I don't think it's quite right in the second example. In your example, where Q are the rationals, the set S of rationals S = {q in Q | -sqrt(5) < q < sqrt(5)} appears to be bounded. The bounds themselves are lower-bound = {q in Q | q < -sqrt(5)}, upper-bound = {q in Q | q > sqrt(5)}. However, the least upper bound sup(s), and the greatest lower bound inf(S) DO NOT EXIST for the following reason: For any rational p < -sqrt(5) in Q supposed as the infima of S one can always find another q in Q where q > p, therefore there is no infima. And similarly for there being no suprema, for any m > sqrt(5) in Q supposed as the as the suprema, we can always find another n != m such that n < m in Q, therefore contrary to our supposition there is no suprema for this set.
DivisionbyZer0 i think there is no upper bound, because there are many infinitely rational number greater than sprt5, as what the video said, but still have sup???
+DivisionbyZer0 Isn't that exactly what he said? He stated that you could say the upper bound is root 5 - 1/10 for example, but that wouldnt be true since you could put 1/100 inbetween those 2 numbers and get a number that's still inside of S. To generalize it, you could put any n from N 1/n and it would still be in between
why cant an upperbound be any rational number greater than root 5? Surely they would still satisfy the definition of an upperbound, i.e. b >= x for any x belonging to set S?
Any number that is greater than or equal to all of the elements of the set is an upper bound. So correct me if am wrong saying that root 5 can be considered as an upper bound.
I think the reason is, root 5 is irrational number but upper bound has to be rational number since u.b. is in set S, set S in a set of rational num, this can be observed from the beginning. altho we surely have smth less than root 5 and within set A, but it can be any number( infinite possibility as long as it is rational number), HOWEVER, if we recall the def, for all x in set A which cant be greater than that upper bound, even if we work out an upper bound, some number x > upper bound still exists since possibility of rational number is infinite long( but def says no x can be greater than upper bound). this is something more important, but he didnt mention in this video. Many may realise some rational number in set A truly exist but cant be counted otherwise something x will be greater than that u.b. This contradicts the definition and cant let upper bound exist.
Absolutely not. That wouldn't be a very insightful question. Anything above sqrt 5 doesnt satisfy the original conditions of the question so it's irrelevant. He said this to point out the incompleteness of the rational numbers
Nice video, i have a question, suppose we have some like this. define h: [0,1] x [0,1] --> R by h (x, y) = 2x + y. the question is determine sup h (x,1) and inf h (1,y). note that the question is already solved, the answer is 3 and 2, but i will like to know step by step guide to the solution. could explain or make two mins video for this please ?
Thanks for the clips. They help a lot ^^
I don't think it's quite right in the second example.
In your example, where Q are the rationals, the set S of rationals
S = {q in Q | -sqrt(5) < q < sqrt(5)}
appears to be bounded.
The bounds themselves are
lower-bound = {q in Q | q < -sqrt(5)},
upper-bound = {q in Q | q > sqrt(5)}.
However, the least upper bound sup(s), and the greatest lower bound inf(S) DO NOT EXIST for the following reason:
For any rational p < -sqrt(5) in Q supposed as the infima of S one can always find another q in Q where q > p, therefore there is no infima.
And similarly for there being no suprema, for any m > sqrt(5) in Q supposed as the as the suprema, we can always find another n != m such that n < m in Q, therefore contrary to our supposition there is no suprema for this set.
DivisionbyZer0 i think there is no upper bound, because there are many infinitely rational number greater than sprt5, as what the video said, but still have sup???
But I think the definition of an upper bound is a number M, in this case in Q, such that for all a in A, a
DivisionbyZer0
noo, it still has... coz supremum is just the least upper bound, but it may not be inside the set..
+DivisionbyZer0 Isn't that exactly what he said? He stated that you could say the upper bound is root 5 - 1/10 for example, but that wouldnt be true since you could put 1/100 inbetween those 2 numbers and get a number that's still inside of S. To generalize it, you could put any n from N 1/n and it would still be in between
why cant an upperbound be any rational number greater than root 5? Surely they would still satisfy the definition of an upperbound, i.e. b >= x for any x belonging to set S?
Any number that is greater than or equal to all of the elements of the set is an upper bound. So correct me if am wrong saying that root 5 can be considered as an upper bound.
I think the reason is, root 5 is irrational number but upper bound has to be rational number since u.b. is in set S, set S in a set of rational num, this can be observed from the beginning. altho we surely have smth less than root 5 and within set A, but it can be any number( infinite possibility as long as it is rational number),
HOWEVER, if we recall the def, for all x in set A which cant be greater than that upper bound, even if we work out an upper bound, some number x > upper bound still exists since possibility of rational number is infinite long( but def says no x can be greater than upper bound). this is something more important, but he didnt mention in this video. Many may realise some rational number in set A truly exist but cant be counted otherwise something x will be greater than that u.b. This contradicts the definition and cant let upper bound exist.
Since S is a subset of the rational numbers, the upper bound of S has to be an element in the rational numbers, which root 5 is not.
Thank you for great help with these!
Bro are you alive?
You just earned yourself one more subscriber
Tell us way to find sup by using epsilon for Other type of question.
These us an inspiration app thanks
Thank God for your life. Can I have a CD for other lessens in mathematical analysis
Why x^2 < 0 has no infimum and supremum? While x^2 < 11 has it in previous video? Please I need your answer...
Cause he defined x to be an element of the rational numbers here. In the previous video, x was in the reals.
i think the question should be " what is the smallest rational number that is greater than sqrt5" something aint right
Absolutely not. That wouldn't be a very insightful question. Anything above sqrt 5 doesnt satisfy the original conditions of the question so it's irrelevant. He said this to point out the incompleteness of the rational numbers
is there any vidoe u have done on real analysis?
good explanation
☺☺☺☺
you are good sir.
You are very welcome! I am glad I could help
you are a legend
These is an inspiration app thanks
You are more than welcome
You are video is nice, but it is represented by using integers.
What is Supremum and infimum for set S={1/m + 1/n | m,n belongs to Z^+}
Thank you thank you :-D
Nice video, i have a question, suppose we have some like this. define h: [0,1] x [0,1] --> R by h (x, y) = 2x + y. the question is determine sup h (x,1) and inf h (1,y). note that the question is already solved, the answer is 3 and 2, but i will like to know step by step guide to the solution. could explain or make two mins video for this please ?
You are welcome :D
Thanks u sir