Is $det(A)=0$ a good indicator to say that a matrix is not invertible?
$begingroup$
In finite elements, for example, appears huge sparce (CRS) matrices (matrices with a lot of zeros). It is possible that matlab (or some other program) calculates $det(A)=0$ even though the matrix is invertible?
matlab numerical-linear-algebra sparse-matrices
asked Dec 5 '18 at 3:06
yeminoyemino
2831314
$endgroup$
add a comment |
$begingroup$
In finite elements, for example, appears huge sparce (CRS) matrices (matrices with a lot of zeros). It is possible that matlab (or some other program) calculates $det(A)=0$ even though the matrix is invertible?
matlab numerical-linear-algebra sparse-matrices
asked Dec 5 '18 at 3:06
yeminoyemino
2831314
$endgroup$
1
$begingroup$
If program doesn't use exact arithmetic but rounds off, then yes possible.
$endgroup$
– coffeemath
Dec 5 '18 at 3:10
add a comment |
$begingroup$
In finite elements, for example, appears huge sparce (CRS) matrices (matrices with a lot of zeros). It is possible that matlab (or some other program) calculates $det(A)=0$ even though the matrix is invertible?
matlab numerical-linear-algebra sparse-matrices
asked Dec 5 '18 at 3:06
yeminoyemino
2831314
$endgroup$
In finite elements, for example, appears huge sparce (CRS) matrices (matrices with a lot of zeros). It is possible that matlab (or some other program) calculates $det(A)=0$ even though the matrix is invertible?
matlab numerical-linear-algebra sparse-matrices
matlab numerical-linear-algebra sparse-matrices
asked Dec 5 '18 at 3:06
yeminoyemino
2831314
asked Dec 5 '18 at 3:06
yeminoyemino
2831314
asked Dec 5 '18 at 3:06
yeminoyemino
2831314
asked Dec 5 '18 at 3:06
yeminoyemino
2831314
asked Dec 5 '18 at 3:06
yeminoyemino
2831314
2831314
1
$begingroup$
If program doesn't use exact arithmetic but rounds off, then yes possible.
$endgroup$
– coffeemath
Dec 5 '18 at 3:10
add a comment |
1
$begingroup$
If program doesn't use exact arithmetic but rounds off, then yes possible.
$endgroup$
– coffeemath
Dec 5 '18 at 3:10
1
1
$begingroup$
If program doesn't use exact arithmetic but rounds off, then yes possible.
$endgroup$
– coffeemath
Dec 5 '18 at 3:10
$begingroup$
If program doesn't use exact arithmetic but rounds off, then yes possible.
$endgroup$
– coffeemath
Dec 5 '18 at 3:10
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The determinant is takes a long time to compute for large matrices. A better way is to look for the smallest singular values of your matrix. If they are 0 or close to machine precision, then it is either not invertible or so poorly conditioned that it probably isn't worth it to invert it. If this is the case, then you can either form a low rank approximation and get an approximate answer or try to reformulate your problem.
answered Dec 5 '18 at 3:12
whpowell96whpowell96
56615
$endgroup$
$begingroup$
The smallest singular value is not enough, it needs to be related to the largest one. Their ratio (or the reciprocal value) of the spectral condition number give you relative distance to the closest singular matrix (measured in the same norm).
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 13:55
$begingroup$
This is true, but in practice your largest singular value is typically much larger than machine precision so singular values on the order of $10^{-16}$ will almost always lead to conditioning problems
$endgroup$
– whpowell96
Dec 5 '18 at 16:31
$begingroup$
I'm not sure you got my point. You need to consider the value of $sigma_min$ relative to $sigma_max$, not its absolute value. Consider $A=10^{-16}I$. The minimal singular value of $A$ is $10^{-16}$ but its perfectly well conditioned.
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 16:37
$begingroup$
I am aware of the definition of the 2-conditoin number. I am saying that in practice, your largest singular value will never be that low because if it is, you are probably losing precision due to the problem being poorly scaled or something.
$endgroup$
– whpowell96
Dec 5 '18 at 16:41
add a comment |
$begingroup$
Computing determinant of a matrix is quite sensitive to round-off. On top of that, it is easy to obtain a zero or infinite determinant as output of computational procedures due to floating precision underflow or overflow.
Consider, e.g., $A_n=0.1times I_n$, where $I_n$ is the $ntimes n$ identity matrix. We have $det(A_n)=10^{-n}$. If $n$ is large enough (324 for double precision), standard techniques to compute the determinant will report you zero although the matrix $A_n$ itself is perfectly conditioned and invertible.
Conditioning of the matrix is a better measure of "(non)singularity" in numerical computations. It gives you information on what is the sensitivity of the matrix "inversion". This is the usual definition of the condition number. Higher the condition number, more sensitive the solution of $Ax=b$ to the perturbations of the input and to round-off.
On top of that, you know how far is the matrix from the nearest singular matrix. If $kappa(A)$ is the condition number of a nonsingular $A$ in some suitable norm (usually one of the three popular $p$-norms), we know that there is a $delta A$ such that $|delta A|/|A|=1/kappa(A)$ is singular. Higher the condition number, closer we are to a singular matrix. Eventually, if $1/kappa(A)approxepsilon$, where $epsilon$ is the machine precision (e.g., $approx 10^{-16}$ for the double precision floating point arithmetic), the matrix is considered numerically singular.
answered Dec 5 '18 at 17:10
Algebraic PavelAlgebraic Pavel
16.3k31840
$endgroup$
add a comment |
$begingroup$
Absolutely.
There’s always round off error, and the numerical stability of algorithms for calculating determinates can be highly unstable.
There are numberical techniques to find inverses for sparse matrices - I don’t know any, but Google will.
answered Dec 5 '18 at 3:40
user458276user458276
31629
$endgroup$
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
The determinant is takes a long time to compute for large matrices. A better way is to look for the smallest singular values of your matrix. If they are 0 or close to machine precision, then it is either not invertible or so poorly conditioned that it probably isn't worth it to invert it. If this is the case, then you can either form a low rank approximation and get an approximate answer or try to reformulate your problem.
answered Dec 5 '18 at 3:12
whpowell96whpowell96
56615
$endgroup$
$begingroup$
The smallest singular value is not enough, it needs to be related to the largest one. Their ratio (or the reciprocal value) of the spectral condition number give you relative distance to the closest singular matrix (measured in the same norm).
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 13:55
$begingroup$
This is true, but in practice your largest singular value is typically much larger than machine precision so singular values on the order of $10^{-16}$ will almost always lead to conditioning problems
$endgroup$
– whpowell96
Dec 5 '18 at 16:31
$begingroup$
I'm not sure you got my point. You need to consider the value of $sigma_min$ relative to $sigma_max$, not its absolute value. Consider $A=10^{-16}I$. The minimal singular value of $A$ is $10^{-16}$ but its perfectly well conditioned.
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 16:37
$begingroup$
I am aware of the definition of the 2-conditoin number. I am saying that in practice, your largest singular value will never be that low because if it is, you are probably losing precision due to the problem being poorly scaled or something.
$endgroup$
– whpowell96
Dec 5 '18 at 16:41
add a comment |
$begingroup$
The determinant is takes a long time to compute for large matrices. A better way is to look for the smallest singular values of your matrix. If they are 0 or close to machine precision, then it is either not invertible or so poorly conditioned that it probably isn't worth it to invert it. If this is the case, then you can either form a low rank approximation and get an approximate answer or try to reformulate your problem.
answered Dec 5 '18 at 3:12
whpowell96whpowell96
56615
$endgroup$
$begingroup$
The smallest singular value is not enough, it needs to be related to the largest one. Their ratio (or the reciprocal value) of the spectral condition number give you relative distance to the closest singular matrix (measured in the same norm).
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 13:55
$begingroup$
This is true, but in practice your largest singular value is typically much larger than machine precision so singular values on the order of $10^{-16}$ will almost always lead to conditioning problems
$endgroup$
– whpowell96
Dec 5 '18 at 16:31
$begingroup$
I'm not sure you got my point. You need to consider the value of $sigma_min$ relative to $sigma_max$, not its absolute value. Consider $A=10^{-16}I$. The minimal singular value of $A$ is $10^{-16}$ but its perfectly well conditioned.
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 16:37
$begingroup$
I am aware of the definition of the 2-conditoin number. I am saying that in practice, your largest singular value will never be that low because if it is, you are probably losing precision due to the problem being poorly scaled or something.
$endgroup$
– whpowell96
Dec 5 '18 at 16:41
add a comment |
$begingroup$
The determinant is takes a long time to compute for large matrices. A better way is to look for the smallest singular values of your matrix. If they are 0 or close to machine precision, then it is either not invertible or so poorly conditioned that it probably isn't worth it to invert it. If this is the case, then you can either form a low rank approximation and get an approximate answer or try to reformulate your problem.
answered Dec 5 '18 at 3:12
whpowell96whpowell96
56615
$endgroup$
The determinant is takes a long time to compute for large matrices. A better way is to look for the smallest singular values of your matrix. If they are 0 or close to machine precision, then it is either not invertible or so poorly conditioned that it probably isn't worth it to invert it. If this is the case, then you can either form a low rank approximation and get an approximate answer or try to reformulate your problem.
answered Dec 5 '18 at 3:12
whpowell96whpowell96
56615
answered Dec 5 '18 at 3:12
whpowell96whpowell96
56615
answered Dec 5 '18 at 3:12
whpowell96whpowell96
56615
answered Dec 5 '18 at 3:12
whpowell96whpowell96
56615
56615
$begingroup$
The smallest singular value is not enough, it needs to be related to the largest one. Their ratio (or the reciprocal value) of the spectral condition number give you relative distance to the closest singular matrix (measured in the same norm).
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 13:55
$begingroup$
This is true, but in practice your largest singular value is typically much larger than machine precision so singular values on the order of $10^{-16}$ will almost always lead to conditioning problems
$endgroup$
– whpowell96
Dec 5 '18 at 16:31
$begingroup$
I'm not sure you got my point. You need to consider the value of $sigma_min$ relative to $sigma_max$, not its absolute value. Consider $A=10^{-16}I$. The minimal singular value of $A$ is $10^{-16}$ but its perfectly well conditioned.
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 16:37
$begingroup$
I am aware of the definition of the 2-conditoin number. I am saying that in practice, your largest singular value will never be that low because if it is, you are probably losing precision due to the problem being poorly scaled or something.
$endgroup$
– whpowell96
Dec 5 '18 at 16:41
add a comment |
$begingroup$
The smallest singular value is not enough, it needs to be related to the largest one. Their ratio (or the reciprocal value) of the spectral condition number give you relative distance to the closest singular matrix (measured in the same norm).
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 13:55
$begingroup$
This is true, but in practice your largest singular value is typically much larger than machine precision so singular values on the order of $10^{-16}$ will almost always lead to conditioning problems
$endgroup$
– whpowell96
Dec 5 '18 at 16:31
$begingroup$
I'm not sure you got my point. You need to consider the value of $sigma_min$ relative to $sigma_max$, not its absolute value. Consider $A=10^{-16}I$. The minimal singular value of $A$ is $10^{-16}$ but its perfectly well conditioned.
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 16:37
$begingroup$
I am aware of the definition of the 2-conditoin number. I am saying that in practice, your largest singular value will never be that low because if it is, you are probably losing precision due to the problem being poorly scaled or something.
$endgroup$
– whpowell96
Dec 5 '18 at 16:41
$begingroup$
The smallest singular value is not enough, it needs to be related to the largest one. Their ratio (or the reciprocal value) of the spectral condition number give you relative distance to the closest singular matrix (measured in the same norm).
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 13:55
$begingroup$
The smallest singular value is not enough, it needs to be related to the largest one. Their ratio (or the reciprocal value) of the spectral condition number give you relative distance to the closest singular matrix (measured in the same norm).
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 13:55
$begingroup$
This is true, but in practice your largest singular value is typically much larger than machine precision so singular values on the order of $10^{-16}$ will almost always lead to conditioning problems
$endgroup$
– whpowell96
Dec 5 '18 at 16:31
$begingroup$
This is true, but in practice your largest singular value is typically much larger than machine precision so singular values on the order of $10^{-16}$ will almost always lead to conditioning problems
$endgroup$
– whpowell96
Dec 5 '18 at 16:31
$begingroup$
I'm not sure you got my point. You need to consider the value of $sigma_min$ relative to $sigma_max$, not its absolute value. Consider $A=10^{-16}I$. The minimal singular value of $A$ is $10^{-16}$ but its perfectly well conditioned.
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 16:37
$begingroup$
I'm not sure you got my point. You need to consider the value of $sigma_min$ relative to $sigma_max$, not its absolute value. Consider $A=10^{-16}I$. The minimal singular value of $A$ is $10^{-16}$ but its perfectly well conditioned.
$endgroup$
– Algebraic Pavel
Dec 5 '18 at 16:37
$begingroup$
I am aware of the definition of the 2-conditoin number. I am saying that in practice, your largest singular value will never be that low because if it is, you are probably losing precision due to the problem being poorly scaled or something.
$endgroup$
– whpowell96
Dec 5 '18 at 16:41
$begingroup$
I am aware of the definition of the 2-conditoin number. I am saying that in practice, your largest singular value will never be that low because if it is, you are probably losing precision due to the problem being poorly scaled or something.
$endgroup$
– whpowell96
Dec 5 '18 at 16:41
add a comment |
$begingroup$
Computing determinant of a matrix is quite sensitive to round-off. On top of that, it is easy to obtain a zero or infinite determinant as output of computational procedures due to floating precision underflow or overflow.
Consider, e.g., $A_n=0.1times I_n$, where $I_n$ is the $ntimes n$ identity matrix. We have $det(A_n)=10^{-n}$. If $n$ is large enough (324 for double precision), standard techniques to compute the determinant will report you zero although the matrix $A_n$ itself is perfectly conditioned and invertible.
Conditioning of the matrix is a better measure of "(non)singularity" in numerical computations. It gives you information on what is the sensitivity of the matrix "inversion". This is the usual definition of the condition number. Higher the condition number, more sensitive the solution of $Ax=b$ to the perturbations of the input and to round-off.
On top of that, you know how far is the matrix from the nearest singular matrix. If $kappa(A)$ is the condition number of a nonsingular $A$ in some suitable norm (usually one of the three popular $p$-norms), we know that there is a $delta A$ such that $|delta A|/|A|=1/kappa(A)$ is singular. Higher the condition number, closer we are to a singular matrix. Eventually, if $1/kappa(A)approxepsilon$, where $epsilon$ is the machine precision (e.g., $approx 10^{-16}$ for the double precision floating point arithmetic), the matrix is considered numerically singular.
answered Dec 5 '18 at 17:10
Algebraic PavelAlgebraic Pavel
16.3k31840
$endgroup$
add a comment |
$begingroup$
Computing determinant of a matrix is quite sensitive to round-off. On top of that, it is easy to obtain a zero or infinite determinant as output of computational procedures due to floating precision underflow or overflow.
Consider, e.g., $A_n=0.1times I_n$, where $I_n$ is the $ntimes n$ identity matrix. We have $det(A_n)=10^{-n}$. If $n$ is large enough (324 for double precision), standard techniques to compute the determinant will report you zero although the matrix $A_n$ itself is perfectly conditioned and invertible.
Conditioning of the matrix is a better measure of "(non)singularity" in numerical computations. It gives you information on what is the sensitivity of the matrix "inversion". This is the usual definition of the condition number. Higher the condition number, more sensitive the solution of $Ax=b$ to the perturbations of the input and to round-off.
On top of that, you know how far is the matrix from the nearest singular matrix. If $kappa(A)$ is the condition number of a nonsingular $A$ in some suitable norm (usually one of the three popular $p$-norms), we know that there is a $delta A$ such that $|delta A|/|A|=1/kappa(A)$ is singular. Higher the condition number, closer we are to a singular matrix. Eventually, if $1/kappa(A)approxepsilon$, where $epsilon$ is the machine precision (e.g., $approx 10^{-16}$ for the double precision floating point arithmetic), the matrix is considered numerically singular.
answered Dec 5 '18 at 17:10
Algebraic PavelAlgebraic Pavel
16.3k31840
$endgroup$
add a comment |
$begingroup$
Computing determinant of a matrix is quite sensitive to round-off. On top of that, it is easy to obtain a zero or infinite determinant as output of computational procedures due to floating precision underflow or overflow.
Consider, e.g., $A_n=0.1times I_n$, where $I_n$ is the $ntimes n$ identity matrix. We have $det(A_n)=10^{-n}$. If $n$ is large enough (324 for double precision), standard techniques to compute the determinant will report you zero although the matrix $A_n$ itself is perfectly conditioned and invertible.
Conditioning of the matrix is a better measure of "(non)singularity" in numerical computations. It gives you information on what is the sensitivity of the matrix "inversion". This is the usual definition of the condition number. Higher the condition number, more sensitive the solution of $Ax=b$ to the perturbations of the input and to round-off.
On top of that, you know how far is the matrix from the nearest singular matrix. If $kappa(A)$ is the condition number of a nonsingular $A$ in some suitable norm (usually one of the three popular $p$-norms), we know that there is a $delta A$ such that $|delta A|/|A|=1/kappa(A)$ is singular. Higher the condition number, closer we are to a singular matrix. Eventually, if $1/kappa(A)approxepsilon$, where $epsilon$ is the machine precision (e.g., $approx 10^{-16}$ for the double precision floating point arithmetic), the matrix is considered numerically singular.
answered Dec 5 '18 at 17:10
Algebraic PavelAlgebraic Pavel
16.3k31840
$endgroup$
Computing determinant of a matrix is quite sensitive to round-off. On top of that, it is easy to obtain a zero or infinite determinant as output of computational procedures due to floating precision underflow or overflow.
Consider, e.g., $A_n=0.1times I_n$, where $I_n$ is the $ntimes n$ identity matrix. We have $det(A_n)=10^{-n}$. If $n$ is large enough (324 for double precision), standard techniques to compute the determinant will report you zero although the matrix $A_n$ itself is perfectly conditioned and invertible.
Conditioning of the matrix is a better measure of "(non)singularity" in numerical computations. It gives you information on what is the sensitivity of the matrix "inversion". This is the usual definition of the condition number. Higher the condition number, more sensitive the solution of $Ax=b$ to the perturbations of the input and to round-off.
On top of that, you know how far is the matrix from the nearest singular matrix. If $kappa(A)$ is the condition number of a nonsingular $A$ in some suitable norm (usually one of the three popular $p$-norms), we know that there is a $delta A$ such that $|delta A|/|A|=1/kappa(A)$ is singular. Higher the condition number, closer we are to a singular matrix. Eventually, if $1/kappa(A)approxepsilon$, where $epsilon$ is the machine precision (e.g., $approx 10^{-16}$ for the double precision floating point arithmetic), the matrix is considered numerically singular.
answered Dec 5 '18 at 17:10
Algebraic PavelAlgebraic Pavel
16.3k31840
answered Dec 5 '18 at 17:10
Algebraic PavelAlgebraic Pavel
16.3k31840
answered Dec 5 '18 at 17:10
Algebraic PavelAlgebraic Pavel
16.3k31840
answered Dec 5 '18 at 17:10
Algebraic PavelAlgebraic Pavel
16.3k31840
16.3k31840
add a comment |
add a comment |
$begingroup$
Absolutely.
There’s always round off error, and the numerical stability of algorithms for calculating determinates can be highly unstable.
There are numberical techniques to find inverses for sparse matrices - I don’t know any, but Google will.
answered Dec 5 '18 at 3:40
user458276user458276
31629
$endgroup$
add a comment |
$begingroup$
Absolutely.
There’s always round off error, and the numerical stability of algorithms for calculating determinates can be highly unstable.
There are numberical techniques to find inverses for sparse matrices - I don’t know any, but Google will.
answered Dec 5 '18 at 3:40
user458276user458276
31629
$endgroup$
add a comment |
$begingroup$
Absolutely.
There’s always round off error, and the numerical stability of algorithms for calculating determinates can be highly unstable.
There are numberical techniques to find inverses for sparse matrices - I don’t know any, but Google will.
answered Dec 5 '18 at 3:40
user458276user458276
31629
$endgroup$
Absolutely.
There’s always round off error, and the numerical stability of algorithms for calculating determinates can be highly unstable.
There are numberical techniques to find inverses for sparse matrices - I don’t know any, but Google will.
answered Dec 5 '18 at 3:40
user458276user458276
31629
answered Dec 5 '18 at 3:40
user458276user458276
31629
answered Dec 5 '18 at 3:40
user458276user458276
31629
answered Dec 5 '18 at 3:40
user458276user458276
31629
31629
add a comment |
add a comment |
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$begingroup$
If program doesn't use exact arithmetic but rounds off, then yes possible.
$endgroup$
– coffeemath
Dec 5 '18 at 3:10