Theorem: if a|c and b|d, then ab|cd Let ak=c and bj=d cd=(ak)(bj)=(ab)(kj) Since k and j are integers, (kj) will also be an integer, which we'll call n. Then the last line can be rewritten as: cd=(ab)n Therefore, ab|cd, which is what was to be proved.
3:28 is correct way ? you mean, 1/(k+j)=a/(b+c) ? i am in confusion in proof methodology.. and in 6:00 there are 3/6, 6/18 and 3/18 is correct i think according to example.
"If a|b then ac=b. Here a and b are integers and c is a positive integer". But c need not be a positive integer always. Eg. 2|(-8), here value of c is -4 which is not positive integer. Correct me if I am wrong.
Sir, if 3|6 and 6|18 then it must have been 3|18 ? you said a|b and b|c then a|c which must have been 3|18 ? why is it 3|(6x18) which means a|(b x c) ? Can you please explain? ua-cam.com/video/K2YmMpulFA4/v-deo.html
Muhammad Farhaan Yes, that was my bad. I took it a step further accidentally. Since a|b and b|c, we get that a|c. However, because a|b and a|c, we can prove a|bc. My mistake. You're correct.
You are mixing up your terms, but the best part of your comment is that he DID make a mistake there and you didn't even see it lol. He pointed at 3 and called it d, but 3 is actually a. Anyways, yes, 3 is odd, but like he said, it is also positive (however it is not even if that's what you are trying to say, although nobody ever said it was) :)
Why c or k or the quotient must be positive integer? How about when a is negatif integer and b is positif integer? c can be negatif integer? Could someone explain this or maybe i miss something rule?
Thank you, but I'd add a small correction. Asymmetric means that 'a' and 'b' are never equal to each other. In this case it would be antisymmetric, because they can still be equal.
so this may be a dumb question, but does it have to false just once for it to be false? of course i can prove it either way but for it to be true does it ALWSAYS have to be true ?
how to prove the following: let a and b be integers with a > 0, and let r be the remainder when b is divided by a. Prove that d is a common divisor of a and b, then d also divides r.
upon division of a by b, remainder is r which means a=bq+r (where q is the quotient). Let d be a common divisor of a and b and let a=dm and b=dn. Substituting in the values, we get dm=dnq+r which means r=d(m-nq). Since m, n , q are integers, m-nq is an integer implying that d divides r
sir you say 3/15 and 3/9 then 3/24 when we solve it by adding 1 and 2 it become 24=24 how we write it ??'we write it in generic form (as a/(b+c))??????
Would this work? Let {a,b,c,k,j,m} all be integers. GIVENS: a|b ak = b a = b/k GIVENS: b|c bj = c PROVE: a|c am=c m = c/a To prove that a|c, we would have to show that m is an integer. We have to show that (c/a) simplifies to be an integer. c/a bj / (b/k) Copy Dot Flip bj * (k/b) j*k Integers Closed Under Multiplication We find that m was an integer. Thus, it is true that a|c. I would really appreciate a response, thank you so much!!
![[Discrete Mathematics] Divisibility Examples](http://i.ytimg.com/vi/uBI6ZHyFq_Y/mqdefault.jpg)
![[Discrete Mathematics] Divisibility Examples](/img/tr.png)
![[Discrete Mathematics] Modular Arithmetic](http://i.ytimg.com/vi/d-n92Ml1iu0/mqdefault.jpg)
Ur videos are whats standing between me and a D grade, many thanks
@1:41 "It'll be fun!"
Narrator, turning to the camera: "It was not, in fact, fun."
(Still grateful!)
At 05:22, we have a|b = 3|6, and b|c = 6|18, wouldn't a|c = 3|18? Why do you suddenly refer to 3|108?
You are correct. 108 is a thought accident. :)
Theorem: if a|c and b|d, then ab|cd
Let ak=c and bj=d
cd=(ak)(bj)=(ab)(kj)
Since k and j are integers, (kj) will also be an integer, which we'll call n. Then the last line can be rewritten as:
cd=(ab)n
Therefore, ab|cd, which is what was to be proved.
For me it was easier to think it this way: ab|cd => (ab)n=cd => (ab)(kj)=(ak)(bj)
Thank you so much for making Discrete Math appear fun and amazing to learn! Your channel is a blessing for all students!
i had no idea where to study discrete math from but then ifound this channel. Many thanks for explaining things so clearly
Hello. On 8:52 it should be 0 ≤ r < d because we divide 1999 by 1000 not 1. Please fix that mistake.
This guy is singlehandedly saving my discrete math grade
3:28 is correct way ? you mean, 1/(k+j)=a/(b+c) ? i am in confusion in proof methodology.. and in 6:00 there are 3/6, 6/18 and 3/18 is correct i think according to example.
I guess he did it like
a/b and b/c, implies a/c(as proven by him) which in turn imply that a/bc
i got sick so i missed a whole week of college But this is actually amazing!! THANK YOU!
I think you meant to say
0
"If a|b then ac=b. Here a and b are integers and c is a positive integer". But c need not be a positive integer always. Eg. 2|(-8), here value of c is -4 which is not positive integer. Correct me if I am wrong.
I was thinking the same
These are extremely helpful when I get stuck on my textbooks! Greetings from Mexico!
at 5:16, I think it should be 3|18 because c=18. If a|b and b|c then a|c. So, if 3|6 and 6|18 the 3|18. Correct me if I am wrong.
You are correct. How did C magically change.
not all superheroes wear capes... TrevTutor is single-handledly saving thousands of students from failing.
Mistake at 5:23. You have C changing from 18 to 108. C is still C which is still 18.
Why'd you take 108 instead of 18 in 6:00
@@karanshirur4190 he said it’s meant to be 18 in another comment, it is a mistake
Thank you.i feel relieved now
I think you just opened doors for me , my future is clear now
thank you for this. Now Im kinda bit ready for my quiz tomorrow hahaha
divide forms a poset so it should be antisymmetric....asymmetric does not holds reflexive property..asymmetric is Antisymmetric + irreflexive
thank you for the dedications you made sir!❤❤
@5:23 where did 108 come from? The proof says "if a|b and b|c then a|c", not "a|b*c".
And thank you for such a good videos 🙂
you are a legend
Sir, if 3|6 and 6|18 then it must have been 3|18 ?
you said a|b and b|c then a|c which must have been 3|18 ? why is it 3|(6x18) which means a|(b x c) ?
Can you please explain?
ua-cam.com/video/K2YmMpulFA4/v-deo.html
Muhammad Farhaan Yes, that was my bad. I took it a step further accidentally.
Since a|b and b|c, we get that a|c. However, because a|b and a|c, we can prove a|bc. My mistake. You're correct.
If a | b, there is c that ac = b. My question is, why a, b are in Z, but c in Z+? if a = -2, b = 8, so c would be -4.
i have the same question after 3 years too lol
Great content man! nice explanation skills u"ve got! me n my friends look forward to watch more engineering topics from you
Awesome 🙌
Exercise: 5:53
Thank you very much!
Which software you use write on screen?
at 8:20 mark you said d was positive and d happens to be 3...last time i checked 3 was odd.......
.......... um .......... yeah...
You are mixing up your terms, but the best part of your comment is that he DID make a mistake there and you didn't even see it lol. He pointed at 3 and called it d, but 3 is actually a. Anyways, yes, 3 is odd, but like he said, it is also positive (however it is not even if that's what you are trying to say, although nobody ever said it was) :)
Why c or k or the quotient must be positive integer? How about when a is negatif integer and b is positif integer? c can be negatif integer? Could someone explain this or maybe i miss something rule?
Thank you, but I'd add a small correction. Asymmetric means that 'a' and 'b' are never equal to each other. In this case it would be antisymmetric, because they can still be equal.
Excellent!
MUCH HELP!!! MUCH APPRECIATION!! MUCH THANKS!!
does this problem use transitive property to prove a/c ?
so this may be a dumb question, but does it have to false just once for it to be false? of course i can prove it either way but for it to be true does it ALWSAYS have to be true ?
Hey Trev, if a relation is reflexive it can't be asymmetric.
how to prove the following:
let a and b be integers with a > 0, and let r be the remainder when b is divided by a. Prove that d is a common divisor of a and b, then d also divides r.
upon division of a by b, remainder is r which means
a=bq+r (where q is the quotient). Let d be a common divisor of a and b and let a=dm and b=dn.
Substituting in the values, we get dm=dnq+r which means r=d(m-nq). Since m, n , q are integers, m-nq is an integer implying that d divides r
God Bless You
thanks for the vedio
sir you say 3/15 and 3/9 then 3/24 when we solve it by adding 1 and 2 it become 24=24 how we write it ??'we write it in generic form (as a/(b+c))??????
is this the same thing as divisors?
it is 3|18 not 3|108. 3|108 is indeed true, but irrelevant to the theorem.
Are you casually explained? You sound exactly like him ;)
Would this work?
Let {a,b,c,k,j,m} all be integers.
GIVENS: a|b ak = b a = b/k
GIVENS: b|c bj = c
PROVE: a|c am=c m = c/a
To prove that a|c, we would have to show that m is an integer.
We have to show that (c/a) simplifies to be an integer.
c/a
bj / (b/k)
Copy Dot Flip
bj * (k/b)
j*k
Integers Closed Under Multiplication
We find that m was an integer.
Thus, it is true that a|c.
I would really appreciate a response, thank you so much!!
I need help with big O, big omega, big theta
thats Algorithm analysis. Search them on youtube u will find tons of videos on them
i dont understand why m=an ⇒a/m ?
I mean actually these are not college courses, our university are teaching the same thing.
Define asymmatric
Confusing still
"If a divides b", as opposed to, "if a is divided by b" . Wish you would have clarified that. This is why people don't like discrete math.
this video was fucking awsome
you my shun
Thank yous
has anyone ever told u your voice sounds like the casually explained guy?
bj nice
You are just assuming everything to prove the solution. Just worst video I seen on this topic
4:15 bj ;)
I dont get where k and j came from
They're just variables, like x and y