VERY disappointed. In the thumbnail, I was promised FIRE. I expected you to hold and bend fire while teaching me pre calculus. I am unsubscribing and disliking this video.
I watched 3 different videos before coming here. Yours by far did the best explaining. I followed along with a similar problem and got my answer successfully :)
The chalk board sounds are so satisfying! I have my P4 A level mocks in a few hours, very well prepped but always found proof difficult, but not so much after this video! Thank you very much
Is this correct for the last example? >> a^2 = 4b +2 and then a = sqrt of 4b +2. and after taking square root by plugging in values for b, a = a decimal number. Hence a is not an integer. So our assumption is not true.
~(p->q) is p and ~q So 3n + 7 is even and n is even (Matches with contrapositive?) 3n + 7 = 2k where k is any odd integer and n = 2m with m being any odd integer 3(2m) + 7 = 2k 6m + 7 = 2k 7 = 2(k - 3m) Spz a number g = k-3m 7=2g Implying that 7 is even, which is a contradiction Hence the original statement is true Would having 2 different odd integers be false? Thanks
I assume you want the proof explained in a little more detail. Here's what the proof says in English. Lets assume that conditions 1 and 2 hold. We use a proof by contradiction that it must be true for all n>=1. As with all proofs by contradiction, we assume the statement is false and then show it leads to a contradiction. So we assume there is some s for which P(s) is false. Lets pick the smallest s where P(s) is false. (We know that there must be a smallest number s where it is false, because any non-empty set of natural numbers must contain a smallest value). But we know that P(s-1) is true (because s is the smallest value where P(s) is false, and s-1 is less than s). But if P(s-1) is true, then P(s-1+1) = P(s) is true, contradicting our assumption that P(s) is false. Therefore there cannot be any s where P(s) is false. The s-1 finds the number immediately before the rule supposedly breaks down. But we know that if P(s-1) is true then P(s) must also be true, because of condition 2. This contradicts our assumption that the rule doesn't hold for P(s). Well-ordering appears explicitly when we use the fact that any non-empty collection of natural numbers (in this case the collection of natural numbers x where P(x) is false) must have a lowest number. More generally, the well-ordering principle states that if you consider 1,2,3 ... then this list will contain all natural numbers; there are no natural numbers that cannot be reached by starting at 1 and adding 1 repeatedly. If there were any missing numbers (ie if the set cannot be well ordered) then proof by induction would not work.
VERY disappointed. In the thumbnail, I was promised FIRE. I expected you to hold and bend fire while teaching me pre calculus. I am unsubscribing and disliking this video.
LOL :D
I watched 3 different videos before coming here. Yours by far did the best explaining. I followed along with a similar problem and got my answer successfully :)
i’m taking a midterm on proofs today so this video was perfect timing
GL! :D
Now that was an effective video. I had almost given up, but this made it so easy. Thank you!
That simplicity in your explanation and the aura on you are gooooood...Thank you...I have subscribed
What an amazing video. Clear and concise. Thank you!
thankyou so much this video helped more than i expected!
The explanation is really simple
Thanks
Hey Andrew keep it up your content heal the pain of abstraction of concept
I personally love this video it was wonderful. I'm liking this video.
this by far the most helpful video. thanks!
Thank you so much this really helped me in my exam
The chalk board sounds are so satisfying! I have my P4 A level mocks in a few hours, very well prepped but always found proof difficult, but not so much after this video! Thank you very much
Thank you chairman
The third and fourth proofs can also be proven by contrapositive
You are the best 🤗🎊
This video helped me a lot. Thanks!
Is this correct for the last example? >> a^2 = 4b +2 and then a = sqrt of 4b +2. and after taking square root by plugging in values for b, a = a decimal number. Hence a is not an integer. So our assumption is not true.
~(p->q) is p and ~q
So 3n + 7 is even and n is even
(Matches with contrapositive?)
3n + 7 = 2k where k is any odd integer and n = 2m with m being any odd integer
3(2m) + 7 = 2k
6m + 7 = 2k
7 = 2(k - 3m)
Spz a number g = k-3m
7=2g
Implying that 7 is even, which is a contradiction
Hence the original statement is true
Would having 2 different odd integers be false?
Thanks
Great video, thanks a lot
SUPER helpful, thank you!!!
Brilliant 🎉
can we assume negation of p to be false and by contradiction run to be true and conclude p is false?
Man you really helped me out a lot bro, you explained really well and I kinda like math a little more now😂👌
That's great to hear!
i like your stuff bro and im really flexing my brain muscles
there may have been a small mistake in the truth table where p is F and q is T. I think (p->q) should be F. and (p^~q) should be T.
Well explained 👏
Well, this was awesome.
Simply amazing!
So helpful..thanks
Nice vid!
the video was very awesome🥰🥰🥰🥰
sir? how about using proof of contradiction, for all real numbers x, if x is irrational, the -x is irrational..
I assume you want the proof explained in a little more detail.
Here's what the proof says in English. Lets assume that conditions 1 and 2 hold. We use a proof by contradiction that it must be true for all n>=1.
As with all proofs by contradiction, we assume the statement is false and then show it leads to a contradiction. So we assume there is some s for which P(s) is false. Lets pick the smallest s where P(s) is false. (We know that there must be a smallest number s where it is false, because any non-empty set of natural numbers must contain a smallest value). But we know that P(s-1) is true (because s is the smallest value where P(s) is false, and s-1 is less than s). But if P(s-1) is true, then P(s-1+1) = P(s) is true, contradicting our assumption that P(s) is false. Therefore there cannot be any s where P(s) is false.
The s-1 finds the number immediately before the rule supposedly breaks down. But we know that if P(s-1) is true then P(s) must also be true, because of condition 2. This contradicts our assumption that the rule doesn't hold for P(s).
Well-ordering appears explicitly when we use the fact that any non-empty collection of natural numbers (in this case the collection of natural numbers x where P(x) is false) must have a lowest number.
More generally, the well-ordering principle states that if you consider 1,2,3 ... then this list will contain all natural numbers; there are no natural numbers that cannot be reached by starting at 1 and adding 1 repeatedly. If there were any missing numbers (ie if the set cannot be well ordered) then proof by induction would not work.
you are the goat
tnx a lot sir it really helped
How can we prove my contradiction that the world is round?
that was helpful
Thanks bro
thank you
hi, do you have videos on predicate logic ?
Man GG the best ty !!!!
I'm not sure all of these need the contradiction formulation to prove
God bless you
Damn right I'm gonna flex em
Awww!! So handsome i watched u throughout the video instead of listening
this is proof by contraposition
انتا مُزه ي استاد
but you are handsome. how is a handsome man like you became so smart ?
This dude thinks he's fancy with his 60 fps
Your explanation is too fast can you slow it down please
wait omg why is he so hot... i dont hate discrete math as much anymore...
thats a 2? nah bro thats a z lol
Please speak in hindi.