Solving polynomial equations by decomposition
$begingroup$
I have a very little background on mathematics. I have a very basic question about solving polynomial equations.
if we have $$P_n(x) = 0$$ where $P_n(x)$ is a polynomial of degree $n$.For example: $$x^5-3x^3+x^2-7x+3 = 0$$
Why we cannot just decompose the polynomial equation to a system of equations solving them separately and the solutions to the original polynomial will be the intersection of the roots of each equation:
$$begin{cases}
x^5-3x^3 = 0 \ text{and} \x^2-7x+3=0
end{cases}$$
There is a lot of possibilities (infinity?) to choose this sub-equations.
I know this seems very stupid but could you tell me which rule of logic I broke.
As far as I know, if $a=0$ and $b=0$ then we can add them to form $a+b=0$ we cannot go the other way, that is $a+b=0implies a=0 quadtext{and}, b=0$ why?
polynomials systems-of-equations
asked Jan 21 at 22:38
IamNotaMathematicianIamNotaMathematician
33
$endgroup$
add a comment |
$begingroup$
I have a very little background on mathematics. I have a very basic question about solving polynomial equations.
if we have $$P_n(x) = 0$$ where $P_n(x)$ is a polynomial of degree $n$.For example: $$x^5-3x^3+x^2-7x+3 = 0$$
Why we cannot just decompose the polynomial equation to a system of equations solving them separately and the solutions to the original polynomial will be the intersection of the roots of each equation:
$$begin{cases}
x^5-3x^3 = 0 \ text{and} \x^2-7x+3=0
end{cases}$$
There is a lot of possibilities (infinity?) to choose this sub-equations.
I know this seems very stupid but could you tell me which rule of logic I broke.
As far as I know, if $a=0$ and $b=0$ then we can add them to form $a+b=0$ we cannot go the other way, that is $a+b=0implies a=0 quadtext{and}, b=0$ why?
polynomials systems-of-equations
asked Jan 21 at 22:38
IamNotaMathematicianIamNotaMathematician
33
$endgroup$
2
$begingroup$
It does not follow from $a+b=0$ that $a=0$ and $b=0$.
$endgroup$
– J. W. Tanner
Jan 21 at 22:40
$begingroup$
@J.W.Tanner $a=0$ and $b=0$ then we can write $a+b=0$ why we cannot go the other way?
$endgroup$
– IamNotaMathematician
Jan 21 at 22:45
1
$begingroup$
It could be, for example, that $a=1$ and $b=-1$.
$endgroup$
– J. W. Tanner
Jan 21 at 22:52
$begingroup$
You have 2 equations for 1 unknown, apart very special cases you will get an incompatible system (no solution). Moreover not all solutions of a+b=0 are of the form a=0 and b=0
$endgroup$
– Picaud Vincent
Jan 21 at 22:53
$begingroup$
@PicaudVincent but we accept only the values that satisfy both equations
$endgroup$
– IamNotaMathematician
Jan 21 at 22:57
add a comment |
$begingroup$
I have a very little background on mathematics. I have a very basic question about solving polynomial equations.
if we have $$P_n(x) = 0$$ where $P_n(x)$ is a polynomial of degree $n$.For example: $$x^5-3x^3+x^2-7x+3 = 0$$
Why we cannot just decompose the polynomial equation to a system of equations solving them separately and the solutions to the original polynomial will be the intersection of the roots of each equation:
$$begin{cases}
x^5-3x^3 = 0 \ text{and} \x^2-7x+3=0
end{cases}$$
There is a lot of possibilities (infinity?) to choose this sub-equations.
I know this seems very stupid but could you tell me which rule of logic I broke.
As far as I know, if $a=0$ and $b=0$ then we can add them to form $a+b=0$ we cannot go the other way, that is $a+b=0implies a=0 quadtext{and}, b=0$ why?
polynomials systems-of-equations
asked Jan 21 at 22:38
IamNotaMathematicianIamNotaMathematician
33
$endgroup$
I have a very little background on mathematics. I have a very basic question about solving polynomial equations.
if we have $$P_n(x) = 0$$ where $P_n(x)$ is a polynomial of degree $n$.For example: $$x^5-3x^3+x^2-7x+3 = 0$$
Why we cannot just decompose the polynomial equation to a system of equations solving them separately and the solutions to the original polynomial will be the intersection of the roots of each equation:
$$begin{cases}
x^5-3x^3 = 0 \ text{and} \x^2-7x+3=0
end{cases}$$
There is a lot of possibilities (infinity?) to choose this sub-equations.
I know this seems very stupid but could you tell me which rule of logic I broke.
As far as I know, if $a=0$ and $b=0$ then we can add them to form $a+b=0$ we cannot go the other way, that is $a+b=0implies a=0 quadtext{and}, b=0$ why?
polynomials systems-of-equations
polynomials systems-of-equations
asked Jan 21 at 22:38
IamNotaMathematicianIamNotaMathematician
33
asked Jan 21 at 22:38
IamNotaMathematicianIamNotaMathematician
33
edited Jan 21 at 23:19
IamNotaMathematician
asked Jan 21 at 22:38
IamNotaMathematicianIamNotaMathematician
33
asked Jan 21 at 22:38
IamNotaMathematicianIamNotaMathematician
33
asked Jan 21 at 22:38
IamNotaMathematicianIamNotaMathematician
33
33
2
$begingroup$
It does not follow from $a+b=0$ that $a=0$ and $b=0$.
$endgroup$
– J. W. Tanner
Jan 21 at 22:40
$begingroup$
@J.W.Tanner $a=0$ and $b=0$ then we can write $a+b=0$ why we cannot go the other way?
$endgroup$
– IamNotaMathematician
Jan 21 at 22:45
1
$begingroup$
It could be, for example, that $a=1$ and $b=-1$.
$endgroup$
– J. W. Tanner
Jan 21 at 22:52
$begingroup$
You have 2 equations for 1 unknown, apart very special cases you will get an incompatible system (no solution). Moreover not all solutions of a+b=0 are of the form a=0 and b=0
$endgroup$
– Picaud Vincent
Jan 21 at 22:53
$begingroup$
@PicaudVincent but we accept only the values that satisfy both equations
$endgroup$
– IamNotaMathematician
Jan 21 at 22:57
add a comment |
2
$begingroup$
It does not follow from $a+b=0$ that $a=0$ and $b=0$.
$endgroup$
– J. W. Tanner
Jan 21 at 22:40
$begingroup$
@J.W.Tanner $a=0$ and $b=0$ then we can write $a+b=0$ why we cannot go the other way?
$endgroup$
– IamNotaMathematician
Jan 21 at 22:45
1
$begingroup$
It could be, for example, that $a=1$ and $b=-1$.
$endgroup$
– J. W. Tanner
Jan 21 at 22:52
$begingroup$
You have 2 equations for 1 unknown, apart very special cases you will get an incompatible system (no solution). Moreover not all solutions of a+b=0 are of the form a=0 and b=0
$endgroup$
– Picaud Vincent
Jan 21 at 22:53
$begingroup$
@PicaudVincent but we accept only the values that satisfy both equations
$endgroup$
– IamNotaMathematician
Jan 21 at 22:57
2
2
$begingroup$
It does not follow from $a+b=0$ that $a=0$ and $b=0$.
$endgroup$
– J. W. Tanner
Jan 21 at 22:40
$begingroup$
It does not follow from $a+b=0$ that $a=0$ and $b=0$.
$endgroup$
– J. W. Tanner
Jan 21 at 22:40
$begingroup$
@J.W.Tanner $a=0$ and $b=0$ then we can write $a+b=0$ why we cannot go the other way?
$endgroup$
– IamNotaMathematician
Jan 21 at 22:45
$begingroup$
@J.W.Tanner $a=0$ and $b=0$ then we can write $a+b=0$ why we cannot go the other way?
$endgroup$
– IamNotaMathematician
Jan 21 at 22:45
1
1
$begingroup$
It could be, for example, that $a=1$ and $b=-1$.
$endgroup$
– J. W. Tanner
Jan 21 at 22:52
$begingroup$
It could be, for example, that $a=1$ and $b=-1$.
$endgroup$
– J. W. Tanner
Jan 21 at 22:52
$begingroup$
You have 2 equations for 1 unknown, apart very special cases you will get an incompatible system (no solution). Moreover not all solutions of a+b=0 are of the form a=0 and b=0
$endgroup$
– Picaud Vincent
Jan 21 at 22:53
$begingroup$
You have 2 equations for 1 unknown, apart very special cases you will get an incompatible system (no solution). Moreover not all solutions of a+b=0 are of the form a=0 and b=0
$endgroup$
– Picaud Vincent
Jan 21 at 22:53
$begingroup$
@PicaudVincent but we accept only the values that satisfy both equations
$endgroup$
– IamNotaMathematician
Jan 21 at 22:57
$begingroup$
@PicaudVincent but we accept only the values that satisfy both equations
$endgroup$
– IamNotaMathematician
Jan 21 at 22:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have that: $$a +b =0 to a=-b$$
Which means it holds for $a=b=0$, but also that you can't go the other way, as you said.
Your system of equations should be: $$begin{cases}
x^5-3x^3 = k \ text{and} \x^2-7x+3=-k
end{cases}$$
We can solve $x^2-7x+(3+k)=0$ for $x=frac{7pmsqrt{37-4k}}{2}$, and then insert than into the quintic for solutions, but this results in:
$$frac{(7pmsqrt{37-4k})^5}{32}-frac{3(7pmsqrt{37-4k})^3}{8}-k=0$$
One question springs to mind, is this really any simpler? To my mind, not at all.
We can substitute $t=7+sqrt{37-4k}$, but this requires the solution of:
$$t^5-12t^3+8t^2-112t+96=0$$
which is more complicated than the polynomial we are trying to solve.
All in all: is this a valid method? Absolutely. Is it practical? Absolutely not in my opinion.
answered Jan 21 at 23:25
Rhys HughesRhys Hughes
6,9011530
$endgroup$
$begingroup$
I think that is what I was looking for thank you
$endgroup$
– IamNotaMathematician
Jan 21 at 23:34
add a comment |
$begingroup$
Note that this is not exactly the same question, for example what if $x^5 - 3x^3 = 1$ and $x^2 - 7x + 3 = -1$? Then the original equation would be satisfied, but neither of the original equations would be satisfied. However, they also take on these values at different values of $x$. Additionally, if you continued to break down the equation into systems of equations, you could eventually get to the case of a $0th$ order polynomial which could never be equal to $0$. In your particular example this would be looking for solutions of $3 = 0$ which clearly doesn't make sense.
One counter-example is enough to show that there are solutions not contained in the intersection of the zeros. As for the proving that the intersection of the zeros is also a solution to the original polynomial do as follows:
Let $P_n(x) = P_k(x) + P_m(k).$ Suppose $P_k(a) = 0,$ and $P_m(a) = 0$. Then, $P_n(a) = P_k(a) + P_m(a) = 0 +0.$
answered Jan 21 at 22:48
Jack PfaffingerJack Pfaffinger
162112
$endgroup$
$begingroup$
Yes I know but why it doesn't work how to prove that
$endgroup$
– IamNotaMathematician
Jan 21 at 22:49
$begingroup$
You are correct however, that if both equations are $0$ for the same value of $x$, then that would be a solution. Consider for example $x^5 - 3x^3+x^2-7x = 0$. You can break it into: $x^5 - 3x^3 = 0,$$ and x^2-7x = 0.$ Both of which have solutions at $x = 0$, and so the original equation also has a solution at $x = 0$.
$endgroup$
– Jack Pfaffinger
Jan 21 at 22:52
$begingroup$
but how could you prove that there is no such decomposition not necessarily 2 terms at a time. in other words if I say for a given polynomial equation there is an equivalent system of simple equations how can you disprove that
$endgroup$
– IamNotaMathematician
Jan 21 at 23:03
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
We have that: $$a +b =0 to a=-b$$
Which means it holds for $a=b=0$, but also that you can't go the other way, as you said.
Your system of equations should be: $$begin{cases}
x^5-3x^3 = k \ text{and} \x^2-7x+3=-k
end{cases}$$
We can solve $x^2-7x+(3+k)=0$ for $x=frac{7pmsqrt{37-4k}}{2}$, and then insert than into the quintic for solutions, but this results in:
$$frac{(7pmsqrt{37-4k})^5}{32}-frac{3(7pmsqrt{37-4k})^3}{8}-k=0$$
One question springs to mind, is this really any simpler? To my mind, not at all.
We can substitute $t=7+sqrt{37-4k}$, but this requires the solution of:
$$t^5-12t^3+8t^2-112t+96=0$$
which is more complicated than the polynomial we are trying to solve.
All in all: is this a valid method? Absolutely. Is it practical? Absolutely not in my opinion.
answered Jan 21 at 23:25
Rhys HughesRhys Hughes
6,9011530
$endgroup$
$begingroup$
I think that is what I was looking for thank you
$endgroup$
– IamNotaMathematician
Jan 21 at 23:34
add a comment |
$begingroup$
We have that: $$a +b =0 to a=-b$$
Which means it holds for $a=b=0$, but also that you can't go the other way, as you said.
Your system of equations should be: $$begin{cases}
x^5-3x^3 = k \ text{and} \x^2-7x+3=-k
end{cases}$$
We can solve $x^2-7x+(3+k)=0$ for $x=frac{7pmsqrt{37-4k}}{2}$, and then insert than into the quintic for solutions, but this results in:
$$frac{(7pmsqrt{37-4k})^5}{32}-frac{3(7pmsqrt{37-4k})^3}{8}-k=0$$
One question springs to mind, is this really any simpler? To my mind, not at all.
We can substitute $t=7+sqrt{37-4k}$, but this requires the solution of:
$$t^5-12t^3+8t^2-112t+96=0$$
which is more complicated than the polynomial we are trying to solve.
All in all: is this a valid method? Absolutely. Is it practical? Absolutely not in my opinion.
answered Jan 21 at 23:25
Rhys HughesRhys Hughes
6,9011530
$endgroup$
$begingroup$
I think that is what I was looking for thank you
$endgroup$
– IamNotaMathematician
Jan 21 at 23:34
add a comment |
$begingroup$
We have that: $$a +b =0 to a=-b$$
Which means it holds for $a=b=0$, but also that you can't go the other way, as you said.
Your system of equations should be: $$begin{cases}
x^5-3x^3 = k \ text{and} \x^2-7x+3=-k
end{cases}$$
We can solve $x^2-7x+(3+k)=0$ for $x=frac{7pmsqrt{37-4k}}{2}$, and then insert than into the quintic for solutions, but this results in:
$$frac{(7pmsqrt{37-4k})^5}{32}-frac{3(7pmsqrt{37-4k})^3}{8}-k=0$$
One question springs to mind, is this really any simpler? To my mind, not at all.
We can substitute $t=7+sqrt{37-4k}$, but this requires the solution of:
$$t^5-12t^3+8t^2-112t+96=0$$
which is more complicated than the polynomial we are trying to solve.
All in all: is this a valid method? Absolutely. Is it practical? Absolutely not in my opinion.
answered Jan 21 at 23:25
Rhys HughesRhys Hughes
6,9011530
$endgroup$
We have that: $$a +b =0 to a=-b$$
Which means it holds for $a=b=0$, but also that you can't go the other way, as you said.
Your system of equations should be: $$begin{cases}
x^5-3x^3 = k \ text{and} \x^2-7x+3=-k
end{cases}$$
We can solve $x^2-7x+(3+k)=0$ for $x=frac{7pmsqrt{37-4k}}{2}$, and then insert than into the quintic for solutions, but this results in:
$$frac{(7pmsqrt{37-4k})^5}{32}-frac{3(7pmsqrt{37-4k})^3}{8}-k=0$$
One question springs to mind, is this really any simpler? To my mind, not at all.
We can substitute $t=7+sqrt{37-4k}$, but this requires the solution of:
$$t^5-12t^3+8t^2-112t+96=0$$
which is more complicated than the polynomial we are trying to solve.
All in all: is this a valid method? Absolutely. Is it practical? Absolutely not in my opinion.
answered Jan 21 at 23:25
Rhys HughesRhys Hughes
6,9011530
answered Jan 21 at 23:25
Rhys HughesRhys Hughes
6,9011530
answered Jan 21 at 23:25
Rhys HughesRhys Hughes
6,9011530
answered Jan 21 at 23:25
Rhys HughesRhys Hughes
6,9011530
6,9011530
$begingroup$
I think that is what I was looking for thank you
$endgroup$
– IamNotaMathematician
Jan 21 at 23:34
add a comment |
$begingroup$
I think that is what I was looking for thank you
$endgroup$
– IamNotaMathematician
Jan 21 at 23:34
$begingroup$
I think that is what I was looking for thank you
$endgroup$
– IamNotaMathematician
Jan 21 at 23:34
$begingroup$
I think that is what I was looking for thank you
$endgroup$
– IamNotaMathematician
Jan 21 at 23:34
add a comment |
$begingroup$
Note that this is not exactly the same question, for example what if $x^5 - 3x^3 = 1$ and $x^2 - 7x + 3 = -1$? Then the original equation would be satisfied, but neither of the original equations would be satisfied. However, they also take on these values at different values of $x$. Additionally, if you continued to break down the equation into systems of equations, you could eventually get to the case of a $0th$ order polynomial which could never be equal to $0$. In your particular example this would be looking for solutions of $3 = 0$ which clearly doesn't make sense.
One counter-example is enough to show that there are solutions not contained in the intersection of the zeros. As for the proving that the intersection of the zeros is also a solution to the original polynomial do as follows:
Let $P_n(x) = P_k(x) + P_m(k).$ Suppose $P_k(a) = 0,$ and $P_m(a) = 0$. Then, $P_n(a) = P_k(a) + P_m(a) = 0 +0.$
answered Jan 21 at 22:48
Jack PfaffingerJack Pfaffinger
162112
$endgroup$
$begingroup$
Yes I know but why it doesn't work how to prove that
$endgroup$
– IamNotaMathematician
Jan 21 at 22:49
$begingroup$
You are correct however, that if both equations are $0$ for the same value of $x$, then that would be a solution. Consider for example $x^5 - 3x^3+x^2-7x = 0$. You can break it into: $x^5 - 3x^3 = 0,$$ and x^2-7x = 0.$ Both of which have solutions at $x = 0$, and so the original equation also has a solution at $x = 0$.
$endgroup$
– Jack Pfaffinger
Jan 21 at 22:52
$begingroup$
but how could you prove that there is no such decomposition not necessarily 2 terms at a time. in other words if I say for a given polynomial equation there is an equivalent system of simple equations how can you disprove that
$endgroup$
– IamNotaMathematician
Jan 21 at 23:03
add a comment |
$begingroup$
Note that this is not exactly the same question, for example what if $x^5 - 3x^3 = 1$ and $x^2 - 7x + 3 = -1$? Then the original equation would be satisfied, but neither of the original equations would be satisfied. However, they also take on these values at different values of $x$. Additionally, if you continued to break down the equation into systems of equations, you could eventually get to the case of a $0th$ order polynomial which could never be equal to $0$. In your particular example this would be looking for solutions of $3 = 0$ which clearly doesn't make sense.
One counter-example is enough to show that there are solutions not contained in the intersection of the zeros. As for the proving that the intersection of the zeros is also a solution to the original polynomial do as follows:
Let $P_n(x) = P_k(x) + P_m(k).$ Suppose $P_k(a) = 0,$ and $P_m(a) = 0$. Then, $P_n(a) = P_k(a) + P_m(a) = 0 +0.$
answered Jan 21 at 22:48
Jack PfaffingerJack Pfaffinger
162112
$endgroup$
$begingroup$
Yes I know but why it doesn't work how to prove that
$endgroup$
– IamNotaMathematician
Jan 21 at 22:49
$begingroup$
You are correct however, that if both equations are $0$ for the same value of $x$, then that would be a solution. Consider for example $x^5 - 3x^3+x^2-7x = 0$. You can break it into: $x^5 - 3x^3 = 0,$$ and x^2-7x = 0.$ Both of which have solutions at $x = 0$, and so the original equation also has a solution at $x = 0$.
$endgroup$
– Jack Pfaffinger
Jan 21 at 22:52
$begingroup$
but how could you prove that there is no such decomposition not necessarily 2 terms at a time. in other words if I say for a given polynomial equation there is an equivalent system of simple equations how can you disprove that
$endgroup$
– IamNotaMathematician
Jan 21 at 23:03
add a comment |
$begingroup$
Note that this is not exactly the same question, for example what if $x^5 - 3x^3 = 1$ and $x^2 - 7x + 3 = -1$? Then the original equation would be satisfied, but neither of the original equations would be satisfied. However, they also take on these values at different values of $x$. Additionally, if you continued to break down the equation into systems of equations, you could eventually get to the case of a $0th$ order polynomial which could never be equal to $0$. In your particular example this would be looking for solutions of $3 = 0$ which clearly doesn't make sense.
One counter-example is enough to show that there are solutions not contained in the intersection of the zeros. As for the proving that the intersection of the zeros is also a solution to the original polynomial do as follows:
Let $P_n(x) = P_k(x) + P_m(k).$ Suppose $P_k(a) = 0,$ and $P_m(a) = 0$. Then, $P_n(a) = P_k(a) + P_m(a) = 0 +0.$
answered Jan 21 at 22:48
Jack PfaffingerJack Pfaffinger
162112
$endgroup$
Note that this is not exactly the same question, for example what if $x^5 - 3x^3 = 1$ and $x^2 - 7x + 3 = -1$? Then the original equation would be satisfied, but neither of the original equations would be satisfied. However, they also take on these values at different values of $x$. Additionally, if you continued to break down the equation into systems of equations, you could eventually get to the case of a $0th$ order polynomial which could never be equal to $0$. In your particular example this would be looking for solutions of $3 = 0$ which clearly doesn't make sense.
One counter-example is enough to show that there are solutions not contained in the intersection of the zeros. As for the proving that the intersection of the zeros is also a solution to the original polynomial do as follows:
Let $P_n(x) = P_k(x) + P_m(k).$ Suppose $P_k(a) = 0,$ and $P_m(a) = 0$. Then, $P_n(a) = P_k(a) + P_m(a) = 0 +0.$
answered Jan 21 at 22:48
Jack PfaffingerJack Pfaffinger
162112
edited Jan 21 at 22:58
answered Jan 21 at 22:48
Jack PfaffingerJack Pfaffinger
162112
answered Jan 21 at 22:48
Jack PfaffingerJack Pfaffinger
162112
answered Jan 21 at 22:48
Jack PfaffingerJack Pfaffinger
162112
162112
$begingroup$
Yes I know but why it doesn't work how to prove that
$endgroup$
– IamNotaMathematician
Jan 21 at 22:49
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You are correct however, that if both equations are $0$ for the same value of $x$, then that would be a solution. Consider for example $x^5 - 3x^3+x^2-7x = 0$. You can break it into: $x^5 - 3x^3 = 0,$$ and x^2-7x = 0.$ Both of which have solutions at $x = 0$, and so the original equation also has a solution at $x = 0$.
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– Jack Pfaffinger
Jan 21 at 22:52
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but how could you prove that there is no such decomposition not necessarily 2 terms at a time. in other words if I say for a given polynomial equation there is an equivalent system of simple equations how can you disprove that
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– IamNotaMathematician
Jan 21 at 23:03
add a comment |
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Yes I know but why it doesn't work how to prove that
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– IamNotaMathematician
Jan 21 at 22:49
$begingroup$
You are correct however, that if both equations are $0$ for the same value of $x$, then that would be a solution. Consider for example $x^5 - 3x^3+x^2-7x = 0$. You can break it into: $x^5 - 3x^3 = 0,$$ and x^2-7x = 0.$ Both of which have solutions at $x = 0$, and so the original equation also has a solution at $x = 0$.
$endgroup$
– Jack Pfaffinger
Jan 21 at 22:52
$begingroup$
but how could you prove that there is no such decomposition not necessarily 2 terms at a time. in other words if I say for a given polynomial equation there is an equivalent system of simple equations how can you disprove that
$endgroup$
– IamNotaMathematician
Jan 21 at 23:03
$begingroup$
Yes I know but why it doesn't work how to prove that
$endgroup$
– IamNotaMathematician
Jan 21 at 22:49
$begingroup$
Yes I know but why it doesn't work how to prove that
$endgroup$
– IamNotaMathematician
Jan 21 at 22:49
$begingroup$
You are correct however, that if both equations are $0$ for the same value of $x$, then that would be a solution. Consider for example $x^5 - 3x^3+x^2-7x = 0$. You can break it into: $x^5 - 3x^3 = 0,$$ and x^2-7x = 0.$ Both of which have solutions at $x = 0$, and so the original equation also has a solution at $x = 0$.
$endgroup$
– Jack Pfaffinger
Jan 21 at 22:52
$begingroup$
You are correct however, that if both equations are $0$ for the same value of $x$, then that would be a solution. Consider for example $x^5 - 3x^3+x^2-7x = 0$. You can break it into: $x^5 - 3x^3 = 0,$$ and x^2-7x = 0.$ Both of which have solutions at $x = 0$, and so the original equation also has a solution at $x = 0$.
$endgroup$
– Jack Pfaffinger
Jan 21 at 22:52
$begingroup$
but how could you prove that there is no such decomposition not necessarily 2 terms at a time. in other words if I say for a given polynomial equation there is an equivalent system of simple equations how can you disprove that
$endgroup$
– IamNotaMathematician
Jan 21 at 23:03
$begingroup$
but how could you prove that there is no such decomposition not necessarily 2 terms at a time. in other words if I say for a given polynomial equation there is an equivalent system of simple equations how can you disprove that
$endgroup$
– IamNotaMathematician
Jan 21 at 23:03
add a comment |
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It does not follow from $a+b=0$ that $a=0$ and $b=0$.
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– J. W. Tanner
Jan 21 at 22:40
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@J.W.Tanner $a=0$ and $b=0$ then we can write $a+b=0$ why we cannot go the other way?
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– IamNotaMathematician
Jan 21 at 22:45
1
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It could be, for example, that $a=1$ and $b=-1$.
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– J. W. Tanner
Jan 21 at 22:52
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You have 2 equations for 1 unknown, apart very special cases you will get an incompatible system (no solution). Moreover not all solutions of a+b=0 are of the form a=0 and b=0
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– Picaud Vincent
Jan 21 at 22:53
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@PicaudVincent but we accept only the values that satisfy both equations
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– IamNotaMathematician
Jan 21 at 22:57