This is an abstract and difficult concept . However , it has been so clearly explained in this great video .Thank you for your great work ❤

@WrathofMath

11 ай бұрын

Thank you!

Great video. Is any course available for " Foundaton of Mathematics"

it's a great video thumbs up. if i may ask, y do u have replace "a" by "b-1/n"

oh my goodness, you filmed this at 2 am? Thanks for your hard work XD

@WrathofMath

5 ай бұрын

Always on the grind! Thanks for watching!

Hey, I love you. Continue the good work.

@WrathofMath

6 ай бұрын

Hey, thanks! I will, until I turn to dust

can we chose m=a-b and n=2 and work from there?

if anyone's curious about the irrationals if a and b are irrational, atleast one of (a+b)/2 or (a+3b)/4 will be irrational, if both were rational, then (a+3b)/2 would be rational and (a+3b)/2 - (a+b)/2 = 2b/2 = b would be rational, which wouldn't make sense if only one of them is irrational, just do (a+b)/2 if both are rational, do (a + pi(b))/(1+pi), since that is just equal to b - (b-a)/(1+pi) and b-a isn't zero, it's irrational

Keep going 🎉🎉🎉🎉thx a lot ❤❤❤

@WrathofMath

8 ай бұрын

Thank you, I will!

Is there a video on dedekind cuts and their use in forming the reals?

@WrathofMath

3 ай бұрын

I haven't made one yet, but it would be a lot of fun to create such a video - maybe later this year when I have some more time. Would like to come back to producing for the Real Analysis playlist but right now it's hard to afford working on just because it's so niche. Lots more topics will be covered eventually though!

One would have to be dense not to be a fan of Wrath of Math! 😂

@shadow15kryans23

5 ай бұрын

Nice. XD

ooh finaaly its in playlist

@WrathofMath

Жыл бұрын

Yes indeed - and you're early, the lesson isn't even out yet! Hope you find the playlist helpful!

@Aman_iitbh

Жыл бұрын

@@WrathofMath yes its life saving playlist . btw how can we show any number is between two consecutive natural number as u used that m-1

@WrathofMath

Жыл бұрын

We can consider the set of all positive integers greater than na. Then the well ordering principle guarantees a smallest member of this set: m, and then m-1 must be

@Aman_iitbh

Жыл бұрын

@@WrathofMath yes ,but this will proof for all positive integer i think , if na is negative like -4.12 then we need to proof its between -4 to -5 .does well ordering can proof this?

@Aman_iitbh

Жыл бұрын

@@WrathofMath and will u make vedio on well ordering too later ? plz

if x is a real number, prove that -x≤|x| can you prove this

@VS-is9yb

9 ай бұрын

One can just consider separate cases: x is positive, x is zero, x is negative In all these cases the inequality is trivial, and there is no other case.