I graduated college last year with a degree in math and the 1 book i kept was baby rudin from my analysis classes. Kind of fun because rudin wrote this at the univeristy of wisconsin and that is where i graduated from
When I read that second page of Ruding for the first time, I was baffled by the following "how did he figure out that particular formula for finding rationals closer to square root 2?" The formula given by Rudin obviously work, but its appearance seemed unmotivated and kind of magic to me. It took me a while but I figured out that we can consider the function f(x)=x^2 - 2 as a function from Q to Q. Now, for every rational number p, we can consider the unique line g that goes through the points (2, f(2)) and (p, f(p)). We can think of the function g also as a function from Q to Q. It is not hard to see that the line g has to cross zero at a rational number (this is because 2, p, f(2) and f(p) are rational numbers... so the slope of the line g and its intercept are rational number). Now, when you solve g(x) = 0, you get Rudin's magic choice... If you do a picture of what this process does, you'll see why that choice should work. By the way, this is similar to the method used to prove the formula for all Pythagorean triples... Actually, when I saw the proof for pythagorean triples, it finally clicked to me how Rudin could've have found the magic formula
Nice work! Rudin's approach here is actually playing off of Dedekind (but is a bit more refined). Dedekind had a little bit more of a complicated function where you had to ultimately factor a cubic, but the results are the same.
That's interesting... I had no idea Rudin took the idea from Dedekind... now I am curious about the function that Dedekind used... thanks for pointing that out!
This is exactly an example that Q is not complete (under the ordinary norm), and is much more insightful and non-trivial. But personally I think it's a shame that Rudin did not state enough of this. He only mentioned that Q is a non-example of complete metric space. In the exercise he composed an exercise on the completion of a metric space then it is followed by the question "What's the completion of Q?" (Refuse to elaborate and leave.) In my opinion the fact that the completion of Q is R should be very explicitly stated and frequently recalled, because it is one of the most examples in analysis to bear in mind. If the student doesn't think that exercise is important or simply did not solve it correctly, then the student will have some trouble.
For an analyst returning to the text, Rudin can be a nice terse treatment of the subject. It gets you to important results fast, but he notoriously glosses over a lot of really important points as you say.
@@JoelRosenfeld for a clueless student, at the point of chapter 3, the student may still have issues with the concept of metric space, hence digging out the closure of Q through an exercise on metric space should be more than challenging imo. I still remember that I could not get the point of "treating A as a subobject of B if there is a (structure preserving) injective map", most notably treating Q as a subfield of the space of Dedekind cuts.
This was all discovered with mathematicians kicking and screaming through the 19th century. You just have to keep asking “why?” And eventually you end up here.
The problem with Rudin’s proof is that for someone who’s never seen a proof (ie. a beginner just taken out of calculus), is pretty useless, because that function seems to be taken out of hat. The whole book is written in the same way, that’s what makes it a bad book to learn from for beginners.
It’s definitely a poor book to start out with especially for self learning. It’s approachable if you have someone there to help then it can be a lot more workable. I hope that these videos can also help fill these gaps.
Consider the function f defined by f(x) = x^2. We want to solve for x when f(x) = 2. We make an initial guess x0 = 2 and observe f(x0) = 4. Now we make our next choice x1 = p ≠ x0, which yields f(x1) = p^2. Now, apply the secant method two these two points to obtain x2. Further, let's use the symbol q instead of x2. Now...
What we are doing here is exploiting a property of rational numbers. If a number is rational, then it can be written as a/b. If an and b had any common factors, we could cancel them in the fraction, and ultimately we get two new number a' and b' with no common factors. So we express the root of 2 as the ratio of that particular a' and b'. The proof proceeds to show that these a' and b' would have to have common factors if root of 2 is rational. But that contradicts our construction.
I've just started Baby Rudin out of sheer interest. You saved my night, thanks a lot. I appreciate it.
I’m glad to hear it! Let me know if there is anything in particular you’d like me to cover.
As someone once said, analysis is proving trivial things until they are not so trivial anymore.
Absolutely! The first chapter of Rudin really exemplifies that.
I graduated college last year with a degree in math and the 1 book i kept was baby rudin from my analysis classes. Kind of fun because rudin wrote this at the univeristy of wisconsin and that is where i graduated from
When I read that second page of Ruding for the first time, I was baffled by the following "how did he figure out that particular formula for finding rationals closer to square root 2?" The formula given by Rudin obviously work, but its appearance seemed unmotivated and kind of magic to me.
It took me a while but I figured out that we can consider the function f(x)=x^2 - 2 as a function from Q to Q. Now, for every rational number p, we can consider the unique line g that goes through the points (2, f(2)) and (p, f(p)). We can think of the function g also as a function from Q to Q. It is not hard to see that the line g has to cross zero at a rational number (this is because 2, p, f(2) and f(p) are rational numbers... so the slope of the line g and its intercept are rational number). Now, when you solve g(x) = 0, you get Rudin's magic choice... If you do a picture of what this process does, you'll see why that choice should work.
By the way, this is similar to the method used to prove the formula for all Pythagorean triples... Actually, when I saw the proof for pythagorean triples, it finally clicked to me how Rudin could've have found the magic formula
Nice work! Rudin's approach here is actually playing off of Dedekind (but is a bit more refined). Dedekind had a little bit more of a complicated function where you had to ultimately factor a cubic, but the results are the same.
That's interesting... I had no idea Rudin took the idea from Dedekind... now I am curious about the function that Dedekind used... thanks for pointing that out!
This is exactly an example that Q is not complete (under the ordinary norm), and is much more insightful and non-trivial. But personally I think it's a shame that Rudin did not state enough of this. He only mentioned that Q is a non-example of complete metric space. In the exercise he composed an exercise on the completion of a metric space then it is followed by the question "What's the completion of Q?" (Refuse to elaborate and leave.) In my opinion the fact that the completion of Q is R should be very explicitly stated and frequently recalled, because it is one of the most examples in analysis to bear in mind. If the student doesn't think that exercise is important or simply did not solve it correctly, then the student will have some trouble.
For an analyst returning to the text, Rudin can be a nice terse treatment of the subject. It gets you to important results fast, but he notoriously glosses over a lot of really important points as you say.
@@JoelRosenfeld for a clueless student, at the point of chapter 3, the student may still have issues with the concept of metric space, hence digging out the closure of Q through an exercise on metric space should be more than challenging imo. I still remember that I could not get the point of "treating A as a subobject of B if there is a (structure preserving) injective map", most notably treating Q as a subfield of the space of Dedekind cuts.
Nice video. Keep creating stuff like this.
I’m a Physics student. We never learnt this type of math. I’m surprised at how much math is there behind such a simple thing .
This was all discovered with mathematicians kicking and screaming through the 19th century. You just have to keep asking “why?” And eventually you end up here.
Can you do a course on Markov / Semi Markov / Hidden Markov / Semi Hidden Markov models please.
The problem with Rudin’s proof is that for someone who’s never seen a proof (ie. a beginner just taken out of calculus), is pretty useless, because that function seems to be taken out of hat. The whole book is written in the same way, that’s what makes it a bad book to learn from for beginners.
It’s definitely a poor book to start out with especially for self learning. It’s approachable if you have someone there to help then it can be a lot more workable.
I hope that these videos can also help fill these gaps.
@@JoelRosenfeld one could say it's designed to require a teacher :P
@@girlsinacoma well, here I am, if you have any questions! Lol
Consider the function f defined by f(x) = x^2. We want to solve for x when f(x) = 2.
We make an initial guess x0 = 2 and observe f(x0) = 4. Now we make our next choice x1 = p ≠ x0, which yields f(x1) = p^2.
Now, apply the secant method two these two points to obtain x2. Further, let's use the symbol q instead of x2. Now...
Fun
Thanks! I’m glad you liked it!
Nobody explains in that proof the condition that square root of two has no common factors.
What we are doing here is exploiting a property of rational numbers. If a number is rational, then it can be written as a/b. If an and b had any common factors, we could cancel them in the fraction, and ultimately we get two new number a' and b' with no common factors.
So we express the root of 2 as the ratio of that particular a' and b'. The proof proceeds to show that these a' and b' would have to have common factors if root of 2 is rational.
But that contradicts our construction.