Exploring D. J. H. Garling's Theorem 3.1.1

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In summary: On... On... On... On...In summary, Garling's book provides a rigorous proof of Theorem 3.1.1. question 2 and question 3 are explained in detail. There is a discussion on rigour.
  • #1
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I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume I: Foundations and Elementary Real Analysis ... ...

I am focused on Chapter 3: Convergent Sequences

I need some help to fully understand some remarks by Garling made after the proof of Theorem 3.1.1 ...Garling's statement and proof of Theorem 3.1.1 (together with the interesting remarks) reads as follows:

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My questions on the remarks after the proof are as follows:
Question 1

In the remarks after the proof of Theorem 3.1.1 we read the following:

" ... ... There exists a least positive integer, \(\displaystyle q_0\), say, such that 1\(\displaystyle /q_0 \lt y - x\) ... ... "Can someone please explain exactly why this is true ... ..

Question 2

In the remarks after the proof of Theorem 3.1.1 we read the following:

" ... ... and there then exists a least integer, \(\displaystyle p_0\), say, such that \(\displaystyle x \lt p_0 / q_0\) ... ..."
Question 3

In the remarks after the proof of Theorem 3.1.1 we read the following:

" ... ... Then \(\displaystyle x \lt p_0 / q_0 \lt y\) and \(\displaystyle r_0 = p_o / q_0\) is uniquely determined ... ... "Can someone please explain exactly why this is true ... ..Help will be appreciated ...

Peter
 

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  • #2
Peter said:
In the remarks after the proof of Theorem 3.1.1 we read the following:

" ... ... There exists a least positive integer, \(\displaystyle q_0\), say, such that 1\(\displaystyle /q_0 \lt y - x\) ... ... "Can someone please explain exactly why this is true ... ..
Because $y-x>0$, so if you go $1,\frac12,\frac13,\ldots$ you will eventually come to an integer $n$ such that $y-x>\frac1n$. $q_0$ is the least such $n$.
Peter said:
Question 2

In the remarks after the proof of Theorem 3.1.1 we read the following:

" ... ... and there then exists a least integer, \(\displaystyle p_0\), say, such that \(\displaystyle x \lt p_0 / q_0\) ... ..."
Similar to above. We go $\frac1{q_0},\frac2{q_0},\frac3{q_0},\ldots$ until we come to an integer $m$ such that $x<\frac m{q_0}$; then $p_0$ is the least such integer $m$.

Peter said:
Question 3

In the remarks after the proof of Theorem 3.1.1 we read the following:

" ... ... Then \(\displaystyle x \lt p_0 / q_0 \lt y\) and \(\displaystyle r_0 = p_o / q_0\) is uniquely determined ... ... "Can someone please explain exactly why this is true ... ..
Because $q_0$ is the least integer $n$ such that $y-x>\frac1n$ and $p_0$ is the least integer $m$ such that $x<\frac m{q_0}$ – and those least integers are unique given $x$ and $y$. Note that this is not saying that there is a unique rational $r$ satisfying $x<r<y$ (which isn’t true since we know there are infinitely many rationals between $x$ and $y$); it is only saying that the process finds one particular rational between $x$ and $y$.
 
  • #3
Olinguito said:
Because $y-x>0$, so if you go $1,\frac12,\frac13,\ldots$ you will eventually come to an integer $n$ such that $y-x>\frac1n$. $q_0$ is the least such $n$.Similar to above. We go $\frac1{q_0},\frac2{q_0},\frac3{q_0},\ldots$ until we come to an integer $m$ such that $x<\frac m{q_0}$; then $p_0$ is the least such integer $m$.

Because $q_0$ is the least integer $n$ such that $y-x>\frac1n$ and $p_0$ is the least integer $m$ such that $x<\frac m{q_0}$ – and those least integers are unique given $x$ and $y$. Note that this is not saying that there is a unique rational $r$ satisfying $x<r<y$ (which isn’t true since we know there are infinitely many rationals between $x$ and $y$); it is only saying that the process finds one particular rational between $x$ and $y$.
Thanks Olinguito ... appreciate your reply and your help ...

But just a discussion point ...

You write:

" ... ...
Because $y-x>0$, so if you go $1,\frac12,\frac13,\ldots$ you will eventually come to an integer $n$ such that $y-x>\frac1n$. $q_0$ is the least such $n$. ... ... "

Now what you have said is definitely very plausible ... but is it rigorous ... what do you think ... can you make a more rigorous argument ... ... does a more rigorous argument exist ... ?

Or am I overplaying and exaggerating (indeed perhaps misrepresenting ...? ) the idea of rigour ... particularly in a case where what is to be proved might be sensibly regarded as obvious ,,, what do you think ...

Indeed ... what is rigour anyway ...

Peter

 
  • #4
Peter said:
You write:

" ... ... Because $y-x>0$, so if you go $1,\frac12,\frac13,\ldots$ you will eventually come to an integer $n$ such that $y-x>\frac1n$. $q_0$ is the least such $n$. ... ... "

Now what you have said is definitely very plausible ... but is it rigorous ... what do you think ... can you make a more rigorous argument ... ... does a more rigorous argument exist ... ?

Alternatively, you can take $q_0$ to be the smallest integer $n$ such that $n>\dfrac1{y-x}$.
 
  • #5
Olinguito said:
Alternatively, you can take $q_0$ to be the smallest integer $n$ such that $n>\dfrac1{y-x}$.
On reflection ... I have a further question ... concerning Garling's remark following the proof concerning statement (iii) ...

Garling's logic proceeds as follows:

We have \(\displaystyle x \gt 0\) and we want to determine a rational \(\displaystyle r\) with \(\displaystyle x \lt r \lt y\).

We first determine the least positive integer \(\displaystyle q_0\) such that \(\displaystyle 1/q_0 \lt y - x\) ...

We then determine the least positive integer such that such that \(\displaystyle x \lt p_0 / q_0\)

Then ...

...
Garling claims we have \(\displaystyle x \lt p_0 / q_0 \lt y\) ... ...


My question is as follows:

In the above process .. how are we sure that \(\displaystyle p_0 / q_0 \lt y\) ... ... ?

Help will be appreciated ...

Peter

 
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  • #6
$p_0$ is the least integer such that $x<\dfrac{p_0}{q_0}$​

$\implies\ \dfrac{p_0-1}{q_0}\ \leqslant\ x$

$\implies\ \dfrac{p_0}{q_0}\ \leqslant\ x+\dfrac{p_0}{q_0}\ <\ x+y-x\ =\ y$.
 
  • #7
Olinguito said:
$p_0$ is the least integer such that $x<\dfrac{p_0}{q_0}$​

$\implies\ \dfrac{p_0-1}{q_0}\ \leqslant\ x$

$\implies\ \dfrac{p_0}{q_0}\ \leqslant\ x+\dfrac{p_0}{q_0}\ <\ x+y-x\ =\ y$.
Thanks for all your help, Olinguito ,,,

Peter
 

FAQ: Exploring D. J. H. Garling's Theorem 3.1.1

1. What is D. J. H. Garling's Theorem 3.1.1?

D. J. H. Garling's Theorem 3.1.1 is a mathematical theorem that states the existence of a unique solution to a system of linear equations. It is often used in linear algebra and is a fundamental concept in mathematics.

2. How is D. J. H. Garling's Theorem 3.1.1 used in scientific research?

This theorem is used in various scientific fields, such as physics, engineering, and economics, to solve complex systems of equations. It allows researchers to find a single, unique solution to a set of equations, making it a powerful tool in problem-solving and data analysis.

3. Can you explain the proof of D. J. H. Garling's Theorem 3.1.1?

The proof of this theorem involves using mathematical techniques, such as Gaussian elimination and matrix operations, to show that a unique solution exists for a system of linear equations. It is a complex and rigorous proof that requires a strong understanding of linear algebra and mathematical reasoning.

4. What are the practical applications of D. J. H. Garling's Theorem 3.1.1?

This theorem has many practical applications, such as in solving optimization problems, modeling physical systems, and analyzing data in various fields. It is also used in computer science for solving systems of linear equations in algorithms and programming.

5. Are there any limitations to D. J. H. Garling's Theorem 3.1.1?

While this theorem is a powerful tool in mathematics and science, it does have some limitations. It only applies to systems of linear equations and cannot be used for non-linear systems. It also assumes that all variables in the equations are known and that the equations are consistent, meaning they have a solution. If these assumptions are not met, the theorem may not be applicable.

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