Basic Mathematics - 2019-12-20

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4:59 AM

@user21820 I read in Abbott's analysis textbook, "A rigorous study of
the natural numbers and the theory of sets is certainly recommended, but only
after we have an understanding of the subtleties of the real number system."

I can't think of any reason why "only after we have ...". Any idea?

5:51 AM

12 hours ago, by user21820

Previously you stated the answer without justification. Now you should justify it, by explaining what each additional factor of 10 does.

@yh05 I suppose Abbott has a similar view to me; there is little point going on to set theory until we have mastered PA and basic real analysis. Not everyone holds this view, but I think it is pedagogically correct.

Yuvraj Singh...

@user21820, I am still figuring out how do I justify that, how a factor 10 changes the answer.

@YuvrajSingh... There are many ways to do it, and I keep telling you to look back at what you did before:

2 days ago, by user21820

Dec 12 at 12:42, by yuvraj singh

9.9.9.9 mod 10 =(-1).(-1).(-1).(-1) mod 10=1

But it seems like you can't be bothered.

Yuvraj Singh...

10^n mod 11=10.10^n-1 mod 11=(11-1).10^n-1 mod 11=-10^n-1 mod 11.......

@user21820

@YuvrajSingh... You got one right idea, but you failed to put the right brackets so what you wrote is wrong.

Yuvraj Singh...

OK I will rewrite it.

5:58 AM

Look, everything I try to teach you is relevant. If I teach you about precedence rules, I expect you to learn it, not forget it just one week later.

Yuvraj Singh...

Oh yes.

Also, if I teach you the definition of exponentiation, and it has two cases, I also expect you to use it properly.

So, try to express your correct idea in correct statements.

@yh05: By the way, I must emphasize that for real analysis one cannot have merely the axiomatization of the reals, since that does not provide any means of constructing any set of reals, which is needed to use the supremum axiom. I mentioned this before:

Yuvraj Singh...

10^n mod 11=(-1).(10^n-1) mod 11=(-1).(-1).(10^n-2) mod 11.............@user21820

Jul 16 at 7:27, by user21820

There are theorems with applications to the real world that rely on the existence of a structure like the reals. But that does not imply that the reals themselves are needed to have physical meaning. For example, almost all real analysis can be proven in a system that assumes PA (the first-order theory of naturals) plus an ordered field (to represent reals) such that any bounded sequence of reals has a supremum.

Jul 16 at 7:32, by user21820

Within a suitable set theory, the last is equivalent to "any bounded set of reals has a supremum". But typically we only need "any bounded sequences of reals that is encoded as a real has a supremum". Here a real is encoded as a Cauchy sequence of rationals, which can be encoded as a func(N,N), so a func(N,real) can be encoded as a func(N,func(N,N)), which can be encoded as a func(N×N,N) and hence a func(N×N).

Jul 16 at 7:39, by user21820

Thus any real can be decoded into a sequence of reals. Let's call these internal sequences of reals. Thus "any bounded internal sequences of reals has a supremum" can in fact be stated as a first-order sentence about reals and naturals. This is unlike the supremum axiom stated in the wikipedia article, which is not first-order and so requires an ambient set theory.

Jul 16 at 7:43, by user21820

The fact is that almost all real analysis can be carried out in PA for N plus ordered field R where every internal sequence from R has a supremum. And this system is essentially as weak as a system called ACA, which has a countable model. This model essentially comprises the naturals plus all programs that can each use some finite Turing jump, where a real is encoded as a program that computes a Cauchy sequence of rationals converging to it. Note that this model includes the computable reals.

Yuvraj Singh...

(-1).(-1)............n times mod 11

@user21820

6:11 AM

Clearly, the real analysis that you are taught in textbooks relies on an ambient set theory. As I mentioned in our previous conversation, in many cases such as IVT we actually can make do with just supremum for internal sequences.

But that is unwieldy. It is better to have higher-order sorts and permit constructing sets involving them, namely that we have base sorts N,R and for any sorts S,T we have the sort S→T representing the functions from S to T. Then IVT is naturally stated as ∀f∈R→R ( f is continuous ∧ f(0) < 0 ∧ f(1) > 0 ⇒ ∃x∈R ( f(x) = 0 ) ), and we can prove it by simply doing as in the linked message.

So in my opinion the right pedagogical approach would be to have PA for N plus the axiomatization of reals for R plus axioms stating that we can construct any set of reals by a defining first-order property over the language. The reason to permit higher-order sorts is so that we can continue to theorems about arbitrary real functions and operations on real functions.

@YuvrajSingh...: Next time, have the courtesy not to interrupt someone in the middle of saying something when you are not contributing anything relevant.

@YuvrajSingh... Nonsense. Actually read the precedence rules again. Use the search box to find them if you didn't save them somewhere.

@YuvrajSingh... In mathematics, "..." is not rigorous. If you cannot express it without "..." then you don't actually know what it is.

Yuvraj Singh...

6:31 AM

@user21820 I think I answered according to precedence rules only!

@YuvrajSingh... Every time I say you made a mistake, you doubt me. You have made this kind of mistake before and wasted hours of my time. Don't ping me until you have read my precedence rules and found your mistake.

Yuvraj Singh...

6:45 AM

10^n mod 11=10.(10^n-1) mod 11=-(1.10^n-1) mod 11=-(1.(10.10^n-2)) mod 11=-(1.((11-1).10^n-2) mod 11=-(1.-(1.10^n-2)) mod 11= so the last term will be - (1.-(1.-(1..........))))))).. Mod 11 =will depends upon what value of n we choose.

@user21820

Wrong again with a sign error.

@YuvrajSingh...: For the last time, stop pinging me in other rooms just because I don't reply instantly. I am not your maid.

> Don't ping me until you have read my precedence rules and found your mistake.

You have not read my rules, nor have you applied them to what you wrote.

I will not respond for the rest of today, so don't disturb me.

Yuvraj Singh...

See the new one.

5 hours later…

Yuvraj Singh...

12:03 PM

It is for me not include in answering It must have matching brackets and to evaluate it you must evaluate what is within matching brackets first before anything outside. Rule 2: Certain operations have precedence over others. Some have the same precedence. For operations of the same precedence, we specify whether we perform them from left to right or from right to left. I will give the precedence table below.
In each row I list the operations of the same precedence between {}, and specify ltr (left-to-right) or rtl (right-to-left). The higher row has higher precedence.

10^n mod 11=10.10^n-1 mod 11

(11-1).(10^n-1) mod 11=(-1).(10^n-1) mod 11

(-1).((-1).(10^n-2)) mod 11 so that last term will be (-1)((-1).(-1).....)))) mod 11=1.

10^n mod 11=(-1)^n mod 11.

So we can prove it using the induction because you have asked justification.

For n=0 1=1 hence it holds.

And for n=1 10 mod 11=1

$$\begin{align*}
(-1)^{n+1} - 10^{n+1} &= -1\left( (-1)^n + 10^{n+1}\right)\\
&= -\left( (-1)^n -10^n + 10^n + 10^{n+1}\right)\\
&= -\left( \Bigl((-1)^n - 10^n\Bigr) + 10^n\Bigl(1 + 10\Bigr)\right)\\
&= -\left( 11k + 10^n(11)\right) &\quad&\text{(by the induction hypothesis)}\\
&= -\left( 11(k+10^n)\right)\\
&= 11\left( -(k+10^n)\right),
\end{align*}$$
which gives that $(-1)^{n+1} - 10^{n+1}$ is a multiple of $11$, as desired, from the assumption that $(-1)^n - 10^n$ is a multiple of $11$..

@user21820, I think it is the best possible I can do.

Yuvraj Singh...

12:31 PM

@user21820 you can scold me!, but please check my answer, put... If my answer is till wrong.

2 hours later…

2:18 PM

← 2 messages moved from Logic

@YuvrajSingh... You copy what I wrote, but you did not even apply it to what you wrote. As I said earlier, you repeatedly make the same mistakes over and over again, not because you cannot fix them but because you don't learn from them.

Dec 12 at 8:21, by user21820

Now look at where "^" and "−" appear in the table...

Yuvraj Singh...

@user21820 is justification using induction is correct or not?

And I give you one more warning; if you ping me in the other room again just to get my attention, I will block you from this room for some time as well.

7 hours ago, by user21820

> Don't ping me until you have read my precedence rules and found your mistake.

There is zero point going on if you cannot get basic arithmetic correct. And that includes getting the precedence rules correct. So I won't respond until you actually put in the effort to learn what you were supposed to.

Fix meaning add brackets to make what you wrote below correct:

8 hours ago, by Yuvraj Singh...

10^n mod 11=10.(10^n-1) mod 11=-(1.10^n-1) mod 11=-(1.(10.10^n-2)) mod 11=-(1.((11-1).10^n-2) mod 11=-(1.-(1.10^n-2)) mod 11= so the last term will be - (1.-(1.-(1..........))))))).. Mod 11 =will depends upon what value of n we choose.

Nothing else!

Abhas Kumar Sinha

2:49 PM

@user21820 Need some help here.

There are three numbers $(p, q, r) \equiv (2^a, 2^b, 2^c)$ where $a, b, c$ are integers. Find $a, b, c$ if $$ (pq - r, qr -p, pr - q) \equiv (2^x, 2^y, 2^z) $$ where $x, y, z \in \mathbb N$

One solution is $a=b=c=2$, what are other solutions?

I've got no clue how to start.

3 hours later…

5:26 PM

@AbhasKumarSinha I've got no clue what your question is. What does "≡" mean?

user image

3

Q: Problem 8, chapter 1 - Rudin's functional analysis

Trying to solve this problem: Problem 8: a) Suppose $\mathcal{P}$ is a separating family of seminorms on a vector space $X$. Let $\mathcal{Q}$ be the smallest family of seminorms on $X$ that contains $\mathcal{P}$ and is closed under max. [This means: if $p1,p2 \in\mathcal{Q}$ and $p=\max(p1,...

functional-analysis proof-writing topological-vector-spaces

kindly, help if you can

How to get |b_n| > |b|/2 ?

@user21820

Abhas Kumar Sinha

5:48 PM

@user21820 $(x,y) \equiv (a,b)$ means $x=a, y=b$

Nevermind, I got it. $|b| \leq |b-b_n| + |b_n| < |b|/2 + |b_n|$

6:20 PM

@AbhasKumarSinha Don't use "≡" for equality; it's wrong. So next time use "=". The standard trick is to break the symmetry (order a,b,c by size) and factorize. See if that works.

Abhas Kumar Sinha

@user21820 What? I didn't get it.

@yh05 Yea, normally I just use my intuition first; |b[n]−b| is distance of b[n] from b. If that is less than |b|/2, then b[n] is closer to b than to 0. Then you know how to apply triangle inequality. This technique works well when things become more complicated and it is less obvious how to use triangle inequality.

@AbhasKumarSinha Didn't get what?

Abhas Kumar Sinha

@user21820 Your approach.

Your equations are symmetric. You can obtain more information by fixing an ordering for a,b,c, which you can by symmetry.

6:34 PM

@user8469759 Hmm this is not "basic mathematics". Why not ask in the main chat-room, or in the Analysis chat-room? =)

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