[tex]n = p + q + 1[/tex]Originally Posted by Subhotosh Khan
[tex]\text{Therefore}[/tex]
[tex]n + 1 =[/tex]
[tex](p + q + 1) + 1 =[/tex]
[tex]p + (q + 1) + 1 =[/tex]
[tex]\text{What did you intend for your conclusion?}[/tex]
[tex]n = p + q + 1[/tex]Originally Posted by Subhotosh Khan
[tex]\text{Therefore}[/tex]
[tex]n + 1 =[/tex]
[tex](p + q + 1) + 1 =[/tex]
[tex]p + (q + 1) + 1 =[/tex]
[tex]\text{What did you intend for your conclusion?}[/tex]
Originally Posted by lookagain
“... mathematics is only the art of saying the same thing in different words” - B. Russell
But (n + 1) cannot be expressed as (a + b + 1), because n already equals (a + b + 1).Originally Posted by Subhotosh Khan
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.
“... mathematics is only the art of saying the same thing in different words” - B. Russell
- this equation can be demonstrated with complete mathematical induction ?
- 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 ?
jhonny9
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 [tex]\geq \[/tex] 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 [tex]\geq \[/tex] 3.
Let a = (n - 2).
So a is a real number [tex]\geq \[/tex]1.
Let b = 1.
So, b is a real number [tex]\geq \[/tex] 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 [tex]\geq \[/tex] 1 [tex]\leq \[/tex] b and (a + b + 1) = n for any real n [tex]\geq \[/tex] 3.
... this demonstration is like wrong ... or why dont reply me nobody nothing ???
sorry but i now i have got that this my problem have two pages ... so this demonstration with math complete induction what i have wrote on below , on end of the first page write me please this demonstration it is acceptable or ... what is your opinion,please ?
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 !
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.Originally Posted by jhonyy9
As I have said twice now, there is a demonstration for any real n [tex]\geq\[/tex] 3 that does not require induction in its proof. Are you required to give a proof by induction?
- sorry - yes this is true ... - so how you think this ,please,will be true if a,b,n are numbers from the set of natural numbers N ?
- in this conditions - if a,b,n are from N - this prove can may be accepted ?
- thank you for your reply
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