Basic Mathematics - 2019-12-11 (page 2 of 2)

If 5 = 0 then 2^5 = 1, otherwise 2^5 = 2^(5−1)·2. 5 ≠ 0. So 2^5 = 2^4·2.
If 4 = 0 then 2^4 = 1, otherwise 2^4 = 2^(4−1)·2. 4 ≠ 0. So 2^4 = 2^3·2.
If 3 = 0 then 2^3 = 1, otherwise 2^3 = 2^(3−1)·2. 3 ≠ 0. So 2^3 = 2^2·2.
If 2 = 0 then 2^2 = 1, otherwise 2^2 = 2^(2−1)·2. 2 ≠ 0. So 2^2 = 2^1·2.
If 1 = 0 then 2^1 = 1, otherwise 2^1 = 2^(1−1)·2. 1 ≠ 0. So 2^1 = 2^0·2.
If 0 = 0 then 2^0 = 1, otherwise 2^0 = 2^(0−1)·2. 0 = 0. So 2^0 = 1.

skillpatrol

All done by definition^

yuvraj singh

Last one look stupid because in detention we already said m^n=1 for n=0

Then why we need otherwise.

user21820

1:11 PM

@yuvrajsingh What you are told in detention may or may not be correct. But what I told you is the true mathematical definition.

It covers all cases. Look carefully at what I just wrote above to see how the definition I gave you fully controls the meaning of 2^5.

Tell me, by the definition, what is 0^0?

yuvraj singh

What next, my question?

skillpatrol

0 = 0 is the only true statement, right?

yuvraj singh

It is undefined? @user21820

user21820

Wrong.

Apply my definition.

yuvraj singh

I am writing but still I am on my words 0^0=1

skillpatrol

1:15 PM

replace "2" with "0" please :-)

yuvraj singh

If 0=0 then 0^0=1 other wise 0^0=0^(0-1).0

user21820

Right. So the point is that by the definition I gave you 0^0 = 1. I don't care what any other teacher says, because you must understand that this 100% precise definition states that 0^0 = 1.

And if you ask any professional logician they will agree with my definition.

Now I want you to understand one more thing, before we get back to your "a^b^c" question.

yuvraj singh

Yes?

user21820

Observe that when you apply the definition to any specific naturals m,n, such as 2 and 5 in the above example, the definition can be expanded 'all the way to the bottom'.

This is what I meant earlier by saying:

3 hours ago, by user21820

I write it this way to make it clear that the definition is well-founded, meaning that it is not circular. Indeed, you can for any m,n expand the definition of m^n all the way until it 'terminates' at the base case.

Do you understand this? This is a non-trivial issue that you must understand if you want to understand definitions in general.

yuvraj singh

Understand.

But what the word trivial means.

Or non trival means.

user21820

1:20 PM

Good. In the future, you might see in textbooks that such definitions are called "recursive definitions" or "inductive definitions". "Trivial" means "insignificant/small". "Non-trivial issue" means "important".

Right now you do not need to know why they are called "recursive" or "inductive", but if you know programming then I can explain it to you.

yuvraj singh

So what we mean when we say solution is trival.

In mathematics.

user21820

It means it is some easy-to-find solution

yuvraj singh

OK.

@user21820 continue.

user21820

Now back to "a^b^c". This is ambiguous, because you must specify whether you mean (a^b)^c or a^(b^c).

yuvraj singh

Second one.

user21820

1:23 PM

You asked about "128^128^128", and later specified "(128^128)^128". I said that it makes no sense, because there is a basic property regarding (a^b)^c.

Right, now you are saying you actually wanted 128^(128^128).

yuvraj singh

Yes.

user21820

You also incorrectly reduced the exponentiation, because you wrote:

6 hours ago, by yuvraj singh

So I write the above part as $(64)^2(64)^2(64^2)

I really have no idea what it means, but it is surely wrong.

Remember, 128^128 is defined exactly as I showed you above, which expands to 128 copies of 128 multiplied together.

2·3^4 is not the same as (2·3)^4.

yuvraj singh

I admit.

user21820

Okay so the question is what is the remainder of 128^(128^128) when divided by 3.

In programming terms, you want 128^(128^128) mod 3.

Are you familiar with "mod"?

yuvraj singh

No.

user21820

1:28 PM

Okay then we build everything from scratch.

For convenience, let k = 128^128. We want remainder of 128^k when divided by 3.

yuvraj singh

OK.

user21820

The phrase "a mod b" is defined to mean "the remainder of a when divided by b".

Save us time writing.

Now, do you agree that (a+b) mod b = a mod b, for any integers a,b where b > 0?

yuvraj singh

Yes.

user21820

Do you agree that (a+b·c) mod b = a mod b, for any integers a,b,c where b > 0?

yuvraj singh

Yes

user21820

1:32 PM

Using this fact, can you simplify (a+b)·c mod b?

(Here "·" has higher precedence than "mod".)

yuvraj singh

OK.

user21820

What do you get?

In case it is not clear, I'm asking you to fill the blank something simple:

> (a+b)·c mod b = _______, for any integers a,b,c where b > 0.

I'll be back later. Take your time, and then do the same for:

yuvraj singh

a mod b. +c

user21820

> (a+b·d)·c mod b = _______, for any integers a,b,c,d where b > 0.

@yuvrajsingh No guessing. Guesses are almost always wrong, especially when you lack mathematical experience.

Use only the facts I have listed above.

yuvraj singh

No I haven't guess.

I am honest.

user21820

1:40 PM

Then provide the calculation using the facts.

I will be back later.

yuvraj singh

Are you going, how much time?

@user21820

yuvraj singh

1:55 PM

So answering the first one.

user21820

2:23 PM

@yuvrajsingh I'm back, but I don't see your answer..

yuvraj singh

OK.

(a+b). c/b.

So we get a mod b +c mod b @user21820

user21820

I don't know what you mean.

yuvraj singh

?.

Sorry. I rewrite them.

user21820

In mathematics, you should write only true statements.

"1+1 = 2" is a true statement. "1+1" is not a statement.

What you wrote is not a statement, so I don't even know what you're claiming.

(And your final answer is wrong, so you definitely made a mistake somewhere.)

user21820

2:40 PM

@yuvrajsingh: If you cannot get it, I can give you a hint.

yuvraj singh

No I will try.

(a+b). c mod b.

Which implies remainder of (a+b). c when divided by b.

So answer is a mode b +c.

@user21820

user21820

No as I said, that answer is wrong.

yuvraj singh

No sir.

Let me write it again.

a mod b+1 @user21820 it is not a guess.

user21820

"a mod b + c" is wrong.

In mathematics, you cannot make any jump in thinking. You must be able to systematically obtain facts based on previous ones, without relying on intuition or guessing. You went straight from "remainder of (a+b)·c when divided by b" to "a mod b + c", without justification.

Err... in my viewpoint, anything that does not have solid logical justification is a guess.

yuvraj singh

@user21820 if my answer is correct, then I have write to justify it. So is my answer correct?

user21820

2:51 PM

No it is wrong. I said three times already that it is wrong.

yuvraj singh

I can, t find where I am getting wrong?

user21820

You can test small numbers to check your own claim. Your claim is that (a+b)·c mod b = a mod b + c. Try substituting c = 1 and seeing whether that contradicts what you previously agreed with.

yuvraj singh

My answer is wrong.

How stupid I am.

Or I just lost my sense?

@user21820

user21820

No you just made a mistake.

Let me show you how to start.

yuvraj singh

OKM

user21820

2:58 PM

(a+b)·c mod b = (a·c+b·c) mod b = ... Can you finish, using the previous facts?

yuvraj singh

ac mod b +bc mod b.

user21820

It's more or less correct, but not quite. 1 mod 3 + 2 mod 3 = 1 + 2 = 3. (1+2) mod 3 = 3 mod 3 = 0.

Why don't you just use the previous facts?

1 hour ago, by user21820

Do you agree that (a+b·c) mod b = a mod b, for any integers a,b,c where b > 0?

yuvraj singh

Oo so ac mod b.

user21820

Right.

So let's write it again.

> (a+b)·c mod b = (a·c) mod b for any integers a,b,c where b > 0.

1 hour ago, by user21820

> (a+b·d)·c mod b = _______, for any integers a,b,c,d where b > 0.

What about the next one?

yuvraj singh

ac mod b

user21820

3:05 PM

Right again.

If you notice, I try to be careful to always state the conditions under which my statements apply. You also should try.

This is how to be mathematically precise. What works under some conditions may not work under other conditions.

Now back to mod.

yuvraj singh

One question from the last?

user21820

The last fact we went through shows that if you have a product "x·y mod b", you can take away any multiple of b from x and it would remain the same.

@yuvrajsingh Go ahead.

yuvraj singh

How bcd mod b become zero.

user21820

b·cd is a multiple of b.

Are you satisfied with the facts so far?

yuvraj singh

Yes.

Can go to have some water.

user21820

3:11 PM

Sure.

When you're back, is it clear that 8·8 mod 3 = 2·8 mod 3 = 2·2 mod 3?

How about 8·8·8 mod 3?

How about 8^n mod 3, for natural n?

yuvraj singh

3:26 PM

First is 3.

For second I can write (9-1)

So remainder be 1.

@user21820

user21820

I don't get your answers.

yuvraj singh

Why?.

user21820

As I said before, when you do not write a statement, it is unclear what you are claiming. And besides, what you write has nothing to do with the answer.

Write true statements.

At least write statements, so that I can give counter-examples to disprove them.

yuvraj singh

Sorry, I promise I won, t make this mistake again.

For first are you asking the value or like this When you're back, is it clear that 8·8 mod 3 = 2·8 mod 3 = 2·2 mod 3?

@user21820

user21820

@yuvrajsingh Yes, and your answer should be "yes" or "no". If it is "yes" then you should be able to explain how to obtain this result from the above facts.

yuvraj singh

3:39 PM

My eye is painting can I take a break for an hour. @user21820

user21820

@yuvrajsingh You can take a much longer break, as I will need to go soon. =)

Basically, you just have to find suitable substitutions to use the above fact:

yuvraj singh

Sorry if you get. Angry.

user21820

> (a+b·d)·c mod b = a·c mod b, for any integers a,b,c,d where b > 0.

Take your time to find those substitutions to show that 8·8 mod 3 = 2·8 mod 3 = 2·2 mod 3.

yuvraj singh

Got it can I answer.

user21820

And then think how you can use the same technique to simplify 8·8·8 mod 3, and in general 8^n mod 3.

yuvraj singh

3:42 PM

Yes.

user21820

Also think about 8·8 mod 3 = (−1)·(−1) mod 3.

That's all I have for you for now. See you another day!

yuvraj singh

64.8 mod 3, and (2+30.2) mod 3 is it right @user21820

Or (2+20.3)mod 3

user21820

I think you should take a break. I can't understand you, and probably you won't understand yourself after you come back to what you just wrote. Bye!

(Mathematics should not be rushed.)

yuvraj singh

OK. Bye.

yuvraj singh

4:15 PM

(4+30.2).8,mod 3 can be thia similar to (a+bc).d mod b @user21820

user21820

@yuvrajsingh Yes so what is the statement that you get?

yuvraj singh

Actually I was writing this only.

32 mod 3.=2

user21820

But you're not using the same technique. You immediately computed 8·8 = 64. It is correct, but you cannot generalize your method to 8^n.

You need to try to do exactly the same as I did with 8·8 mod 3, but now with 8·8·8 mod 3.

More 8s, but same trick.

yuvraj singh

8^n can be (9-1)^n.

user21820

Yes, and then what? I want to see you do 8·8·8 mod 3 correctly first.

@yuvrajsingh And avoid imprecise language like "can be". 8^n = (9−1)^n.

yuvraj singh

4:20 PM

OK.

8.8.8.mod 3 is 8.2.8mod 3=2.2.8mod 3=2.2.2mod3=8mod3

=2

@user21820

user21820

Good good.

Do the same trick, but using 8 = 9−1.

So that you can handle 8^n mod 3 easily.

Same for 128^n.

yuvraj singh

That will get me to (9-1)mod 3.

-1mod 3

That will be equal to 2.if power n will not be there.

2^n will be the remainder @user21820

user21820

Third time I'm telling you to write statements. Nothing you just wrote is a statement.

I cannot accurately assess whether you are correct if you don't write full statements.

yuvraj singh

8^mod 3=(9-1)^n mod 3=-1^n mod 3=2^n.@user21820

user21820

4:35 PM

Excellent. It is all correct (except missing brackets). But notice that you don't actually want the last equality!

yuvraj singh

OK.

user21820

"−1^n" in mathematics is always interpreted as "−(1^n)" because of precedence rules. So that is not what you should have written.

Write it again with the correct brackets, and don't go to "= 2^n mod 3", and think of another easier final step.

yuvraj singh

- (1^n)mod 3.

user21820

I suspect you have problems with arithmetic...

Substitute exactly into the above facts, if you want to ensure correctness.

yuvraj singh

-(3-2)^n mod 3

user21820

4:41 PM

No. That's why I asked you to do the following first:

58 mins ago, by user21820

Also think about 8·8 mod 3 = (−1)·(−1) mod 3.

You did not do it. And your guessing is incorrect.

yuvraj singh

What wrong in that I get 2^n mod 39

I haven't guess, and I assure, in the example you have given.

user21820

You wrote "(9-1)^n mod 3=-1^n mod 3". That is wrong. You later wrote "-(1^n)mod 3" with no statement. Also wrong.

yuvraj singh

OK so I should write statement. Yes?

user21820

Yes, and do the simple example 8·8 mod 3 = (−1)·(−1) mod 3 first using the fact.

And then repeat for 8^n mod 3.

yuvraj singh

8^n mod 3=(9-1)^n mod 3=-(1^n)mod 3=-(3-2)^n mod 3=2^n mod 3.

user21820

4:47 PM

I asked you to use the fact by exact substitution. You are not following instructions again.

yuvraj singh

8.8mod 3=8.(3.2+2)mod 3=2.2 mod 3.

user21820

1 hour ago, by user21820

> (a+b·d)·c mod b = a·c mod b, for any integers a,b,c,d where b > 0.

@yuvrajsingh Yes that is right, to get 8·8 mod 3 = 2·2 mod 3.

But you did not do it right for the other one.

So do the actual substitution!

yuvraj singh

Got it.

user21820

That is what I mean by "you guessed", because you guessed how it would turn out without actually manually doing the substitution.

yuvraj singh

8^n mod 3=(3.2+2)^n mod 3=2^n mod 3.

@user21820

user21820

4:52 PM

That is correct, but I also asked you to do...

1 hour ago, by user21820

Also think about 8·8 mod 3 = (−1)·(−1) mod 3.

yuvraj singh

I already did it 2 times and you said OK too?

user21820

Nothing in what you wrote includes "(−1)·(−1) mod 3."!!

Don't you see a big fat negative one?!

yuvraj singh

OK (3.3-1)(3.3-1) mod 3=(-1)(-1)mod 3

@user21820

user21820

Aha finally. Though you should have written "(3·3+(−1))", otherwise it is not exactly a rigorous use of the fact.

Now simplify 8^n mod 3 using the same technique.

yuvraj singh

(3.3+(-1))^n mod 3=-(1) ^n mod 3 -(3-2)^n mod 3=2^n mod 3.

@user21820

user21820

5:02 PM

Wrong.

yuvraj singh

Now what?

user21820

Don't anyhow move brackets wherever you like. Substitute properly.

yuvraj singh

I haven, t put any extra Bracket.

user21820

Look, you clearly have a problem with basic arithmetic. One way to avoid your very frequent arithmetic errors is to substitute exactly carefully. Copy-paste the following fact and substitute each variable with the exact same expression.

1 hour ago, by user21820

> (a+b·d)·c mod b = a·c mod b, for any integers a,b,c,d where b > 0.

yuvraj singh

As I said I am not using PC, I have old broken screen phone.

user21820

5:06 PM

You can manually copy-paste. You don't have a choice, because you are making these basic errors that you shouldn't be making.

yuvraj singh

OK.

user21820

((−1)+3·3)·8 mod 3 = ...

yuvraj singh

From next time, I won, t make this error promise.

Next?

user21820

1 min ago, by user21820

((−1)+3·3)·8 mod 3 = ...

Maybe you didn't read what I wrote properly:

34 mins ago, by user21820

"−1^n" in mathematics is always interpreted as "−(1^n)" because of precedence rules. So that is not what you should have written.

yuvraj singh

((−1)+3·3)^n. 8 mod 3 = ...-(3+(-2))^n mod 3=2^n mod 3.

user21820

5:13 PM

I can't believe that you still do not follow instructions. Why waste our time with typing so many other wrong things, instead of copying exactly and substituting?

For the last time, I ask you to substitute the a,b,c,d in "(a+b·d)·c mod b = a·c mod b" with appropriate values to get "((−1)+3·3)·8 mod 3 = ...". Fill it in.

Transcript for

♦ $Mathematics$ Basic Mathematics

Transcript for

♦ Basic Mathematics

♦ $Mathematics$ Basic Mathematics