IMHO it is better to introduce first the notion of inverse element. Then you can say that when an element has an inverse element then you can divide. In this way you may have a self contained video where you can explain why you need gcd(k,n)=1. Just my two cents :)
Thanks to talk about this subject, I study congruences at school and sometimes it's a bit difficult, so you help me to understand with your videos, thanks and continue you're great !
Your video entitled "weekly brilliant problem#3, solving integral by symmetry, IIT JEE mains 2015" has over 20k views, and you promised to do the integral in the vid when it reached that number. Can't wait ;)
Looking forward to your next "This is IT!" videos after your creative break... ;-) Are you planning for some complex analysis? More videos with Dr. Peyam?
Works iff divisor and modulus are relatively prime. Note: Whenever you see, a ≡ b (mod n), you should immediately think, a-b = mn, where m is an integer. This is the *definition* of congruence. That will make this result more obvious. Fred
No it isn't! The question is, when does the following implication hold? ka ≡ kb (mod n) ⇒ a ≡ b (mod n) If GCD(k,n) > 1, there will be some examples (a,b), that make the implication false; the "division" won't work. The proposition is that for some particular k and n, the congruence follows for *all* (a,b). And this is true only if k and n are relatively prime. If you narrow it down to individual (a,b) cases, you can just use a = b, and it will always work. Fred
Your above comment: "Works iff divisor and modulus are relatively prime." is *wrong* since, for example: 2x5=2x1(mod4) and 5=1(mod4) are true, but 2 and 4 are not relatively prime. So, if ka ≡ kb (mod n) and a ≡ b (mod n) are true for some a and b, this doesn't necessarily mean that k and n are relatively primes.
What works iff divisor (k) and modulus (n) are relatively prime, is preserving congruence when dividing by k, *regardless of a and b.* Sorry if I didn't spell that out. I guess I assumed that since it should be obvious that you can always make the division work when a=b, the question becomes, what must you demand of k and n to make it *always* work? For your example, k=2, n=4, division by 2 will not work for a=5, b=3 2·5 ≡≡ 2·3 (mod 4), but 5 ≡/≡ 3 (mod 4). The statement I was trying to make, in full formality, is: ∀(a,b) [ka ≡ kb (mod n)] ⇒ [a ≡ b (mod n)] iff GCD(k,n) = 1 where a, b, k, n are integers, and k ≠ 0. Fred
Proxava Complete the square inside the log, by rewriting the whole log as log(a)+log((x+b/2a)^2 + c-(b/2a)^2)... I think it is now obvious enough to solve.
Actually, adding one point, we can actually divided anything somehow but with a little condition. ka=kb(mod n), then we have [k/gcd(k,n)]a=[k/gcd(k,n)]b(mod (n/gcd(k,n))) and hence "a=b(mod (n/gcd(k,n)))" which is easy to see by just manipulate the equation somehow (k(a-b)=sn,for some integer s and this implies (a-b)|(n/gcd(k,n)). Things in this video is just only a special case of this.
Hi , I have a suggestion for 2 videos First , please make a video about the origin of sine and cosine , I mean how did their function came from Secondly , how to get sine and cosine of π Thx buddy , u r one of the best😍
Hello blackpenredpen I don't know how to solve the matrix problems. Could you make some video to teach about matrix lessons? I hope you help me... thanks
I have general simplification for ka congruence kb (mod n). I don't know whether anyone else has do it (make it patent) but it's quite easy to do so. By definition, a congruence b (mod n) means there exists at least one integer c which makes a - b = cn. So ka congruence b means there exists at least one integer c which makes ka - kb = cn. Of course, we also can write it as k(a-b) = cn Let gcd(k,n) = g. -> ka congruence kb (mod n) -> There exist an integer c, which makes ka - kb = cn -> k(a - b) = cn -> (k/g)(a - b) = c(n/g) (by dividing both side with g, the result is guaranteed to be integer) -> (k/g)a - (k/g)b = c(n/g) -> (k/g)a congruence (k/g)b (mod n/g) Of course, this simplification only helps if gcd(k, n) not equals to 1. If it equals to 1, your method works better. In the example above: 30 congruence 42 (mod 4) -> 6*5 congruence 6*7 (mod 4) -> gcd(6,4) = 2 -> (6/2)*5 congruence(6/2)*7 (mod 4/2) -> 3*5 congruence 3*7 (mod 2) -> 15 congruence 21 (mod 2)
When the gcd is not 1 you can simply divide the modulo by the gcd. So gcd(6,4) is 2 so you can divide the 4 in mod 4 by 2 to get mod 2. So 5 = 7 (mod 2) which is true.
hey bprp, i tried to integrate sqrt(sinx),like u did with tanx (similar steps) but online there are too many answers and some say that its not possible.plz help.I can tell u the steps if u want.thnx.
The thing is that Zn is not Q. You can't just plug in 1/m as if it's the inverse of m. Your rule of thumb (gcd of q,n must be 1) can be understood in light of the fact that the multiplicative group of Zn contains the elements that are relatively prime to n. Or at least that's the case for Zp (with p being a prime number), not sure about Zn in general (I've been taking a group theory course this semester and there your modular groups are almost always modulu a prime..).
If you plan to proceed with doing NT videos and don't want to write (mod n) every time, you can write small letter n just after the conruence sign: 12 Ξn 5, iff n = 7
It's not a *necessarily* condition to have the number that you divide by and the "mod" number to be relatively prime: For example: 10=2(mod4) Divide by 2 5=1(mod4) Although 2 and 4 are not relatively prime
Yeah, in the video he does a proof with two assumptions: that *k·b≡k·a mod(n)* AND that *gcd(k,n)=1* ; then he says the conclusion would be that *a≡b mod(n)* is true. So he is proving the conclusion once he stated two assumptions, therefore he is just proving that *a≡b mod(n)* is always true if those two assumptions are made. But that's not the same as saying that the conclusion is true if and only if those assumptions are made. (The grass can be wet because of the rain, but it could be wet cause we watered it). So, in order to really proof that the assumption *gcd(k,n)=1* is strictly necessary, I think the proof would have to be like this: we assume that *k·a≡k·b mod(n)* AND that *a≡b mod(n)* , so we claim that, with those two assumptions, then *gcd(k,n)=1* is always true. (Which isn't, since you just showed a counterexample).
I Know that he proved the *sufficient* condition (which is fine). But what I'm referring to is his claim at 1:50 that "if gcd is not 1 then we are not allowed to divide" (which is not true, by my counter example)
so a simple way to think about why you have to be so careful is to consider the reverse. note, I will use "=" as if it's the congruence symbol. So let's say we start out with a false statement, such as 5 = 8 (mod 4) Now, what can we do to make this true? well, if we multiply both sides by 4, then no matter what we started with, they will be congruent to 0 (mod 4). That's because with integer n, 4n is always divisible by 4. So if we were to multiply our previous equation by 4, we'd get 20 = 32 (mod 4) which is indeed true. Since we can make something true from something false using multiplication, doing the opposite, i.e. division, we can make something false from something true.
That would just be a proof by counter example. 6=30 (mod 12) and 6 shares factors with 12. 1 isn't congruent with 5 (mod 12). After what the first statement means is 6=30+12a. Dividing both sides by 6 means 1=5+2a. That means exactly 1 is congruent with 5 (mod 2), a strictly weaker claim. 21=69 (mod 4) on the other hand means 21=69+4a. Dividing both sides by 3 means 7=23+4(a/3).
I think the reverse is true, just without gcd(k, n)=1 a=b (mod n) => a-b = 0 (mod n) => n | (a-b) => n | k(a-b), since multiplying by k doesn't change the fact that n|(a-b) => k(a-b)=0 (mod n) => ka = kb (mod n) So, a = b (mod n) => ka = kb (mod n) ka = kb (mod n) and gcd(k, n) = 1 => a = b (mod n)
Didn't realize the reverse was just multiplication. My mind goes different places and doesn't realize its just division and multiplication when working with these things.
First, you set that expression equal to some variable (Z, in this case). Then you apply the original tan function on both sides of the equation, giving you the following thing: i=tan(z) Next, you can express the tan function in terms of exponentials using the euler's formula [e^(ix)=cos(x)+isin(x)]. By doing that, you can set that expression to the value you're looking for (in this case, i). Finally, solve for Z. Bprp did a similar thing with sin(z)=2. Look up that video if you want a more graphic explanation.
If I made a mistake tell me where: n/2=n*1/2 n/3=n*1/3 n/m=n*1/m Now where the mistake comes in if there is one; 0/0=0*1/0, 1/0 is undefined but 0 times anything is 0; therefore 0*1/0=0, which means 0/0=0 Tell me where I messed up cuz mathematicians should've totally figured this out by now.
Can you prove this for me please?; n is a prime number greater than 5. Prove that the value of n^4 has its last digit ending in 1. I was lost, I dont know how to represent a prime number algebraically
Prime numbers greater than 1 can only end in 1,3,7, or 9. The last digit of a number is equivalent to the number modulo 10. 1^4, 3^4, 7^4, and 9^4 are all equal to 1 modulo 10. (Look up modular arithmetic if you need to).
I think it's (mod (4÷gcd(6, 4))), not (mod gcd(6, 4)), although since gcd(6, 4) = 2, the two coincidentally give the same answer in this case. But consider if you have 28 ≡ 658 (mod 105), and you want to divide both sides by 14. Since gcd(105, 14) = 7, and 105÷7 = 15, you end up with 2 ≡ 47 (mod 15) as a true statement.
IMHO it is better to introduce first the notion of inverse element. Then you can say that when an element has an inverse element then you can divide. In this way you may have a self contained video where you can explain why you need gcd(k,n)=1. Just my two cents :)
holy crap dude 110k?! nice!! you've come a long way
pennyy
Yea, I still can't believe it!! Thank you pennyy!!! And thank you so much again for your art work.
of course :)
I don't know about English notations but in France we usually write a=b[n] instead of a=b (mod n), that's faster lol.
drapsag91 😃😃😃😃 nice!! This will save so much time from writing the mod
:D
I have also see:
a ≡_n b
where "_n" is a little n as subscript
Better yet, declare at the start that you're in the modulu group/ring Zn and then just write a=b like the regular notation.
If you know the context It is good, or if you are used to it, but I would find it confusing at the start atleast
Thanks to talk about this subject, I study congruences at school and sometimes it's a bit difficult, so you help me to understand with your videos, thanks and continue you're great !
hey man, everything ok?
Heyyy AndDiracisHisProphet, yes, things are okay. I just haven't had a chance to make videos. Thanks for asking!!
good to hear
AndDiracisHisProphet thanks for the support, as always!
"support"^^
I just love the way he speaks "Mod" , :-)
Instablaster.
Can you please post soon? I can't go this long without your amazing videos.
Also, can you do some more of those live videos? Those are always fun.
Brilliantly explained !
Your video entitled "weekly brilliant problem#3, solving integral by symmetry, IIT JEE mains 2015" has over 20k views, and you promised to do the integral in the vid when it reached that number. Can't wait ;)
Yes. I noticed. But Oon Han did that for me already : )
when will you make another video?? love them to bits!
You are a legend. Clearly explained, straight to the point. Good vid
Can you please do Bode plots ?
Looking forward to your next "This is IT!" videos after your creative break... ;-) Are you planning for some complex analysis? More videos with Dr. Peyam?
Great video, Legend!!
Works iff divisor and modulus are relatively prime.
Note: Whenever you see, a ≡ b (mod n), you should immediately think, a-b = mn, where m is an integer. This is the *definition* of congruence.
That will make this result more obvious.
Fred
Wrong! It might work even if divisor and modulus are not relatively prime.
No it isn't! The question is, when does the following implication hold?
ka ≡ kb (mod n) ⇒ a ≡ b (mod n)
If GCD(k,n) > 1, there will be some examples (a,b), that make the implication false; the "division" won't work.
The proposition is that for some particular k and n, the congruence follows for *all* (a,b). And this is true only if k and n are relatively prime.
If you narrow it down to individual (a,b) cases, you can just use a = b, and it will always work.
Fred
Your above comment: "Works iff divisor and modulus are relatively prime." is *wrong* since, for example: 2x5=2x1(mod4) and 5=1(mod4) are true, but 2 and 4 are not relatively prime.
So, if ka ≡ kb (mod n) and a ≡ b (mod n) are true for some a and b, this doesn't necessarily mean that k and n are relatively primes.
What works iff divisor (k) and modulus (n) are relatively prime, is preserving congruence when dividing by k, *regardless of a and b.*
Sorry if I didn't spell that out. I guess I assumed that since it should be obvious that you can always make the division work when a=b, the question becomes, what must you demand of k and n to make it *always* work?
For your example, k=2, n=4, division by 2 will not work for a=5, b=3
2·5 ≡≡ 2·3 (mod 4), but 5 ≡/≡ 3 (mod 4).
The statement I was trying to make, in full formality, is:
∀(a,b) [ka ≡ kb (mod n)] ⇒ [a ≡ b (mod n)] iff GCD(k,n) = 1
where a, b, k, n are integers, and k ≠ 0.
Fred
Now it's clear. Thanks
sir, can you please make a video on derivation of a formula to find product of all binomial coefficients
Could you show and demostrate the derivative and integral of x!, please? Thank u.
@blackpenredpen I think you are missing a video with the square trick that can be found in the integral of e^(x) * e^-0.5(x^2) / sqrt(2pi)
We want new videos! Almost two weeks since this one.
Hi Crazy Drummer, thank you for asking! Hiii! Videos will come soon.
the symptoms are starting... I cant go this long without a bprp video
What about dividing the modulo number by gdc(k,n)? Like 30 = 42 (mod 4) , 5=7 (mod 2).
Integral(log(ax^2+bx+c)dx)
Proxava Complete the square inside the log, by rewriting the whole log as log(a)+log((x+b/2a)^2 + c-(b/2a)^2)... I think it is now obvious enough to solve.
Please do proof by induction. Like 3n
Not possible
Actually, adding one point, we can actually divided anything somehow but with a little condition.
ka=kb(mod n), then we have
[k/gcd(k,n)]a=[k/gcd(k,n)]b(mod (n/gcd(k,n))) and hence "a=b(mod (n/gcd(k,n)))" which is easy to see by just manipulate the equation somehow (k(a-b)=sn,for some integer s and this implies (a-b)|(n/gcd(k,n)).
Things in this video is just only a special case of this.
It has been about a month, everything okay? hope you get all of these stuff and work done as quickly as possible
At least there are many vids which people could watch :)
I am okay, thank you for asking. : )
Hi , I have a suggestion for 2 videos
First , please make a video about the origin of sine and cosine , I mean how did their function came from
Secondly , how to get sine and cosine of π
Thx buddy , u r one of the best😍
can you make videos on logical (bitwise) arithmetics that will also help some of us.
Hey blackpenredpen YAY, I’m a high school math student and I wanted to know: what is 1 / (sqrt(1+sqrt(1+sqrt(1+sqrt(... infinitely? Thanks!!!
2/(1+√6) or 2/(1_√6)
plz ans, is it possible that 2a^2+2b^2 = c^2 where a,b,c are natural number.
Hey could you make a video on finding the derivative of a factorial function, namely y = x!
Hello blackpenredpen
I don't know how to solve the matrix problems.
Could you make some video to teach about matrix lessons?
I hope you help me...
thanks
How do you find x if you have cos(x) = x ?
It's been over one month since last video, have you started your summer vacation early?
in the first case we can divide by 6 if we also multiply the 4 in mod by 2
I have general simplification for ka congruence kb (mod n). I don't know whether anyone else has do it (make it patent) but it's quite easy to do so.
By definition, a congruence b (mod n) means there exists at least one integer c which makes a - b = cn. So ka congruence b means there exists at least one integer c which makes ka - kb = cn. Of course, we also can write it as k(a-b) = cn
Let gcd(k,n) = g.
-> ka congruence kb (mod n)
-> There exist an integer c, which makes ka - kb = cn
-> k(a - b) = cn
-> (k/g)(a - b) = c(n/g) (by dividing both side with g, the result is guaranteed to be integer)
-> (k/g)a - (k/g)b = c(n/g)
-> (k/g)a congruence (k/g)b (mod n/g)
Of course, this simplification only helps if gcd(k, n) not equals to 1. If it equals to 1, your method works better.
In the example above: 30 congruence 42 (mod 4)
-> 6*5 congruence 6*7 (mod 4)
-> gcd(6,4) = 2
-> (6/2)*5 congruence(6/2)*7 (mod 4/2)
-> 3*5 congruence 3*7 (mod 2)
-> 15 congruence 21 (mod 2)
Hey BPRP, loving the variety at the moment. Is it possible that you could do some statistics?
Matthew Stevens like what in stats?
When the gcd is not 1 you can simply divide the modulo by the gcd. So gcd(6,4) is 2 so you can divide the 4 in mod 4 by 2 to get mod 2. So 5 = 7 (mod 2) which is true.
Hey bprp! Can you integrate sin (2*pi*sqrt (1-x^2)) ?
Thanks👍
hey bprp, i tried to integrate sqrt(sinx),like u did with tanx (similar steps) but online there are too many answers and some say that its not possible.plz help.I can tell u the steps if u want.thnx.
Only when you have an integral domain
A commutative ring with no zero divisors so that cancellation law holds
Yup, we will get there one day. I am trying to start from the basic first.
Why don't you upload now?????
Hey...ur channel is great...I just loved it..!! Please make a video finding cube root of 20√2 + 12√6
It’s been a whole month since this upload.. Pls come back ):
So if a == b (mod c), then a/k = b/k (mod c) if gcd(k, c) = 1?
Oon Han if a/k and b/k are still integers. Then yes.
Oon Han if a/k and b/k are still integers. Then yes.
The thing is that Zn is not Q. You can't just plug in 1/m as if it's the inverse of m.
Your rule of thumb (gcd of q,n must be 1) can be understood in light of the fact that the multiplicative group of Zn contains the elements that are relatively prime to n. Or at least that's the case for Zp (with p being a prime number), not sure about Zn in general (I've been taking a group theory course this semester and there your modular groups are almost always modulu a prime..).
If you plan to proceed with doing NT videos and don't want to write (mod n) every time, you can write small letter n just after the conruence sign: 12 Ξn 5, iff n = 7
It saves A LOT of time, I give you my word!
Can i get help with this question? "if a + b = k what is the maximum value of a^b. (a and b expressed in terms of k)"
somebody know how to integrate e^e^x
Can so you solve integral e^x^2 x'3 please bro??
whats up with the modulo videos recently? :-D
also have a nice day
ansishihi I like the topic.
blackpenredpen, why havent you uploaded in a while?
It's been nearly 2 weeks.
Hiii! Thanks for asking! Videos will come soon.
I miss you
It's not a *necessarily* condition to have the number that you divide by and the "mod" number to be relatively prime:
For example:
10=2(mod4)
Divide by 2
5=1(mod4)
Although 2 and 4 are not relatively prime
Yeah, in the video he does a proof with two assumptions: that *k·b≡k·a mod(n)* AND that *gcd(k,n)=1* ; then he says the conclusion would be that *a≡b mod(n)* is true. So he is proving the conclusion once he stated two assumptions, therefore he is just proving that *a≡b mod(n)* is always true if those two assumptions are made. But that's not the same as saying that the conclusion is true if and only if those assumptions are made. (The grass can be wet because of the rain, but it could be wet cause we watered it). So, in order to really proof that the assumption *gcd(k,n)=1* is strictly necessary, I think the proof would have to be like this: we assume that *k·a≡k·b mod(n)* AND that *a≡b mod(n)* , so we claim that, with those two assumptions, then *gcd(k,n)=1* is always true. (Which isn't, since you just showed a counterexample).
I Know that he proved the *sufficient* condition (which is fine).
But what I'm referring to is his claim at 1:50 that "if gcd is not 1 then we are not allowed to divide" (which is not true, by my counter example)
Heliocentric I should have said "be careful if the gcd isn't 1".
so a simple way to think about why you have to be so careful is to consider the reverse. note, I will use "=" as if it's the congruence symbol.
So let's say we start out with a false statement, such as 5 = 8 (mod 4)
Now, what can we do to make this true? well, if we multiply both sides by 4, then no matter what we started with, they will be congruent to 0 (mod 4).
That's because with integer n, 4n is always divisible by 4.
So if we were to multiply our previous equation by 4, we'd get 20 = 32 (mod 4) which is indeed true. Since we can make something true from something false using multiplication, doing the opposite, i.e. division, we can make something false from something true.
Great one anyway
Could you please prove its converse? If ka is congruent to kb mod n, and gcd(k,n) ≠ 1, then, a is not necessarily congruent to b mod n
That would just be a proof by counter example. 6=30 (mod 12) and 6 shares factors with 12. 1 isn't congruent with 5 (mod 12). After what the first statement means is 6=30+12a. Dividing both sides by 6 means 1=5+2a. That means exactly 1 is congruent with 5 (mod 2), a strictly weaker claim.
21=69 (mod 4) on the other hand means 21=69+4a. Dividing both sides by 3 means 7=23+4(a/3).
Bruno Andrades yup. I did that already in the beginning.
2x = 6(mod 20)
x= 3 (mod 20)
gcd(2,20)=2
is this valid?
I think the reverse is true, just without gcd(k, n)=1
a=b (mod n)
=> a-b = 0 (mod n)
=> n | (a-b)
=> n | k(a-b), since multiplying by k doesn't change the fact that n|(a-b)
=> k(a-b)=0 (mod n)
=> ka = kb (mod n)
So,
a = b (mod n) => ka = kb (mod n)
ka = kb (mod n) and gcd(k, n) = 1 => a = b (mod n)
Didn't realize the reverse was just multiplication. My mind goes different places and doesn't realize its just division and multiplication when working with these things.
Pls tell me if we can evaluate tan^-1(i)
First, you set that expression equal to some variable (Z, in this case). Then you apply the original tan function on both sides of the equation, giving you the following thing:
i=tan(z)
Next, you can express the tan function in terms of exponentials using the euler's formula [e^(ix)=cos(x)+isin(x)]. By doing that, you can set that expression to the value you're looking for (in this case, i). Finally, solve for Z.
Bprp did a similar thing with sin(z)=2. Look up that video if you want a more graphic explanation.
Blackpenredpen + ime = you are a radical teacher
i=√-1
what happened to your videos?? did you retire..?
hope to see more :)
Hi Raziel, I haven't retired yet... Videos will come soon.
You`re the best :)
Oscar David Alarcon thank you!!!
But it us true that if ka=kb (mod n) then a=b (mod n/gcd(n,k)) ISN'T IT?
Are you still active?
Find three consecutive integers such that their product is equal to their sum.
Mohamed Alnajjar 1,2,3
I miss your videos so much...
sqrt(-1) miss you guys (BlackPenRedPen and Dr. Peyam' Show)😘😘
BTW Could you do (i)! And (i)!! Please 😍
If I made a mistake tell me where:
n/2=n*1/2
n/3=n*1/3
n/m=n*1/m
Now where the mistake comes in if there is one;
0/0=0*1/0, 1/0 is undefined but 0 times anything is 0; therefore 0*1/0=0, which means 0/0=0
Tell me where I messed up cuz mathematicians should've totally figured this out by now.
everything Ok?
Yes, thank you for asking. Things just have been busy for me lately. : )
glad to hear so
Oh so this time you have started to dive into number theory after calculus..... well good job
Actually you can write (mod n) as [n] 🙂
Can you make a video to prove that 1 is not a prime number ?
Trivial but wonderful, I love it.
What is your name?
log0 = ?
do i factorial
Hey, where are you?
Sorry, things are busy for me lately.
So good....
Can you prove this for me please?;
n is a prime number greater than 5. Prove that the value of n^4 has its last digit ending in 1.
I was lost, I dont know how to represent a prime number algebraically
Prime numbers greater than 1 can only end in 1,3,7, or 9. The last digit of a number is equivalent to the number modulo 10. 1^4, 3^4, 7^4, and 9^4 are all equal to 1 modulo 10. (Look up modular arithmetic if you need to).
So goood so gooooooooooood
Thank you!!!!!!!!
If you divide both numbers and the n by the gcd...
Plz help: sec (x)=arccos (x) solve for x
How can one david when one's name isn't David?
Try to start in introduction
u can divide by 4 at the beginning, it's just won't be mod 4 but instead mod(gcd(6,4))
I think it's (mod (4÷gcd(6, 4))), not (mod gcd(6, 4)), although since gcd(6, 4) = 2, the two coincidentally give the same answer in this case.
But consider if you have 28 ≡ 658 (mod 105), and you want to divide both sides by 14. Since gcd(105, 14) = 7, and 105÷7 = 15, you end up with 2 ≡ 47 (mod 15) as a true statement.
ur absolutely right, number theory was last semester :D
Quote the theorems! Gauss I love you
upload plz?
Very nice👍
08:55 I have a theory that squirrels actually teleport through space-time.
What happened to videos :(
Profile picture changed. which means... Come again please, will you?
It's been too long since you posted a video... You're worrying us, sir.
Is everything okay?
Hi Reeta, thank you so much for asking! Things are okay. I will upload videos soon.
blackpenredpen Thank goodness! I'll be intently awaiting the upcoming videos.
Reeta Singh :)
We write [n] instead of (mod n)😐
A corollary of this is if ak = bk (mod n), then a = b (mod n/gcd(k,n))
Miss u, please submit videoooo
Sorry... I will try to make more soon. : )
blackpenredpen I'll been waiting 💕 love u
+
Yaiiii
i am the suprem electric motor
...divide
Sir please help me do this
$ ((x^(x-1)/lnx) + x^(x-1)) dx
Please help me sir,