- let a and b graters or equal 1 from the set of real numbers R,

- we have the equation : n=a+b+1

- what will be one demonstration that for every value of number n always will be,always there two numbers a and b for this equation is true ?

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- let a and b graters or equal 1 from the set of real numbers R,

- we have the equation : n=a+b+1

- what will be one demonstration that for every value of number n always will be,always there two numbers a and b for this equation is true ?

- a and b are graters or equals with 1 from R,

- and we have the equation n=a+b+1

- what we have to demonstrate is that always,for every value of n will be one number a and one number b so that the equation will be true

- if a and b are graters or equals with 1 ,we can writing that a+b is greater or equal with 2

- from this results that n-1=a+b ie where n=a+b+1

- this demonstration may be accepted ?

J

If n is a real number greater than or equal to 3,

Then there exists at least one pair of real numbers, a and b, each greater than or equal to 1, such that (a + b + 1) = n.

Is that the problem?

How have real numbers been defined for you? What can you assume?

IF YOU ARE IN BEGINNING ALGEBRA I see nothing wrong with a proof like:

Let a = n - 2.

So, a is a real number and a >= (3 - 2).

So, a is a real number >= 1.

Let b = 1.

So, b is a real number >= 1.

So, (a + b + 1) = (a + 1 + 1) = (a + 2) = (n - 2) + 2 = n.

(n - 2) and 1 are a pair of real numbers that meet the requirements.

So there is at least one such pair of real numbers.

IF YOU ARE IN SOME ADVANCED MATH COURSE AND MUST PROVE THAT THE REAL NUMBERS ARE CLOSED UNDER ADDITION, YOU ARE WAY BEYOND ME AND NEED TO POST THIS PROBLEM UNDER ADVANCED MATH.

J

I gave you a constructive proof of what I THOUGHT was your problem. Clearly, I did not understand it. Sorry to be of no help.jhonyy9 said:- this equation can be demonstrated with complete mathematical induction ?

- if so,how it can be demonstrated ?

PS. As you have defined the problem SO FAR, n is to be a REAL number >= 3. The proof I gave was for any such real number. There is no need for mathematical induction in the problem as posed.

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n= a+b+1 is true for n=3jhonyy9 said:

- let a and b graters or equal 1 from the set of real numbers R,

- we have the equation : n=a+b+1

- what will be one demonstration that for every value of number n always will be,always there two numbers a and b for this equation is true ?

assume that it is true for an arbitrary 'n'.

n = p + q + 1

Then:

n+1 = (p+q+1) + 1 = p + (q+1) + 1 = a + b + 1

This is what JeffM had presented. It is surprising that you could not rewrite that in this form!!!

You can state your message without yelling it by not using all caps, such as the above.JeffM said:IF YOU ARE IN SOME ADVANCED MATH COURSE AND MUST PROVE THAT THE REAL NUMBERS ARE CLOSED UNDER ADDITION, YOU ARE WAY BEYOND ME AND NEED TO POST THIS PROBLEM UNDER ADVANCED MATH.

I have been corrected on this same technique before.

Sir, please let the moderators do any reprimanding.lookagain said:You can state your message without yelling it by not using all caps, such as the above.JeffM said:IF YOU ARE IN SOME ADVANCED MATH COURSE AND MUST PROVE THAT THE REAL NUMBERS ARE CLOSED UNDER ADDITION, YOU ARE WAY BEYOND ME AND NEED TO POST THIS PROBLEM UNDER ADVANCED MATH.

And the all caps also shows a poor response to your impatience and frustration with

the user.

It is the reason why they are moderators.

We thank you for your cooperation.

\(\displaystyle n = p + q + 1\)Subhotosh Khan said:n= a+b+1 is true for n=3

assume that it is true for an arbitrary 'n'.

n = p + q + 1

Then:

n+1 = (p+q+1) + 1 = p + (q+1) + 1 = \(\displaystyle \text{ Error here:}\)a + b + 1

\(\displaystyle \text{Therefore}\)

\(\displaystyle n + 1 =\)

\(\displaystyle (p + q + 1) + 1 =\)

\(\displaystyle p + (q + 1) + 1 =\)

\(\displaystyle \text{What did you intend for your conclusion?}\)

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lookagain said:\(\displaystyle n = p + q + 1\)Subhotosh Khan said:n= a+b+1 is true for n=3

assume that it is true for an arbitrary 'n'.

n = p + q + 1

Then:

n+1 = (p+q+1) + 1 = p + (q+1) + 1 = \(\displaystyle \text{ Error here:}\)a + b + 1

\(\displaystyle \text{Therefore}\)

\(\displaystyle n + 1 =\)

\(\displaystyle (p + q + 1) + 1 =\)

\(\displaystyle p + (q + 1) + 1 =\)

\(\displaystyle \text{What did you intend for your conclusion?}\)

a = p

b = q+1

thus

n+1 = a + b + 1

Thus if (n+1) can be expressed as a+b+1.

Thus anunumber can be expressed as a+b+1

But (n + 1) cannot be expressed as (a + b + 1), because n already equals (a + b + 1).Subhotosh Khan said:a = p

b = q+1

thus

n+1 = a + b + 1

Thus if (n+1) can be expressed as a+b+1.

Thus anunumber can be expressed as a+b+1\(\displaystyle You \ can't \ use \ the \ same \ variables\)

\(\displaystyle here \ as \ were \ given \ in \ the \ problem.\)

You have to use at least one different variable (~ letter) for your point.

n and (n + 1) cannot both be equal to (a + b + 1).

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Thank you for the correction.

- n=a+b+1 --- 3=1+1+1 --- is true

--- 4=2+1+1 --- is true

--- 5=2+2+1 --- is true

- if n=m --- m=a+b+1 --- suppose is true,

then m+1=a+b+1+1

- m+1=a+1+b+1 --- is true -- commutativity

- m+1=a+(b+1)+1 --- is true --- associativity

- results that this equation n=a+b+1 -- is true

- this demonstration may be accepted ?

J

In your first post, you said "let n greater or equal 3 from the set of real numbers R."

The proof you give in your last post is of a type that is valid for integers. If it is true for some integer k, it is true for the integer (k + 1). It is true for some integer m. Therefore it is true for any integer n \(\displaystyle \geq \\) m. That precise form of proof is NOT valid for real numbers.

If we understand what you are being asked to prove (please see my first post under JeffM), it can be proved for any real number WITHOUT mathematical induction. If you are required to prove the statement for a real number using some form of mathematical induction, please let us know. I am sure that someone other than me can show you how that is done. We need you to tell us what is to be proved and by what method.

Again a valid CONSTRUCTIVE proof (treating as axiomatic that the system of real numbers is closed under addition) of what I THINK you are being asked is:

Let n be an ARBITRARY real number \(\displaystyle \geq \\) 3.

Let a = (n - 2).

So a is a real number \(\displaystyle \geq \\)1.

Let b = 1.

So, b is a real number \(\displaystyle \geq \\) 1.

(a + b + 1) = (a + 1 + 1) = (a + 2) = [(n - 2) + 2] = n.

So there is at least one pair of real numbers a and b such that a \(\displaystyle \geq \\) 1 \(\displaystyle \leq \\) b and (a + b + 1) = n for any real n \(\displaystyle \geq \\) 3.

for JefM what i have asked is how is possible writing a demonstration that for every value of n,when n is greater or equal 3 from R will be always one number a and one number b , every two graters or equals 1 from R,so that the equation n=a+b+1 is true

- for example : 3=1+1+1 , 4=2+1+1 , 5=2+2+1 ,...

- check please my last demonstration ... and i wait your reply !

thank you very much !

J

In my opinion, your demonstration is not acceptable. You are effectively assuming that n is an integer, but you said that n is a real number.jhonyy9 said:

- n=a+b+1 --- 3=1+1+1 --- is true

--- 4=2+1+1 --- is true

--- 5=2+2+1 --- is true

- if n=m --- m=a+b+1 --- suppose is true,

then m+1=a+b+1+1

- m+1=a+1+b+1 --- is true -- commutativity

- m+1=a+(b+1)+1 --- is true --- associativity

- results that this equation n=a+b+1 -- is true

- this demonstration may be accepted ?

As I have said twice now, there is a demonstration for any real n \(\displaystyle \geq\\) 3 that does not require induction in its proof. Are you required to give a proof by induction?