I believe there's something wrong... You must prove that 1. Induction basis (one k' which holds P(k')) (which is okay) then: 2. If P(k) holds, then P(k+1) holds i.e: P(k) => P(k+1). So you shouldn't start your proof assuming P(k+1)... You have proven that P(k+1) => P(k), so...it doesn't seem correct to me. One way, being simplistic on notation here: 5^k + 3 = 4P, P must be integer, our k must be non-negative integer (assumed true)(our P(k)) so we must manipulate this initial expression so that P(k+1) holds: 5^k+3 = 4P 5^k = 4P - 3 5^(k+1) = 20P - 15 P is some integer, so: 5P is also some integer P 5^(k+1) = 4P - 12 - 3 5^(k+1) + 3 = 4(P-3) But P-3 is also some integer P 5^(k+1) + 3 = 4P then, we got that: P(k) => P(k+1)
@Mateus Marcuzzo: I disagree. It was shown that 5^(k+1)+3 is a multiple of 4 if 5^k+3 is as well. So P(k) => P(k+1) was indeed proven. You don't have to manipulate the assumption to get to your statement to prove. You just have to use it to show that the left-hand side of the statement to prove is indeed equal to the right-hand side. Also: In your proof you change the value of P multiple times. You should introduce a new variable instead, as was done in the video. For example, the statement "5P is some integer P" implies P=0, which is not what you mean.
hypercent thanks for the answer. I've been thinking on this, and my conclusion is that I've learned something new today. About P: its okay, since I just want to represent that P is some integer. So some integer *4 is a multiple of 4. I could do P, P' and so on, as you suggest.
Brilliant and highly engaging. A natural. BTW. I do wish at the end of every maths lesson, everywhere, everyone stops, takes a breath and asks...and answers the question......how is this applicable in everyday activity. eg writing code for ......
Well sometimes people do maths purely because it's interesting (pure mathematics), so it may not be the best idea to force students to only care about applications of math, when the math itself can be beautiful without necessarily having a purpose.
![[UNCUT]“ฅนตื่นธรรม”ปะทะ“หมอปลา-ปู่มหามุนี-อั๋นโอกิ”สะเทือนกระทรวงเวทมนตร์ Iคนดังนั่งเคลียร์I10 ตค.67](http://i.ytimg.com/vi/peMD_SAkwNc/mqdefault.jpg)
Different method to how I learnt it, but still interesting to watch.
E.W., you just keep getting better and better.
Thank you for posting this... very helpful!
I believe there's something wrong...
You must prove that
1. Induction basis (one k' which holds P(k')) (which is okay)
then:
2. If P(k) holds, then P(k+1) holds
i.e:
P(k) => P(k+1).
So you shouldn't start your proof assuming P(k+1)...
You have proven that P(k+1) => P(k), so...it doesn't seem correct to me.
One way, being simplistic on notation here:
5^k + 3 = 4P, P must be integer, our k must be non-negative integer (assumed true)(our P(k))
so we must manipulate this initial expression so that P(k+1) holds:
5^k+3 = 4P
5^k = 4P - 3
5^(k+1) = 20P - 15
P is some integer, so: 5P is also some integer P
5^(k+1) = 4P - 12 - 3
5^(k+1) + 3 = 4(P-3)
But P-3 is also some integer P
5^(k+1) + 3 = 4P
then, we got that: P(k) => P(k+1)
He also showed that P(k)=>P(k+1) is true.
@Mateus Marcuzzo:
I disagree. It was shown that 5^(k+1)+3 is a multiple of 4 if 5^k+3 is as well. So P(k) => P(k+1) was indeed proven. You don't have to manipulate the assumption to get to your statement to prove. You just have to use it to show that the left-hand side of the statement to prove is indeed equal to the right-hand side.
Also: In your proof you change the value of P multiple times. You should introduce a new variable instead, as was done in the video. For example, the statement "5P is some integer P" implies P=0, which is not what you mean.
hypercent thanks for the answer. I've been thinking on this, and my conclusion is that I've learned something new today.
About P: its okay, since I just want to represent that P is some integer. So some integer *4 is a multiple of 4. I could do P, P' and so on, as you suggest.
thank you for the explanations my online class never provided!
Wisely man❤✊
Hi, May you do cardinality and infinite sets exercises??
Brilliant and highly engaging. A natural. BTW. I do wish at the end of every maths lesson, everywhere, everyone stops, takes a breath and asks...and answers the question......how is this applicable in everyday activity. eg writing code for ......
Well sometimes people do maths purely because it's interesting (pure mathematics), so it may not be the best idea to force students to only care about applications of math, when the math itself can be beautiful without necessarily having a purpose.
I didn't quite understand 3:15, where you wrote 5*5^k +3 = 4*5^k + 5^k + 3 ?
Multiplication is just simplified addition. 5*5^k is just 5^k+5^k+5^k+5^k+5^k, or 5 sets of 5^k. So using algebra you can change it to 4*5^k + 5^k.
@@nathandam6415 thank god for you buddy! that cleared it up for me completely!
@@nathandam6415 Thank you, for your time
THANKYOU
Cheers mate!
thanks man already knew this but i really appreciate your channel
I don't if it's correct but I learned that Z+ means all non-negative integers instead of positive integers. Can someone confirm it?
that means the same thing
Can someone tell me how is 5^k + P = Q?
Thanks
Why is it 4Q
How old are the students learning this
JackKnowsIt NoHeDoesn't yes
16 years old
16-18 years old
@@praisewinston770 might just want to know the level
Can you please make vid about sin(x)÷x=1
Joseph its a limit..
I know it's the limit but I mean the prove of sin(x)÷x=1
www.khanacademy.org/math/ap-calculus-ab/ab-derivative-rules/ab-derivtive-rules-opt-vids/v/sinx-over-x-as-x-approaches-0
He has already made a video on it. It is split into 3 videos and it will all make sense by the third: th-cam.com/video/R1mEeQhgNuM/w-d-xo.html
Thank you🥰✌🏽🤓❤️🩹❤️🩹