Constant term

[Dr.Peterson]

Free bound see When I saw the terms I went to given link page to see the definition .
After seeing the eg of bound variable especially this
For all x,
(x + 1)^2 = x^2 + 2x + 1 "
Then suddenly independent and dependent as I knew these before hit my mind so are free and bound are same as independent and dependent variable respectively.

No, you are jumping to conclusions. Free variables are not the same thing as independent variables.

In that example, what makes x a bound variable is the phrase "for all x", which is called a quantifier, and specifies which values x can take (even though in this case there is no restriction at all). In order to understand this, you would need to understand the concept of quantifiers, in the context of logic.

The article, as usual, is putting too many different contexts together (that should be obvious just from scanning it!), and saying too little to make the meaning clear.

Then I checked study.com
Where they are saying
f(x)=3x-1 --> 'x' is free variable

That page is not equating "free" to "independent", either. Yes, an independent variable, in the absence of further context, is free; but that is not the point they are making. They are distinguishing this x from the i in a summation, the former referring to any possible value, the latter only to one of the values identified in the limits of the summation.

But go back to this eg : for all x, (x+1)^2=x^2+2x+1 (In Wiki page)

writing the wiki eg in terms of function : f(x)=x^2+2x+1
so x must be free variable in this case (acc to study.com) as i can put any value in place of x , isn't it ? There are no limitations on x

CONTRADICTION isn't it ?

What's the problem in my doubt

You totally missed what the Wikipedia page is saying, and then totally changed it by dropping the words "for all x" that make x bound, and putting it into a function, where it is an independent variable, which you had wrongly equated to "free".

You caused the contradiction by playing with ideas you don't understand.

Once again, stop doing this. Don't read a Wikipedia article that condenses huge ideas, and think you understand everything. I'm not sure I understand all that they are saying there!

 

[Saumyojit]

I saw about the " integer multiple" Vs multiple .
I knew that everything has to be a integer
If a is a multiple of b then a=b*n
'n' should be a integer .
8=4*2 ( 8 is a multiple of 4)
But
12=1.5*8
Can 12 be a multiple of 1.5 ? Even though 1.5 is not a integer .
Is this something to do with "Euclid's Division always deals with integer values only"

 

[Dr.Peterson]

I saw about the " integer multiple" Vs multiple .
I knew that everything has to be a integer
If a is a multiple of b then a=b*n
'n' should be a integer .
8=4*2 ( 8 is a multiple of 4)
But
12=1.5*8
Can 12 be a multiple of 1.5 ? Even though 1.5 is not a integer .
Is this something to do with "Euclid's Division always deals with integer values only"

What are you responding to? Did you put this in the wrong thread? This seems to be about division, not constants or free variables.

 

[Saumyojit]

What are you responding to? Did you put this in the wrong thread? This seems to be about division, not constants or free variables.

Yes I had only this small doubt.
So I thought you might respond here Only

 

[Dr.Peterson]

I saw about the " integer multiple" Vs multiple .
I knew that everything has to be a integer
If a is a multiple of b then a=b*n
'n' should be a integer .
8=4*2 ( 8 is a multiple of 4)
But
12=1.5*8
Can 12 be a multiple of 1.5 ? Even though 1.5 is not a integer .
Is this something to do with "Euclid's Division always deals with integer values only"

Yes, we can say that 12 is a multiple of 1.5, or more specifically an integer multiple, if you want to make sure you are clear.

This is not Euclidean division, though. It is simply multiplication by an integer.

 

[Saumyojit]

But can we say 1.5 is a Divisor of rs 12
I think No acc to wiki
In wiki they say that a Divisor has to be a integer m when mutlpiled by another integer gives us a product which is also a integer.

 

[JeffM]

But can we say 1.5 is a Divisor of rs 12
I think No acc to wiki
In wiki they say that a Divisor has to be a integer m when mutlpiled by another integer gives us a product which is also a integer.

Because you jump from topic to topic, you create all sorts of confusion for yourself.

When we talk about integers, division is not a closed operation. In the realm of integers, where a and b are both integers, we say that a is a divisor of b if and only if there is another integer c such that a * c = b. "Divisor" in that context is with reference to a specific dividend and implies the quotient is an integer.

When we talk about real numbers, division by all reals except zero is a closed operation. Therefore every number (except zero) is a divisor of any real number.

 

[Saumyojit]

every number (except zero) is a divisor of any real number.

Is 8 a Divisor of 12 then .I think yes

 

[pka]

Is 8 a Divisor of 12 then .I think yes

Is there an integer \(K\) such that \(8\cdot K=12~?\)
Your answer to that will tell us all if you are a troll or not.

 

[Saumyojit]

Is there an integer \(K\) such that \(8\cdot K=12~?\)
Your answer to that will tell us all if you are a troll or not.

By making this mistake I understood that the Divisor has to completely Divide " Dividend" in both the context of "integrs" and " real no's" .(the Quotient has to be a integer)

Now if we are talking in terms of " integers" only then the Divisor and Dividend has to be both integers.

Or in case of real no's both Divisor and Dividend can be real no making sure that quotient is a integer.

Quotient has to be a integer

 

[pka]

By making this mistake I understood that the Divisor has to completely Divide " Dividend" in both the context of "integrs" and " real no's" .(the Quotient has to be a integer) Now if we are talking in terms of " integers" only then the Divisor and Dividend has to be both integers. Or in case of real no's both Divisor and Dividend can be real no making sure that quotient is a integer. Quotient has to be a integer

You are making what a logician would say is a category error. That is, speaking of divisors one is comparing two integers,
It is true that \(12\cdot\frac{1}{2}=6\) but we don't say that one-half is a divisor of \(12\) Look at this LINK!
Look on than link at the divisors of \(12\).

 

[JeffM]

12[MATH]\div 8 = 1.5.[/MATH]
12 is the dividend.

8 is the divisor.

1.5 is the quotient.

[MATH]7.2 \div 14.4 = 0.5.[/MATH]
7.2 is the dividend.

14.4 is the divisor.

0.5 is the quotient.

That is ONE meaning of the word ”divisor” in mathematics. “Divisor” is simply the name of one of the operands in the operation of division. If we are talking about real numbers, any number whatsoever except zero can be a divisor, any number can be a dividend, and there is no implication that the quotient must be an integer.

There is ANOTHER meaning of the word ”divisor” in mathematics. If we are talking only about integers, we say that integer a is a divisor of integer b if and only if there exists integer c such that a multiplied by c equals b. In that sense, 8 is not a divisor of 12, even though both are integers, because there is no integer that you can multiply by 8 and get a result of 12.

You cannot blend these two usages to make a consistent synthesis. It is a word with two distinct meanings.

 

[Saumyojit]

@Dr.Peterson

A polynomial in one indeterminate is called a univariate polynomial, a polynomial in more than one indeterminate is called a multivariate polynomial. A polynomial with two indeterminates is called a bivariate polynomial.[3] These notions refer more to the kind of polynomials one is generally working with than to individual polynomials; for instance, when working with univariate polynomials, one does not exclude constant polynomials (which may result from the subtraction of non-constant polynomials)

https://en.m.wikipedia.org/wiki/Po
Here I was reading this paragraph of which I have doubts


I know what is a constant polynomial

I don't understand the below two quotated lines
" kind of polynomials one is generally working with"

" individual polynomials"

I know that univariate polynomials =one variable polynomial

Eg: 2x or 6x-2 or 3x^2

constant polynomials =3,5 any real no

for instance, when working with univariate polynomials, one does not exclude constant polynomials (which may result from the subtraction of non-constant polynomials)

What does the paragraph mean
Why did it say this

 

[Dr.Peterson]

I don't understand the below two quotated lines
" kind of polynomials one is generally working with"

" individual polynomials"

I know that univariate polynomials =one variable polynomial

Eg: 2x or 6x-2 or 3x^2

constant polynomials =3,5 any real no

They are saying that when you are working with bivariate polynomials, you would not refuse to work with 3x+2, for example, because y doesn't happen to appear in that polynomial. (That might arise from subtracting bivariate polynomials (xy+3x) - (xy-1), for example.) The set of bivariate polynomials contains the univariate polynomials as a subset.

But outside of that context, if you were asked what 3x+2 is, you would call it a univariate polynomial, because that is what it is in itself -- as an individual expression. As we've said, context determines meaning.

Similarly, a constant would be considered a polynomial (and therefore an algebraic expression), when your context is working with polynomials, even though, in another context, you might call it merely "a number" and not mention algebra at all.

for instance, when working with univariate polynomials, one does not exclude constant polynomials (which may result from the subtraction of non-constant polynomials)

What does the paragraph mean
Why did it say this

This is what I just said above. If you subtract two polynomials (3x+5) - (3x+2), you would get 3, and constant; in doing do, you have not left the world of polynomials because the variable disappeared; you just have a very simple polynomial.

 

[JeffM]

@Dr.Peterson

https://en.m.wikipedia.org/wiki/Po
Here I was reading this paragraph of which I have doubts


I know what is a constant polynomial

I don't understand the below two quotated lines
" kind of polynomials one is generally working with"

" individual polynomials"

I know that univariate polynomials =one variable polynomial

Eg: 2x or 6x-2 or 3x^2

constant polynomials =3,5 any real no


What does the paragraph mean
Why did it say this

Why do you persist in reading articles that are trying to cover a huge swath of mathematics in very concise form when you are trying to learn a basic but relatively simple branch of mathematics. When you read a word like "ring," an alarm bell should go off in your head: you have gone far beyond basic algebra.

All that is being said in that paragraph is this: if we are talking about n-variable polynomials we can deal with polynomials in fewer than n variables.

[MATH]x^2 + 2x - 15 - (x^2 + 2x - 24) = 9 = 0 * x^2 + 0 * x + 9[/MATH]
We can always restate that constant in an expression that contains the variables.

And Merry Christmas.

PS I have suddenly realized that when you say "doubts," you do not mean to imply that someone is lying to you, that what they say is not to be trusted. You mean to say that you do not understand what they are getting at, what they mean exactly, why they say something in a particular way.

 

[Saumyojit]

you would get 3, and constant; in doing do, you have not left the world of polynomials because the variable disappeare

Are you saying that the writer thought the readers who are not that mature in maths might mistook that if there is no variable attached to a number but merely a numeral present then if the reader doesn't know that there is a thing called constant polynomial he might think that after subtraction the result 3 has left the world of polynomial.
Right?

You would not refuse to work with 3x+2

Why should I refuse ; even if I did .
thats because my original two polynomial was bivariate that's why u are saying.

But outside of that context, if you were asked what 3x+2

Can I not say by subtracting two non constant bivariate polynomials we got univariate polynomial i.e 3x+2
Why it's need to be outside of the context?


As it is being said
"These notions refer more to the kind of polynomials one is generally working"

What is the relationship the writer is trying to draw between notion and 'kind of polynomial working on"

As I feel although there is a name for polynomials with one variable still I cannot call it univariate polynomial or 3x+2 cannot be referred to by its corresponding notion
as I am getting it from the subtraction of two bivariate polynomials in this case which are the polynomial I am working on.
right?

(xy+3x) - (xy-1) this eg I am referring

 

[Dr.Peterson]

Are you saying that the writer thought the readers who are not that mature in maths might mistook that if there is no variable attached to a number but merely a numeral present then if the reader doesn't know that there is a thing called constant polynomial he might think that after subtraction the result 3 has left the world of polynomial.
Right?

How would I know what the author(s) thought? But, no, I think they wanted to clarify what they mean, because in mathematics definitions need to be precise.

You do realize, I hope, that Wikipedia articles have typically been modified many times by many individuals with different perspectives, and one sentence may contain mixed information. I suspect that this part has been handled by both people trying to speak to mathematicians and others trying to clarify for beginners.

As for the rest, I don't think you can understand it until you have actually done some abstract algebra so that you would know what it means to be precise, and to work within a particular context. Yes, contexts matter.

 

[Saumyojit]

As for the rest,

Why should I refuse ; even if I did .
thats because my original two polynomial was bivariate that's why u are saying
Can I not say by subtracting two non constant bivariate polynomials we got univariate polynomial i.e 3x+2
Why it's need to be outside of the context?

As it is being said
"These notions refer more to the kind of polynomials one is generally working"

What is the relationship the writer is trying to draw between notion and 'kind of polynomial working on"

As I feel although there is a name for polynomials with one variable still I cannot call it univariate polynomial or 3x+2 cannot be referred to by its corresponding notion
as I am getting it from the subtraction of two bivariate polynomials in this case which are the polynomial I am working on.
right?

are u referring to this that falls under abstract algebra

 

[JeffM]

@Saumyojit

We have perhaps reached the height of absurdity: in your last post, you ask Dr. Peterson to explain what you yourself wrote.

I agree that what you wrote is far from clear, but I do not agree that anyone else can possibly tell you what you really meant.

 

[Saumyojit]

@Saumyojit

We have perhaps reached the height of absurdity: in your last post, you ask Dr. Peterson to explain what you yourself wrote.

I agree that what you wrote is far from clear, but I do not agree that anyone else can possibly tell you what you really meant.

Dr p said "as for the rest " so I thought rest meant or implied to the bunch of questions that I asked in post no 57 Which Dr p kind of Gave me a hint by
saying that my question's answers will take me to Field of abstract algebra.thats why to drop it.