back in school I hated math... you, sir, among others, reignited my interest in the subject by explaining things so well and clearly that it's really easy to follow and I'm not even an native English speaker. you really have a way with words. Keep up the good work. I'm sure you are inspiring lots of people out there. Thank you. I really mean it.
Eddie's proof contains several logical errors (negating the given claim wrongly; using Part(a) wrongly by mistaking P⇒Q for its converse Q⇒P; most egregiously, proving the *weaker* claim "....must have at least one factor..." instead of the given, *stronger* claim "....must have at least one _prime_ factor..."). Here's a more rigorous presentation: [ Let p be a composite number, suppose that it is of the form 4m-1, and *_assume_* that it has no prime factor of the form 4n-1. Now, p, being of the form 4m-1, is odd, so its prime factorisation contains only odd numbers, so either it has a prime factor of the form 4n-1 or, by Part (a), its prime factors are all of the form 4n+1. The latter, by Part (b), is false; thus, p must have a prime factor of the form 4n-1. This *contradicts* our assumption; therefore, p _has_ a prime factor of the form 4n-1. Hence, every composite number of the form 4m-1 has a prime factor of the form 4n-1. ] ----------- PS. For reference, this is my previous proof, which is invalid because ∀n[P(n) or Q(n)] of course doesn't imply [∀nP(n) or ∀nQ(n)] (the converse is true though; thanks to Oisin Carroll for highlighting this crucial error): [ Let m be a positive integer, suppose that 4m-1 is composite, and *assume* that for each n, 4n-1 is non-prime or not a factor of 4m-1. Now, 4m-1 is odd, so must be a product of two odd numbers, so either has some factor 4k-1 or, by Part (a), equals (4p+1)(4q+1) for some p & q. The latter, by Part (b), is false; thus, some some 4k-1 must be a factor of 4m-1, i.e., it is false that for each k, 4k-1 is not a factor of 4m-1. Therefore-continuing from the first paragraph-for each n, 4n-1 is non-prime; in particular, 4(1)-1=3 is non-prime; this is a *contradiction*; therefore, our assumption must be false; i.e., for some n, 4n-1 is a prime factor of 4m-1. Hence, for each positive integer m, if 4m-1 is composite, then there is some n for which 4n-1 is its prime factor, as required.]
I believe the step in the first paragraph of your proof is incorrect. You can't redefine forall. n and have the inequality hold. It's equilivent to saying... forall n. n is even OR n is odd => (forall n. n is even) OR (forall n. n is odd) You can do a similar rearrangement keeping a single forall n. up front but of course it breaks your contradiction at the end.
@@ryang628 Yes I was talking about line B. [q OR (p AND r)] doesn't follow from the line above. For e.g. q=true, p,r=false satisfy that but not the previous. There's a similar assignment of variables that work for the first and not second.
@@oisincarroll1171 Line(B) is *not* intended to be equivalent to Line(A); it just needs to be a *consequence* of Line(A). Here, P=false is not a possible assignment, because the proof-by-contradiction argument is based on the premise that THERE DOES EXIST a composite number 4m-1), i.e., that P=true! I've just condensed the proof (it remains equivalent to the original one) to make it easier to parse; perhaps give that a read?
@@ryang628 I admit I haven't heard the term consequence used in logic, but I understood it to mean line A implies line B, which it does not. This is because there is an assignment that satisfies A but not B - giving True implies False, which is false. Edit: whoops, no, you're right. Had the logic backwards here.
Can you do truth trees? I like them as an effective means of solving knights and knaves and other propositional logic problems. Vry good tools for discrete mathematics.
What if 4n-1 has a factor of the form 4m-1, but that factor is a composite number? Since the question says that it needs to have a prime factor in the form 4m-1, should we do extra steps to show that that number is prime as well?
its day 45 after this post, my hubby left me, the house is gone, I barely have access to these videos anymore. Im on the fucking streets cuuuunnntt. Im on my last gigabyte of data cuzz. This is my goodbye eddie woo. BTW Nick Hatziliradis wants to fight you
@@geza9580 meet me at broady station bruhhh, Ill sell you 1 extra gig of data for a bag lad. You need you ataday lad? meet me at the station now habibi shuu
I've really enjoyed these, I'm not sure I agree with some logic here. k could be a product of the form (4n-1)(4n+1) which would still satisfy the negation as long as 4n-1 is not prime (i.e. is composite). The full proof would need to contain some recursive logic to further factor (4n-1) as (4s-1)(4s+1) etc, and see you can't factor indefinately and reach some contradiction there... Maybe.
Yes, I also believe there is a step missing but you don’t need to use recursive logic to resolve the flaw. We know that no PRIME factor can be of the form (4n-1) by the assumption but there might be an odd COMPOSITE factor of the form (4n-1). This cannot be the case however because all the prime factors are odd and they are all of the form (4n+1) by the assumption. Using part 2, we know that the product of any (4n+1) numbers is also (4n+1). So this means that all the odd composite factors must also be of the form (4n+1). There cannot be an odd COMPOSITE factor of the form (4n-1) either. This was just assumed in the original proof but it needs to be proved in this way.
@@tidder1066 Ahh, very good! Much neater. Your logic can be easily stretched to do the full proof in one step too. Consider the (unique) prime factorization of 4m-1: if there are none of the form 4n-1, and 4m-1 is odd (it is), then all prime factors must be of the form 4n+1. That means 4m-1 itself must be of the form 4n+1, since it is just the product of these numbers (!). That's a contradiction, and so it must have at least one prime factor of the form 4n-1.
![ILLSLICK X YOUNGOHM - We're The same [Official Video]](http://i.ytimg.com/vi/yUJNkobYR0E/mqdefault.jpg)
back in school I hated math... you, sir, among others, reignited my interest in the subject by explaining things so well and clearly that it's really easy to follow and I'm not even an native English speaker. you really have a way with words. Keep up the good work. I'm sure you are inspiring lots of people out there. Thank you. I really mean it.
Great teachers are so easy to follow. Good vid.
You are best math teacher in this world woo sir ❤️
"Matematiğin Fatihi" is best math teacher in world. He is a Turk man :)))
one of the best teachers
explanations are so succinct
Amazing teacher🙏
Eddie's proof contains several logical errors (negating the given claim wrongly; using Part(a) wrongly by mistaking P⇒Q for its converse Q⇒P; most egregiously, proving the *weaker* claim "....must have at least one factor..." instead of the given, *stronger* claim "....must have at least one _prime_ factor..."). Here's a more rigorous presentation:
[ Let p be a composite number, suppose that it is of the form 4m-1, and *_assume_* that it has no prime factor of the form 4n-1.
Now, p, being of the form 4m-1, is odd, so its prime factorisation contains only odd numbers, so either it has a prime factor of the form 4n-1 or, by Part (a), its prime factors are all of the form 4n+1. The latter, by Part (b), is false; thus, p must have a prime factor of the form 4n-1. This *contradicts* our assumption; therefore, p _has_ a prime factor of the form 4n-1.
Hence, every composite number of the form 4m-1 has a prime factor of the form 4n-1. ]
-----------
PS. For reference, this is my previous proof, which is invalid because ∀n[P(n) or Q(n)] of course doesn't imply [∀nP(n) or ∀nQ(n)] (the converse is true though; thanks to Oisin Carroll for highlighting this crucial error):
[ Let m be a positive integer, suppose that 4m-1 is composite, and *assume* that for each n, 4n-1 is non-prime or not a factor of 4m-1.
Now, 4m-1 is odd, so must be a product of two odd numbers, so either has some factor 4k-1 or, by Part (a), equals (4p+1)(4q+1) for some p & q. The latter, by Part (b), is false; thus, some some 4k-1 must be a factor of 4m-1, i.e., it is false that for each k, 4k-1 is not a factor of 4m-1.
Therefore-continuing from the first paragraph-for each n, 4n-1 is non-prime; in particular, 4(1)-1=3 is non-prime; this is a *contradiction*; therefore, our assumption must be false; i.e., for some n, 4n-1 is a prime factor of 4m-1.
Hence, for each positive integer m, if 4m-1 is composite, then there is some n for which 4n-1 is its prime factor, as required.]
I believe the step in the first paragraph of your proof is incorrect. You can't redefine forall. n and have the inequality hold. It's equilivent to saying...
forall n. n is even OR n is odd
=>
(forall n. n is even) OR (forall n. n is odd)
You can do a similar rearrangement keeping a single forall n. up front but of course it breaks your contradiction at the end.
@@oisincarroll1171 Did you mean Line(A) "Suppose the contrary,..." or Line(B) "Thus, ∃m such that" ?
@@ryang628 Yes I was talking about line B.
[q OR (p AND r)] doesn't follow from the line above. For e.g. q=true, p,r=false satisfy that but not the previous. There's a similar assignment of variables that work for the first and not second.
@@oisincarroll1171 Line(B) is *not* intended to be equivalent to Line(A); it just needs to be a *consequence* of Line(A).
Here, P=false is not a possible assignment, because the proof-by-contradiction argument is based on the premise that THERE DOES EXIST a composite number 4m-1), i.e., that P=true!
I've just condensed the proof (it remains equivalent to the original one) to make it easier to parse; perhaps give that a read?
@@ryang628 I admit I haven't heard the term consequence used in logic, but I understood it to mean line A implies line B, which it does not. This is because there is an assignment that satisfies A but not B - giving True implies False, which is false.
Edit: whoops, no, you're right. Had the logic backwards here.
Can you do truth trees? I like them as an effective means of solving knights and knaves and other propositional logic problems. Vry good tools for discrete mathematics.
What if 4n-1 has a factor of the form 4m-1, but that factor is a composite number? Since the question says that it needs to have a prime factor in the form 4m-1, should we do extra steps to show that that number is prime as well?
i trying to understand , prof
#GazaUnderAttack
#SaveAlAqsa
#FreePalestine
#IsraeliCrimes
Like this!!!!!!!
Gday Eddie how are ya champ? I've personally quite exquisite
its day 45 after this post, my hubby left me, the house is gone, I barely have access to these videos anymore. Im on the fucking streets cuuuunnntt. Im on my last gigabyte of data cuzz. This is my goodbye eddie woo. BTW Nick Hatziliradis wants to fight you
@@geza9580 meet me at broady station bruhhh, Ill sell you 1 extra gig of data for a bag lad. You need you ataday lad? meet me at the station now habibi shuu
Good job turning math into English
I've really enjoyed these, I'm not sure I agree with some logic here. k could be a product of the form (4n-1)(4n+1) which would still satisfy the negation as long as 4n-1 is not prime (i.e. is composite). The full proof would need to contain some recursive logic to further factor (4n-1) as (4s-1)(4s+1) etc, and see you can't factor indefinately and reach some contradiction there... Maybe.
Yes, I also believe there is a step missing but you don’t need to use recursive logic to resolve the flaw. We know that no PRIME factor can be of the form (4n-1) by the assumption but there might be an odd COMPOSITE factor of the form (4n-1). This cannot be the case however because all the prime factors are odd and they are all of the form (4n+1) by the assumption. Using part 2, we know that the product of any (4n+1) numbers is also (4n+1). So this means that all the odd composite factors must also be of the form (4n+1). There cannot be an odd COMPOSITE factor of the form (4n-1) either. This was just assumed in the original proof but it needs to be proved in this way.
@@tidder1066 Ahh, very good! Much neater. Your logic can be easily stretched to do the full proof in one step too. Consider the (unique) prime factorization of 4m-1: if there are none of the form 4n-1, and 4m-1 is odd (it is), then all prime factors must be of the form 4n+1. That means 4m-1 itself must be of the form 4n+1, since it is just the product of these numbers (!). That's a contradiction, and so it must have at least one prime factor of the form 4n-1.
This proof is a bit meh. Too many words...
Still interesting, at least.
Proofs need words though
Everything is great, but the posture of your neck while you’re writing on ipad is totally wrong