Why we need the same number of equations and variables?

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I once read that to solve a system of equations, I need $n$ equations for $n$ variables.



I think I read it in a work by Euler. However, I have never managed to find a demonstration of this.
So, why is this true? What is the demonstration for any system of equations? I would appreciate a formal proof and another intuitive one, thanks in advance.










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    The answer to your question for linear equations is the substance of the first part of most standard linear algebra courses. An answer here would just echo that material. You can start at the wikipedia page en.wikipedia.org/wiki/System_of_linear_equations
    – Ethan Bolker
    Sep 8 at 13:00














up vote
2
down vote

favorite
2












I once read that to solve a system of equations, I need $n$ equations for $n$ variables.



I think I read it in a work by Euler. However, I have never managed to find a demonstration of this.
So, why is this true? What is the demonstration for any system of equations? I would appreciate a formal proof and another intuitive one, thanks in advance.










share|cite|improve this question



















  • 3




    The answer to your question for linear equations is the substance of the first part of most standard linear algebra courses. An answer here would just echo that material. You can start at the wikipedia page en.wikipedia.org/wiki/System_of_linear_equations
    – Ethan Bolker
    Sep 8 at 13:00












up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





I once read that to solve a system of equations, I need $n$ equations for $n$ variables.



I think I read it in a work by Euler. However, I have never managed to find a demonstration of this.
So, why is this true? What is the demonstration for any system of equations? I would appreciate a formal proof and another intuitive one, thanks in advance.










share|cite|improve this question















I once read that to solve a system of equations, I need $n$ equations for $n$ variables.



I think I read it in a work by Euler. However, I have never managed to find a demonstration of this.
So, why is this true? What is the demonstration for any system of equations? I would appreciate a formal proof and another intuitive one, thanks in advance.







algebra-precalculus systems-of-equations






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edited Sep 10 at 12:09









Harry Peter

5,52111439




5,52111439










asked Sep 8 at 12:48









Mattiu

745317




745317







  • 3




    The answer to your question for linear equations is the substance of the first part of most standard linear algebra courses. An answer here would just echo that material. You can start at the wikipedia page en.wikipedia.org/wiki/System_of_linear_equations
    – Ethan Bolker
    Sep 8 at 13:00












  • 3




    The answer to your question for linear equations is the substance of the first part of most standard linear algebra courses. An answer here would just echo that material. You can start at the wikipedia page en.wikipedia.org/wiki/System_of_linear_equations
    – Ethan Bolker
    Sep 8 at 13:00







3




3




The answer to your question for linear equations is the substance of the first part of most standard linear algebra courses. An answer here would just echo that material. You can start at the wikipedia page en.wikipedia.org/wiki/System_of_linear_equations
– Ethan Bolker
Sep 8 at 13:00




The answer to your question for linear equations is the substance of the first part of most standard linear algebra courses. An answer here would just echo that material. You can start at the wikipedia page en.wikipedia.org/wiki/System_of_linear_equations
– Ethan Bolker
Sep 8 at 13:00










1 Answer
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up vote
2
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It's not true in general.



Let $x_i in mathbbR$ where $i in 1, ldots, 100$.



$$sum_i=1^100x_i^2 =0.$$



We have $100$ variables and only one equation and we can solve it uniquely. That is $forall i in 1, ldots, 100, x_i=0$.






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  • Well, the solution is unique in $mathbbR$, but not in $mathbbC$...
    – Daniel Robert-Nicoud
    Sep 8 at 12:53






  • 3




    @DanielRobert-Nicoud Just change $x_i^2$ to $left|x_iright|$.
    – user202729
    Sep 8 at 12:57











  • @user202729 There's a deep difference between equations involving "nice" algebraic functions such as polynomials and those involving functions like taking the absolute value and similar functions. In the first case, the "$n$ equations for $n$ variables" can be formally understood in terms of algebraic geometry.
    – Daniel Robert-Nicoud
    Sep 8 at 14:01










  • Siong, can you explain me that, please?
    – Mattiu
    Sep 8 at 15:32










  • For the example that I constructed for real numbers, we know that for real number $x_i$, we have $x_i^2 ge 0$. If the sum of the squared terms is zero, then each term must be zero. I just constructed an example where we have one equation and multiple variables and yet it can be solved uniquely. In terms of linear algebra, if the $n$ equations are independent, then the coefficient matrix is invertible and we can invert a matrix.
    – Siong Thye Goh
    Sep 8 at 15:39











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













It's not true in general.



Let $x_i in mathbbR$ where $i in 1, ldots, 100$.



$$sum_i=1^100x_i^2 =0.$$



We have $100$ variables and only one equation and we can solve it uniquely. That is $forall i in 1, ldots, 100, x_i=0$.






share|cite|improve this answer




















  • Well, the solution is unique in $mathbbR$, but not in $mathbbC$...
    – Daniel Robert-Nicoud
    Sep 8 at 12:53






  • 3




    @DanielRobert-Nicoud Just change $x_i^2$ to $left|x_iright|$.
    – user202729
    Sep 8 at 12:57











  • @user202729 There's a deep difference between equations involving "nice" algebraic functions such as polynomials and those involving functions like taking the absolute value and similar functions. In the first case, the "$n$ equations for $n$ variables" can be formally understood in terms of algebraic geometry.
    – Daniel Robert-Nicoud
    Sep 8 at 14:01










  • Siong, can you explain me that, please?
    – Mattiu
    Sep 8 at 15:32










  • For the example that I constructed for real numbers, we know that for real number $x_i$, we have $x_i^2 ge 0$. If the sum of the squared terms is zero, then each term must be zero. I just constructed an example where we have one equation and multiple variables and yet it can be solved uniquely. In terms of linear algebra, if the $n$ equations are independent, then the coefficient matrix is invertible and we can invert a matrix.
    – Siong Thye Goh
    Sep 8 at 15:39















up vote
2
down vote













It's not true in general.



Let $x_i in mathbbR$ where $i in 1, ldots, 100$.



$$sum_i=1^100x_i^2 =0.$$



We have $100$ variables and only one equation and we can solve it uniquely. That is $forall i in 1, ldots, 100, x_i=0$.






share|cite|improve this answer




















  • Well, the solution is unique in $mathbbR$, but not in $mathbbC$...
    – Daniel Robert-Nicoud
    Sep 8 at 12:53






  • 3




    @DanielRobert-Nicoud Just change $x_i^2$ to $left|x_iright|$.
    – user202729
    Sep 8 at 12:57











  • @user202729 There's a deep difference between equations involving "nice" algebraic functions such as polynomials and those involving functions like taking the absolute value and similar functions. In the first case, the "$n$ equations for $n$ variables" can be formally understood in terms of algebraic geometry.
    – Daniel Robert-Nicoud
    Sep 8 at 14:01










  • Siong, can you explain me that, please?
    – Mattiu
    Sep 8 at 15:32










  • For the example that I constructed for real numbers, we know that for real number $x_i$, we have $x_i^2 ge 0$. If the sum of the squared terms is zero, then each term must be zero. I just constructed an example where we have one equation and multiple variables and yet it can be solved uniquely. In terms of linear algebra, if the $n$ equations are independent, then the coefficient matrix is invertible and we can invert a matrix.
    – Siong Thye Goh
    Sep 8 at 15:39













up vote
2
down vote










up vote
2
down vote









It's not true in general.



Let $x_i in mathbbR$ where $i in 1, ldots, 100$.



$$sum_i=1^100x_i^2 =0.$$



We have $100$ variables and only one equation and we can solve it uniquely. That is $forall i in 1, ldots, 100, x_i=0$.






share|cite|improve this answer












It's not true in general.



Let $x_i in mathbbR$ where $i in 1, ldots, 100$.



$$sum_i=1^100x_i^2 =0.$$



We have $100$ variables and only one equation and we can solve it uniquely. That is $forall i in 1, ldots, 100, x_i=0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 8 at 12:51









Siong Thye Goh

82.8k1456104




82.8k1456104











  • Well, the solution is unique in $mathbbR$, but not in $mathbbC$...
    – Daniel Robert-Nicoud
    Sep 8 at 12:53






  • 3




    @DanielRobert-Nicoud Just change $x_i^2$ to $left|x_iright|$.
    – user202729
    Sep 8 at 12:57











  • @user202729 There's a deep difference between equations involving "nice" algebraic functions such as polynomials and those involving functions like taking the absolute value and similar functions. In the first case, the "$n$ equations for $n$ variables" can be formally understood in terms of algebraic geometry.
    – Daniel Robert-Nicoud
    Sep 8 at 14:01










  • Siong, can you explain me that, please?
    – Mattiu
    Sep 8 at 15:32










  • For the example that I constructed for real numbers, we know that for real number $x_i$, we have $x_i^2 ge 0$. If the sum of the squared terms is zero, then each term must be zero. I just constructed an example where we have one equation and multiple variables and yet it can be solved uniquely. In terms of linear algebra, if the $n$ equations are independent, then the coefficient matrix is invertible and we can invert a matrix.
    – Siong Thye Goh
    Sep 8 at 15:39

















  • Well, the solution is unique in $mathbbR$, but not in $mathbbC$...
    – Daniel Robert-Nicoud
    Sep 8 at 12:53






  • 3




    @DanielRobert-Nicoud Just change $x_i^2$ to $left|x_iright|$.
    – user202729
    Sep 8 at 12:57











  • @user202729 There's a deep difference between equations involving "nice" algebraic functions such as polynomials and those involving functions like taking the absolute value and similar functions. In the first case, the "$n$ equations for $n$ variables" can be formally understood in terms of algebraic geometry.
    – Daniel Robert-Nicoud
    Sep 8 at 14:01










  • Siong, can you explain me that, please?
    – Mattiu
    Sep 8 at 15:32










  • For the example that I constructed for real numbers, we know that for real number $x_i$, we have $x_i^2 ge 0$. If the sum of the squared terms is zero, then each term must be zero. I just constructed an example where we have one equation and multiple variables and yet it can be solved uniquely. In terms of linear algebra, if the $n$ equations are independent, then the coefficient matrix is invertible and we can invert a matrix.
    – Siong Thye Goh
    Sep 8 at 15:39
















Well, the solution is unique in $mathbbR$, but not in $mathbbC$...
– Daniel Robert-Nicoud
Sep 8 at 12:53




Well, the solution is unique in $mathbbR$, but not in $mathbbC$...
– Daniel Robert-Nicoud
Sep 8 at 12:53




3




3




@DanielRobert-Nicoud Just change $x_i^2$ to $left|x_iright|$.
– user202729
Sep 8 at 12:57





@DanielRobert-Nicoud Just change $x_i^2$ to $left|x_iright|$.
– user202729
Sep 8 at 12:57













@user202729 There's a deep difference between equations involving "nice" algebraic functions such as polynomials and those involving functions like taking the absolute value and similar functions. In the first case, the "$n$ equations for $n$ variables" can be formally understood in terms of algebraic geometry.
– Daniel Robert-Nicoud
Sep 8 at 14:01




@user202729 There's a deep difference between equations involving "nice" algebraic functions such as polynomials and those involving functions like taking the absolute value and similar functions. In the first case, the "$n$ equations for $n$ variables" can be formally understood in terms of algebraic geometry.
– Daniel Robert-Nicoud
Sep 8 at 14:01












Siong, can you explain me that, please?
– Mattiu
Sep 8 at 15:32




Siong, can you explain me that, please?
– Mattiu
Sep 8 at 15:32












For the example that I constructed for real numbers, we know that for real number $x_i$, we have $x_i^2 ge 0$. If the sum of the squared terms is zero, then each term must be zero. I just constructed an example where we have one equation and multiple variables and yet it can be solved uniquely. In terms of linear algebra, if the $n$ equations are independent, then the coefficient matrix is invertible and we can invert a matrix.
– Siong Thye Goh
Sep 8 at 15:39





For the example that I constructed for real numbers, we know that for real number $x_i$, we have $x_i^2 ge 0$. If the sum of the squared terms is zero, then each term must be zero. I just constructed an example where we have one equation and multiple variables and yet it can be solved uniquely. In terms of linear algebra, if the $n$ equations are independent, then the coefficient matrix is invertible and we can invert a matrix.
– Siong Thye Goh
Sep 8 at 15:39


















 

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