25:04 How is that we can say \forall \epsilon > 0, we have \alpha^2 + \epsilon^2 - 2 \alpha \epsilon \leq 2 ? 22:16 we have a condition on \epsilon that \epsilon < \alpha I think it should be for any 0 < \epsilon < \alpha, \alpha^2 + \epsilon^2 - 2 \alpha \epsilon \leq 2 . In which case, we can't conclude that \alpha^2 \leq 2. Right? What am I missing?
@@gokulakrishnancandassamy4995 this statement is written from LUB axiom, which states that for all \epsilon>0, if alpha is an LUB then, \alpha-\epsilon < x < \alpha. and since \alpha^2 + \epsilon^2 - 2 \alpha \epsilon
sir...if you consider 2^N set can't dat have any element which is very bigger than N...( then how does this statement N is bounded above will be valid )..
sir while proving alpha^2 less than or equal to 2 ,the choice of hepsilon cannot be arbitrary to conclude it.so one should be careful about it so please clarify
I have the same question. We chose \epsilon such that \alpha - \epsilon is strictly positive. So, the statement \alpha^2 - 2 * \alpha * \epsilon + \epsilon^2
@@shashvatshukla oh sorry about that. I just got here, looking for motivation every step of the way. And sorry I didn’t remember the name from the other lecture. Sorry bout that
@@jbm5195 No problem. All the best! Real Analysis comes up all across mathematics, so I think you will only be thankful for the effort you put in now. But don't be afraid to stop if there is a better learning resource for you, I think it is good to learn from multiple sources.
i have a better one, naiming .... (j,k,l,m,) all member of N p and q member of Q p < q p = j/k q = l/m n = k (l+1) now we can write n * p = k (l+1) * j/k =( l+1) * j > (l+1) > l/m = q
@@antoniomantovani3147 your proof is incomplete since you have only proved the statement for rationals, but the Archemedian property is valid FOR ALL real numbers...
19:39 wow. super logic sir. wonderfull. Thanks aloooooooot sir.
@25:00 Shouldn't there be a strict inequality? Cause alpha-epsilon is strictly less than 2.
Sir, what is the difference between Maximum and Supremum
supremum is same as maximum but if it belongs to the set. i.e. if supremum belongs to the set then it is called maximum element.
25:04 How is that we can say \forall \epsilon > 0, we have \alpha^2 + \epsilon^2 - 2 \alpha \epsilon \leq 2 ?
22:16 we have a condition on \epsilon that \epsilon < \alpha
I think it should be for any 0 < \epsilon < \alpha, \alpha^2 + \epsilon^2 - 2 \alpha \epsilon \leq 2 .
In which case, we can't conclude that \alpha^2 \leq 2. Right?
What am I missing?
I too have the same question. Have you figured it out?
@@gokulakrishnancandassamy4995 this statement is written from LUB axiom, which states that for all \epsilon>0, if alpha is an LUB then, \alpha-\epsilon < x < \alpha. and since \alpha^2 + \epsilon^2 - 2 \alpha \epsilon
Superb explanation.
sir...if you consider 2^N set can't dat have any element which is very bigger than N...( then how does this statement N is bounded above will be valid )..
N is NOT bounded above
Does anyone have a link to the paper mentioned? "The role of LUB axiom in real analysis"
wont 1/b-a > 1/b given 0 < a < b?
LUB Completeness Axiom: Foundation for RA. Consequence of it are, Archimedean property of reals, Q dense in R, unique x in R for x^n=y.
sir
while proving alpha^2 less than or equal to 2 ,the choice of hepsilon cannot be arbitrary to conclude it.so one should be careful about it
so please clarify
I have the same question. We chose \epsilon such that \alpha - \epsilon is strictly positive. So, the statement \alpha^2 - 2 * \alpha * \epsilon + \epsilon^2
@@gokulakrishnancandassamy4995 you're right....
He did provide a choice for epsilon (which was 2-alpha^2/(2alpha+1) ) after the proof.
Difference bw n ,N . ????
Sir itshard forme to understand from 29 th minute
Lecture 7 and still going!
I just got here. Did you finish?
@@jbm5195 this was 4 years ago mate, and we discussed already on the other lecture's comments
@@shashvatshukla oh sorry about that. I just got here, looking for motivation every step of the way. And sorry I didn’t remember the name from the other lecture. Sorry bout that
@@jbm5195 No problem. All the best! Real Analysis comes up all across mathematics, so I think you will only be thankful for the effort you put in now. But don't be afraid to stop if there is a better learning resource for you, I think it is good to learn from multiple sources.
Proof of Archimedean property is wrong
It is absolutely correct. In fact, it is BEAUTIFUL!
It is wrong becouse the contradiction part is wrong. Alpha can be equal to n
@@Ayan-rh6gj I don't see how it is wrong? Could you elaborate?
i have a better one, naiming ....
(j,k,l,m,) all member of N
p and q member of Q
p < q
p = j/k
q = l/m
n = k (l+1)
now we can write
n * p = k (l+1) * j/k =( l+1) * j > (l+1) > l/m = q
@@antoniomantovani3147 your proof is incomplete since you have only proved the statement for rationals, but the Archemedian property is valid FOR ALL real numbers...