awesome

Thanks for making the time for this vid. Really helped💪

An alternate non contradiction based proof: Suppose A, B, and C satisfy the premise A-B \subseteq C. Consider an arbitrary x in A-C, so that by definition, x is in A but is not in C. Since x is not in C, x can not be in any subset of C, so in particular, x is not in A-B. Recalling that x is in A and pairing with the new fact that x is not in A-B, it must be the case that x is in B. Since x was arbitrary, we have shown A-C \subseteq B.

Do you have a proof for the multiplication principle in counting? If so I’d love to see it. Thanks for making your proofs!

Here before you're famous jokes aside, your explanations are really clear! Many teachers will only tell students certain formulas but won't explain how they're are derived

2:11 Repetition does *NOT* make an explanation. Either it's already clear since you already said it, either it's confusing again.

1:34 Shouldn't it be XOR rather than OR, based on what you are trying to explain?

பொருள் அந்த மாதிரி இருக்கு வர்மா..!

பொருள் அந்த மாதிரி இருக்கு வர்மா..!

JEE students attendance here

posted 23 hours ago and has 23 views

what if johnny is partially in the kitchen

👍

Thanks, nice clear video

Hey, this is actually a good video, it's easy to understand, can you do me a favor of making a lesson for derivatives? either u can or not I'll sub for notif

I don’t think it was actually a mistake. Defining a|b as “b=ak for some integer k” is perfectly valid. Your argument for why it’s wrong is using a very specific solution technique that is not required. It’s similar to saying that a=√b means “a>0 and a•a=b”. Some numbers satisfy this relation, like a=3,b=9 or a=1.5, b=2.25 and some numbers don’t, like a=1,b=2 or a=9,b=0. And in my example, a=5, b=-37 also does not satisfy this relation, but we don’t have to redefine it as “a>0 and a•a=b and if a=5, then b is not -37”. They already don’t satisfy this relation. Similarly, you don’t have to say b>0 in my example, even though no pair (a,b) with negative b satisfies it. But we don’t have to change the definition because of it. Even though one of the solution techniques can be to calculate √b and compare it to a (which would break if b is negative), this doesn’t mean we have to set restrictions based on what my calculator can do. If that was the case, we must also exclude numbers above 10^60, because my calculator cannot work with them. In your case, expression “0|a” is not malformed, it’s just false. That doesn’t mean you have to exclude it. And the equation a=0k is a valid algebraic expression, that can be solved numerically. The only issue is that you cannot use the technique of dividing by the number in front of k. But there are other solution techniques that work perfectly fine. Nothing in the original definition dictated that you have to solve it by dividing. I can prove that 2 does not divide 3 without using your technique. 3=2k. The left side is odd, the right side is even. Therefore, they cannot be equal. Therefore, no integer k exists. Therefore, 2 does not divide 3. And for 0 and 2. 2=0k. The left side is 2, the right side is 0. Two is not zero. Therefore, there is no integer k. Therefore, 0 does not divide 2.

Shouldn't it be true in a factorization ring if we exclude the unit group of it? Since the only numbers this wouldn't work for (my intuition at least) would be units. 1 and -1 are the units of Z.

I would have liked to see a proof of the 2^n

But I thought 1x1=2. 🤣

A bit trivial since it's easy to find a counter example, so might be fun to do something like: a != b, a*m = b, a = b*n a*m = b => a = b/m a = b*n => b/m = b*n => 1/m = 1*n => 1 = n*m Then instead find the case where this is true, namely n = m = -1

I think a better proof would be to write a = nb and b = ma for a,b,n,m integers. So... ab = nmab. So 1 = nm. n and m are plus or minus 1.

The theorem is true in N, but not true In Z

What if a and b are natural numbers? Contrapositive?

Can you do videos about math olympiads problems

But, is it true for even integers >0?

x^2 -6x will always give an even number because x is an even number but the equation x^2 - 6x + 5 will be odd as an odd number is added to an even number

Amazing

n even -> n^2+2n even, n odd -> n^2+2n odd

Do you have to explain distribution law?