The proof is a bit misleading as it uses an additional property of the integers that is not mentioned here. Namely that for any real number x there exists a minimal integer k such that k - 1
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
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
@@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?
Same question here. I figured that showing that m is less than or equal to a + 1/n is enough to show that m/n is less than b. Did you every find an answer?
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 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.
The proof is a bit misleading as it uses an additional property of the integers that is not mentioned here. Namely that for any real number x there exists a minimal integer k such that k - 1
oh my goodness, you filmed this at 2 am? Thanks for your hard work XD
Always on the grind! Thanks for watching!
This is an abstract and difficult concept . However , it has been so clearly explained in this great video .Thank you for your great work ❤
Thank you!
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
Hey, I love you. Continue the good work.
Hey, thanks! I will, until I turn to dust
Great video. Is any course available for " Foundaton of Mathematics"
can we chose m=a-b and n=2 and work from there?
ooh finaaly its in playlist
Yes indeed - and you're early, the lesson isn't even out yet! Hope you find the playlist helpful!
@@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
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
@@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?
@@WrathofMath and will u make vedio on well ordering too later ? plz
Keep going 🎉🎉🎉🎉thx a lot ❤❤❤
Thank you, I will!
it's a great video thumbs up. if i may ask, y do u have replace "a" by "b-1/n"
Same question here. I figured that showing that m is less than or equal to a + 1/n is enough to show that m/n is less than b. Did you every find an answer?
Is there a video on dedekind cuts and their use in forming the reals?
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! 😂
Nice. XD
if x is a real number, prove that -x≤|x| can you prove this
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.