so amazing bro

i like this so much

I think the first formula is quite easy to use than the short method.

@BrainGainzOfficial Thank you for this video and series; I'm an older student getting back into math who never had experience in proof writing, so I very much appreciate your style and flow of presenting this subject.

Hello sir! May I ask if proof by cases is the same as the choose method? I’m confused about the two. The same goes for the Construction method and direct proofs. Are those two the same as well? Please enlighten me. Thanks.

dude this is extremely funny but very very helpful thank you!

Thanks❤❤

Ok so it's been decided, Anthony Tucci is my actual math professor and I'm just paying to take exams

Best teacher in the universe ❤

To finish the proof you just have to observe that the epsilon>0 is ARBITRARY: If the absolute disfference of two numbers is less than an ARBITRARY epsilon>0, then the two numbers must necessarily be equal.

Brilliant 🎉

Display isnt clear

thank you so much for doing this

This video was great!

What other universities offer a math heavy math education programmes?

Thank you. I have watched 6 videos on this topic, and this is the first one that explained these concepts in a way I could understand. Much appreciated!

If you think this is hard, try discreet math.

That’s not the TI inspire, and no, you would not be satisfied with a five dollar calculator. You obviously don’t know much about math otherwise you would know the difference.

thanks for this

At Q 6.b i'm just confused at the period being 2pi/3 but you drew half of it, shouldn't there be a part of the graph going under the midline as well? Everything else was very helpful thanks!

The explanation is clear and straightforward to understand. Thank you so much ! 😊

I adore u , you're my hero!

This is a really positive video, I too am a mathematics student, and recently I've been hit with the pretty common thought in society that studying maths isn't really useful...("should just be an engineer or computer scientist, etc,...".) Seeing you talk about doing what makes you happy and how important that is really hit home to me. Studying mathematics is the future that I want, and it's a future that I see myself being the happiest in. Yes, it is true that we will probably not be the richest, but being rich is truly defined as being happy and proud of the life that you're living. Thank you, Brain Gainz.

I was wondering if this is another valid way to write the proof: Prove the transitive property: If a|b and b|c, then a|c for all positive integers a,b,c. Let’s analyze our first given. We know that a|b. What does it mean for a to divide b? Let l be any positive integer. lℤ+. This means that a times some integer l equals b. Algebraically, we can say: al=b Manipulating this expression to solve for a, we find that: a=b/l Let’s analyze our second given: We know that b|c. What does it mean for b to divide c? Let m be any positive integer. mℤ+. This means that b times some integer m equals c. Algebraically, we can say: bm=c What are we trying to prove? We are trying to prove that a|c. What does it mean for a to divide c? Let n be any positive integer. nℤ+. This means that a times some integer n equals c. Algebraically, we can say: an=c We are trying to prove that a|c. We are trying to prove that an=c where n is some integer. This means we are trying to demonstrate that n=c/a is an integer. We have n=c/a. Recall that c=bm and a=b/l. Let’s replace c with bm. Let’s replace a with (b/l). n=(bm)/(b/l) We multiply bm by the reciprocal of (b/l). n=bm*(l/b) The bs here cancel, leaving us with: n=ml Recall that m is a positive integer. Recall that l is a positive integer. Positive integers are closed under multiplication. This means that the product of two positive integers yields a positive integer. Thus, n must be a positive integer. Since we have confirmed that n is a positive integer, we have confirm that a indeed divides c. Can you let me know whether or not this proof is valid? Thank you :)

repeated this for one hour on an 8sec loop now it's burned in my brain

I’m getting 16.206 for #71. How am I barely off? Only typing the right side of the equation in the calculator.

2:36 🤣 I'm definitely gonna remember to foil in the future LOL

I've learned far more from this review than I have in class😅. This man has the best most in depth explanations for every problem. Great job 💪

Thank you so much!

pretty good singing voice 😂

Bar far the best review on youtube for college level algebra

AMAZING

Sir brian help on that one

How can you solve this this wuestions:Define a relation on R square\(0,0) by letting (x1,y1)equivalence (x2,y2)if there exists a nonzero real number m such that (mx2,my2).prove that relation defines an eqivalence relation on R square \(0,0).What are the corresponding eqivalence classes ?This equivalence relation defines the projective line,denoted by P(R), which is very important in geometry.

Great

Excellent

A relation R on a set of integers defined as aRb if and only if (a - b) is divisible by m, is a +ve integers. Show that R is an equivalence relation.

Thank u

Amazing it take me time until I came across this video 😂😂😂😂😂❤❤

great video! Your introduction is really clear and insightful. A lot of times in math, we're just thrown problems and are expected to know the set-up every time with no context. This is where the beauty of math comes. My only contribution to the video would be to show how vertex form would look like...etc. as that is becoming a standard in school, but I honestly follow this method so much better. I think you mentioned that this was a geometric approach, so maybe I'm being too algebraic in my thinking. Thanks for the really great videos. I enjoy your complex polar-coordinate video as well, and it is a new concept for me that I am really intrigued by.

Epic intro

everyone: calc 3 is easier mfs that fw algebra: 🤬🤬🤬

swear bro you turned my grades around. God bless you

this is proof by contraposition

Thank you ! I have finals coming up, this will help me.

Shit makes me cry

Absolute legend.

The explanation is really simple Thanks

is the first proof biconditional? or are any of these proofs biconditional?

Thank you chairman