth-teller" and "3 truth tellers" are not contradictory.
It doesn’t contradict ¬Q. (Or ¬P) only R. Well, I suppose you could say it contradicts ¬P because ¬P implies R.
@@jeremypnet True, it doesn't literally contradict the content of ¬Q. What I meant was it contradicts the other implication of ¬Q, that B is a liar.
@@mxlexrd I didn't mean to correct you, quite the opposite. You're right and I wanted to clarify it further. I'm sorry if I've misslead you.
Poor Gamma!!!
Without saying anything, he is a liar!
😂
Amazing eddie!!!!!!!!!!!!!!!!!!!!!!!! i wish i got u as my math teacher :(
*Yes I was right* ☺️
Thanks 😊
What does he use to write on for technology
Notability app on iPad
14:30 I guess here we should write that "there are 2 truth-tellers" contradicts the combination of R and "β is a liar". It is not contradicting R by itself.
15:28 "there are 3 thruth-tellers" is not at all contradicting ¬P ("there is at least one truth-teller"). It is contradicting the statement "β is a liar" (as well as the previously established statement R that α is a liar).
The discussion forced Q's perspective onto R & assumes that Q would not trap themselves in a lie when Q may in fact say many other distinctly deceptive things that may be falsely assumed as true around our aloof figure R
Amazing style of online lessons! Can I copy this style and make my own videos on TH-cam, but in Russian?
yes. well you need a digital pen
Thank you👍
Hello...Eddie what's the difference between between a statement is assumed and as statement is asserted?
Hi. I don't get why false on Q cannot mean 2 truth tellers.
It's because If alpha is a liar and beta is a liar, then there can be maximally 1 truth teller
Sir make videos on abstract algebra.
I deeply enjoy these videos & learning together, I must say that the equation cannot be solved from Q's oblique obfuscating statement. Parallel conditions presented by Q are assumption. If Q wished to not be deceptive, but forthright & truthful Q would have made a succinct reference to P & be known as true.
Q may say a number of oblique statements so what Q said is conditional truth with aloof R.
If Star Trek taught me anything it's that Q, oh Q must say something distinct for who knows what people say when we also cannot attest to truth
I like the way you talk. Are you perhaps a law student or in the said profession?
on the board: solution to the reimann hypothesis
How can Gamma be a liar if they didn’t say anything?
because we already have enough evidence on our hands to conclude that there's one and only one truth-teller, and that is beta or the second person.
So we don't need gamma to say anything for us to decide which tribe they belong to, considering we've already made a decision on who's who.
Hope this helped :).
Ahmad, I still don’t follow. Holding silence is not a lie in this case.
@@tomreingold4024 In the real world we wouldn't know what γ would be. But in this hypothetical scenario, we know that people MUST be either truth-tellers or liars. It is a binary scenario, there is no in-between. Since β is the one and only truth-teller, that means that γ is a liar.
@@tomreingold4024 Have you watched part 2?
@Function Overload, thank you.
Teacher Eddie: in the end you said 3 truth tellers contradicts notP statement. Thats not true, the right conclusion is 3 truth tellers contradicts statement R
The first native was a liar, the second was a truth-teller, and the third was a liar.
If the first native was telling the truth, then there is a problem with his statement, "we are all liars." So, he must be a liar. That means that his statement is false, and there must be at least one person telling the truth.
If the second native is a liar, then either there are two people telling the truth or none. But we know that they can't all be liars, and there can't be two truth-tellers since the first one is known to be a liar.
Hence, the second native is telling the truth, and that means he is the only truth-teller.
So the third native is a liar (even though he didn't speak).