That moment when you realize you've been incorrectly assuming what the equivalence symbol means related to mod. arithmetic for the entire semester. XD Thanks so much for the video! Great stuff!
after reading the book named Discrete mathematics and its application from Rosen, this video makes sense, even though there are some mistakes that trev has made haha Thank you
It's congruent though to all other numbers that also mod 3 = 0. All of these numbers are congruent and form a "congruence class". Also note that the union of all of the congruence classes partitions the original set in this case it partitions the integers
at 4:09 you say "one divided by four is going to have a remainder of 1." Given the discussion after, and that (0 * 4) + 1 = 1, I am guessing you meant to phrase that differently since 1/4 does not have a remainder of 1. {EDIT} Oops! I see no that 4 goes into 1 zero times but that leaves a remainder of 1!
I'v been following all your cources and i love your way of teaching in all of your videos. But i can sadly say that this video was'nt as good as you usually are. I know you can do much better man :)
Ok so when you say a- b and you assign those letters by defining it as 3/(18-12) you make 18 a. But when you converted it to modulus form you put 12 where a should go?
Unlike most, I loved this one, kinda helped me a lot to understand how we calculate remainders in big cases, I'll just make the ap, and find the number I need eehehe
hey Trev, we did congruent mods in our course is there a quick video on that? or is it a natural extension of what you've said in this video? We do a few proofs using congruent mods. Love the videos :)
Hey, If anyone has an answer, please let me know, but the question is this: When solving a linear congruence, how do you determine how many solutions it has? Someone told me that if we have something of the form: ax =(pretend this is the equivalence sign) b (mod m) then the amount of solutioms was determined by the gcd of a and m. Is that right?
I don't really know but with an example I found, it seems it is wrong, again not sure though, take a=5, b=3 and m as 2, like u see, a and m hav gcd=1 Now, m|ax-b for the solution, means 2| 5x-3, now this is true for all odd values of x (obviously, since odd-odd is even) Now, if it is like gcd≠1, then let a= 10, b=7 and m=4 So, 4| 10x-7, now this is false for all integer values of x, since 10x-7 will always give a number ending with 3 or with 7 (for x as -ve) So, with 2 very basic examples I think your statement is wrong, maybe it was something different but similar
I know I'm late, but... Proposition: Let ax ≡ b (mod n) be a linear congruence. ax ≡ b (mod n) has solution ⟺ GCD(a, n) | b Proof: The congruence ax ≡ b (mod n) has solution ⟺ ∃ x ∈ Z such that ax ≡ b (mod n) ⟺ ∃ x ∈ Z such that n | ax - b ⟺ ∃ x ∈ Z such that nh = ax - b ⟺ ∃ x, h ∈ Z such that ax + n(-h) = b ⟺ the equation ax + by = b has solution ⟺ MCD(a, n) | b □ Notes: - Every solution of a linear congruence is also called "particular solution". - If a particular solution exists, then an infinite amount of solutions exists.
Now this is all very easy examples. Let me hit you with the question from my exam: (5+55+101)*(576*555 + 100000000002) mod 5 is equal to? Possible answers 0, 1, 2, 3, 4. I cannot use a calculator, only use pen and paper. What to do?
The answer is 2 You can apply the (mod 5) to each of the terms. So: ( 5 (mod 5) + 55 (mod 5) + 101 (mod 5) ) * ( 576*555 (mod 5) + 100000000002 (mod 5) ) and then equate each of the terms' values ( 0 + 0 + 1) * ( 0 + 2 ) which equals 2 576*555 is one term, so the modulus is applied to both as a whole. Multiplying the last two digits, will give the last digit in the product, in this case it is zero (6*5 is 30). When there's mod 5, you just have to look at the last digit, and if it ends in a multiple of 5 (5 or 0).
This question is also as easy as the examples😂, see 567*555 gives a number with last digit = 5, now if 1000....2 is added, then last digit comes out to be 7, now then u multiplied a number 55+5+101, so last digit is 1, now two numbers multiplied, one with last digit 7, one with last digit 1, so last digit comes out 7 again, we divide by 5, then remainder will come 7-5=2 :) So answer is 2
thank u loved the way you explained 9 mod 4. i wish if you had shown anything for negative numbers too. say -97 mod 11. also if we have to find a huge number like in thousands or hundreds, writing all numbers will take forever. is there a quicker way?
this is not an introductory video on modular arithmetic lmao, but I se it's video 78 in the playlist, so I'm not sure if you need to watch all 78 videos before this. but you prob should have mentioned that at the beginning
Sorry but this was very confusing. I had to find another explaination on Google to finally get it. Maybe you should have started with the modulo operator ? Then you could have drawn the parallel between a (mod b) and a mod b. From your example, 9 (mod 4) ≡ [1], I would also have 9 mod 4 = 1
After 3:47 cant understand anything. Man why did you asume we already know this? Seriously this lecture was bad for me. Had to learn basics from somewhere else then came here again to learn the rest.
Jonatham Vicens Either "The Book of Proof" which you can find on google, or a text written by a dude named Grimaldi which you can probably find illegally as well. Not sure where I mentioned it, but those are the books I'm using for this series.
Hey hope you are doing alright just I wanna say that GOD loved the world so much he sent his only begotten son Jesus to die a brutal death for us so that we can have eternal life and we can all accept this amazing gift this by simply trusting in Jesus, confessing that GOD raised him from the dead, turning away from your sins and forming a relationship with GOD. Nice vid btw
![[Discrete Mathematics] Congruency Proof Examples](http://i.ytimg.com/vi/a6eSpr_EgcY/mqdefault.jpg)
![[Discrete Mathematics] Congruency Proof Examples](/img/tr.png)
This is not a bad video. It just requires a little bit of prior knowledge. Very useful for reviewing. Thank you
"If you learn anything in Discrete Math, make sure its this." *throws out terms and formulas without going over first*. Now majorly confused.
I think the last two minutes are the most valuable, at least for me, that's when I actually got it. Thanks for the video!!
That moment when you realize you've been incorrectly assuming what the equivalence symbol means related to mod. arithmetic for the entire semester. XD
Thanks so much for the video! Great stuff!
Sometimes even the textbook makes more sense
not so clear
Generalization : a mod b = a - b.floor(a/b); works even for a < b
You are litteraly saving my semester
this is the first time your lecture is as bad as my professor's hahaha
false
after reading the book named Discrete mathematics and its application from Rosen, this video makes sense, even though there are some mistakes that trev has made haha Thank you
very confused. isn't 18 mod 3 equal to 0?
It's congruent though to all other numbers that also mod 3 = 0. All of these numbers are congruent and form a "congruence class". Also note that the union of all of the congruence classes partitions the original set in this case it partitions the integers
He was mistaken. 18 is CONGRUENT to 12(mod3) meaning 3|(18-12), which is denoted by a triple-dash sign rather than an equals sign.
Yeah this explanation is VERY imprecise. Don't take this video very seriously. The video from MIT open courseware is better.
Ya it also is 0, it's a class of 0
You say equal, but you mean are equivalent to right? They don't equal each other but have equivalent congruencies.
1:08 should be 18 is CONGRUENT to 12(mod3)
at 4:09 you say "one divided by four is going to have a remainder of 1." Given the discussion after, and that (0 * 4) + 1 = 1, I am guessing you meant to phrase that differently since 1/4 does not have a remainder of 1. {EDIT} Oops! I see no that 4 goes into 1 zero times but that leaves a remainder of 1!
I'v been following all your cources and i love your way of teaching in all of your videos. But i can sadly say that this video was'nt as good as you usually are. I know you can do much better man :)
Exactly!
I never found his videos helpful anyway. He never goes to the concept. He just explains how to use the rules.
I am confuse on 4:32, for -3/4, isn't the remainder is -3? Why it is under set [1] ?
This video was so helpful, thank you so much
Thank you so much! Clearly explained! Great!
My professor's explanation of this is still way worse than this, if it makes you feel better, TrevTutor
At 1:00 it should be 18 = 12 mod 3
Ikr?
It can be what he wrote as well, cuz even if there's that -ve sign, it'll still be divisible by 3, cuz 3(-2)=-6
ok i thought i was the only one who found this vid super confusing and not helpful until i read the comments lol . jfc.
Ok so when you say a- b and you assign those letters by defining it as 3/(18-12) you make 18 a. But when you converted it to modulus form you put 12 where a should go?
Never mind, my impatient brain couldnt wait the 2 seconds to 1:16 to see you correct it lol
Unlike most, I loved this one, kinda helped me a lot to understand how we calculate remainders in big cases, I'll just make the ap, and find the number I need eehehe
Very well explained
hey Trev, we did congruent mods in our course is there a quick video on that? or is it a natural extension of what you've said in this video? We do a few proofs using congruent mods. Love the videos :)
Hey,
If anyone has an answer, please let me know, but the question is this:
When solving a linear congruence, how do you determine how many solutions it has? Someone told me that if we have something of the form:
ax =(pretend this is the equivalence sign) b (mod m) then the amount of solutioms was determined by the gcd of a and m. Is that right?
I don't really know but with an example I found, it seems it is wrong, again not sure though,
take a=5, b=3 and m as 2, like u see, a and m hav gcd=1
Now, m|ax-b for the solution, means 2| 5x-3, now this is true for all odd values of x (obviously, since odd-odd is even)
Now, if it is like gcd≠1, then let a= 10, b=7 and m=4
So, 4| 10x-7, now this is false for all integer values of x, since 10x-7 will always give a number ending with 3 or with 7 (for x as -ve)
So, with 2 very basic examples I think your statement is wrong, maybe it was something different but similar
I know I'm late, but...
Proposition:
Let ax ≡ b (mod n) be a linear congruence.
ax ≡ b (mod n) has solution ⟺ GCD(a, n) | b
Proof:
The congruence ax ≡ b (mod n) has solution ⟺
∃ x ∈ Z such that ax ≡ b (mod n) ⟺
∃ x ∈ Z such that n | ax - b ⟺
∃ x ∈ Z such that nh = ax - b ⟺
∃ x, h ∈ Z such that ax + n(-h) = b ⟺
the equation ax + by = b has solution ⟺
MCD(a, n) | b □
Notes:
- Every solution of a linear congruence is also called "particular solution".
- If a particular solution exists, then an infinite amount of solutions exists.
Now this is all very easy examples. Let me hit you with the question from my exam: (5+55+101)*(576*555 + 100000000002) mod 5 is equal to?
Possible answers 0, 1, 2, 3, 4. I cannot use a calculator, only use pen and paper. What to do?
The answer is 2
You can apply the (mod 5) to each of the terms.
So:
( 5 (mod 5) + 55 (mod 5) + 101 (mod 5) ) * ( 576*555 (mod 5) + 100000000002 (mod 5) )
and then equate each of the terms' values
( 0 + 0 + 1) * ( 0 + 2 )
which equals 2
576*555 is one term, so the modulus is applied to both as a whole.
Multiplying the last two digits, will give the last digit in the product, in this case it is zero (6*5 is 30).
When there's mod 5, you just have to look at the last digit, and if it ends in a multiple of 5 (5 or 0).
@@n0handles thx I tried solving it without your answer but I got tricked by the '576*555' your explanation were a blessing
This question is also as easy as the examples😂, see 567*555 gives a number with last digit = 5, now if 1000....2 is added, then last digit comes out to be 7, now then u multiplied a number 55+5+101, so last digit is 1, now two numbers multiplied, one with last digit 7, one with last digit 1, so last digit comes out 7 again, we divide by 5, then remainder will come 7-5=2
:) So answer is 2
the video wasnt as clear as all your other ones :(
Our lord and savior during these times of corona virus making discrete math classes online when you have a bad teacher
So I’m more confused now than I was before...
How does 1/4 have remainder 1? 4:10
4 divides 1 zero times with a remainder of 1.
essentially 4 cannot go into 1. hope that makes sense
more clarity please, you can do better
I think at 4:50 you want [1] {-5, -3, 3, 5}
the one shouldn't be there?
It is supposed to be there. 1 = 4*0 + 1, so 1 is the remainder.
thank u loved the way you explained 9 mod 4. i wish if you had shown anything for negative numbers too. say -97 mod 11. also if we have to find a huge number like in thousands or hundreds, writing all numbers will take forever. is there a quicker way?
Thank you so much, you cleared it up for me!
this is not an introductory video on modular arithmetic lmao, but I se it's video 78 in the playlist, so I'm not sure if you need to watch all 78 videos before this. but you prob should have mentioned that at the beginning
Sorry but this was very confusing. I had to find another explaination on Google to finally get it. Maybe you should have started with the modulo operator ?
Then you could have drawn the parallel between a (mod b) and a mod b.
From your example, 9 (mod 4) ≡ [1], I would also have 9 mod 4 = 1
Thank you very much...
12 = 18 mod 3? Am I missing something?????
Yes, ig u r missing-ve sign
Instead of equal it should be a congruent sign
3 divides (18-12) is true because 6/3 = Real number
thanks, loved the explanation
Isn't q = r in most, if not all, cases?
Not at all, he just happened to pick exaples where q=r
Thank you
very confusing how you introduce modular in the beginning
I can't realize where u r writing
After 3:47 cant understand anything. Man why did you asume we already know this? Seriously this lecture was bad for me. Had to learn basics from somewhere else then came here again to learn the rest.
12 is certainly not 18 mod 3, nor the other way round.
he meant congurent simple writng mistake
Went from 0 to 100
What's the name of the book you mentioned?
Jonatham Vicens Either "The Book of Proof" which you can find on google, or a text written by a dude named Grimaldi which you can probably find illegally as well. Not sure where I mentioned it, but those are the books I'm using for this series.
+shadowofforms I know. I said "or" :)
sorry.....but I can't understood.
Useless and not clear.
i do not catch it
Am I the only one here who hates mods.. -_-
Anwar Al-Amrad Yes, you are the only one :-)
No, you aren't :3
I have no idea what your saying
Actual MATH is most important for us, keep your proofs to your "heavenly" demigods residing in untouchable Olymp called Professors...
Huh? This is math...??? I'm confused...
this video confused me more. i have to watch another video
tangina wala ako naintindihan
what
now
sorry
Not one of your better videos you kinda rushed through the topic w/o much explanation
not explained very well at all and full of errors
ok.... and this is important WHY?!
in CS, binary/Hex
Hey hope you are doing alright just I wanna say that
GOD loved the world so much he sent his only begotten
son Jesus to die a brutal death for us so that we can have eternal life and we can all accept this amazing gift this by simply trusting in Jesus, confessing that GOD raised him from the dead, turning away from your sins and forming a relationship with GOD. Nice vid btw
not so clear..