What to be chosen c so that all elements of matrix B will be a round number. B=A.c; where A is the matrix of non integers.

Hello All,

Please guide me , if there is any system of equation s.t B=A.c; I need to find a fix value of c such that all elements of matrix B will be a round number. B=A.c; where A is the matrix of non integers numbers.

For example; B=[2.05;4.876,1.455] *c

I want to get value of c so that matrix B will be matrix of integer numbers.

[Rik]

I presume you want to find the smallest value of c that does this? Otherwise you could just choose c=1e100.

What have you done so far?

[Dr. Anju Chaurasia]

I think this will work for rational numbers only, but not for irrational numbers.

[Dr. Anju Chaurasia]

Is there any matlab code?

[Stephen23]

"I think this will work for rational numbers only, but not for irrational numbers."

All binary floating point numbers are rational numbers.

I doubt that there is any solution for irrational numbers in general.

Answer 1

[John D'Errico]

You want to find some scalar value c, such that for some given vector A, you want the product

A*c = B

to be purely a vector of integers? For some vectors, this is provably impossible, IF you allow the components of A to be irrational or even transcendental, AND you insist on an exact solution. For example just choose:

A = [sqrt(2) sqrt(3)]

There is no scalar constant c such that both c*sqrt(2) AND c*sqrt(3) will be integer. If you are willing to accept an approximate solution, then we could surely offer a solution, valid to within some tolerance. (I even recall writing code once to find a minimal solution for a similar APPROXIMATE problem.) But you explicitly said that is not acceptable, since @Rik suggested an approximate solution that will be extremely accurate, and you disagreed.

And of course, if some of the elements of A were possibly transcendental, then things get nastier yet, since there are some numbers where we don't even know their status.

So, no. You cannot solve the problem if you insist on an exact solution. And since some such problems may be indeterminate, then no solution can exist in general. Conversely, if you provide floating point numbers, then an answer is trivial. For example, in the problem you offer:

B=[2.05,4.876,1.455] *c

If you want an exact solution, then we could just start with c=1000. Once you did that, you would find:

A = [2050 4876 1455];

But then you could compute the gcd of the elements of that vector.

gcd(gcd(A(1),A(2)),A(3))

ans = 1

So the three integers in A are collaboratively co-prime. And that tells us the SMALLEST value of c to solve the problem is indeed 1000.

However, had you given the problem as:

B=[2.05,4.876,1.454]*c

Then now we see the solution can be reduced by the global gcd.

A = [2050 4876 1454];
gcd(gcd(A(1),A(2)),A(3))

ans = 2

And that means c = 1000/2 = 500 is a minimal solution.

Finally, IF you provide floating point numbers of that form, we would be forced to write the numbers in their binary form, because those numbers are in fact NOT stored as decimal numbers, but as binary numbers. The n we could form them as a sum of positive and negative powers of 2. Having done that, the solution would again be easy, but not quite so clean.

Again, I'll go back to your example. Having provided the numbers as:

A = [2.05,4.876,1.455];

Now I'll use my num2bin tool. (It should be posted on the file exchange.) And we can see that 2.05, as stored in MATLAB, can be represented as the sum:

format long g
A1 = [1 -5 -6 -9 -10 -13 -14 -17 -18 -21 -22 -25 -26 -29 -30 -33 -34 -37 -38 -41 -42 -45 -46 -49 -50];
sum(2.^A1)

ans = 
                      2.05

Similarly, we can write each of those numbers as an expansion of positive an negative powers of 2.

A2 = [2 -1 -2 -3 -10 -16 -17 -21 -24 -27 -28 -30 -31 -32 -34 -37 -39 -40 -41 -42 -46 -47 -49 -50];
sum(2.^A2)

ans = 
                     4.876

A3 = [0 -2 -3 -4 -6 -10 -11 -12 -13 -15 -17 -18 -19 -24 -26 -30 -31 -32 -33 -35 -37 -38 -39 -44 -46 -49];
sum(2.^A3)

ans = 
                     1.455

Based on that expansion, we can see the number c = 2^50 could be applied to all three of this numbers, to now result in an integer. If we did that, then

Aapprox = [sum(2.^(A1 + 50)),sum(2.^(A2 + 50)),sum(2.^(A3 + 50))]

Aapprox = 1×3

1.0e+00 * 2.30809480902738e+15 5.48988794576464e+15 1.63818436445602e+15

And converting them to symbolic form, gcd tells us they are relatively co-prime.

gcd(sym(Aapprox))

ans = 

1

So the SMALLEST possible number you could use to scale all three of those numbers, having represented them in double precision and stored them as doubles is 2^50.

[Walter Roberson]

   A = [sqrt(2) sqrt(3)]

Expanding on that case. Suppose sqrt(2)*c = b for integer b. Then c=sqrt(2)*r for some rational r. For example if b=7 then c=sqrt(2)*7/2. In particular, c=sqrt(2)*(b/2), which you could also get easily by algebraic manipulation. But there is no other possibility: for every integer b, c is a rational multiple of sqrt(2)

If we then multiply this by sqrt(3) we get sqrt(2)*sqrt(3)*some rational = sqrt(6)*rational. Is that ever an integer? Only if the rational is 0.

So this has solutions for b=0, c=0, but not otherwise for irrational numbers.

[Dr. Anju Chaurasia]

Thanks for your suggestion. But, if my matrix is

A= [-3.9743; -3.8653;-2.9765;-1.8776,-0.8765]; then how to find the common factors of these numbers?

[Rik]

You are ignoring the point. The point is that what you ask is mathematically impossible to answer in general.

But for your specific example you can attempt to determine the number of digits, after which 10^NumberOfDigits is your starting point. Perform the multiplication and use gcd in a loop to determine by what you should divide your c. You may want to check how gcd deals with negative numbers.

[Walter Roberson]

Question:

is -3.9743 exactly - 39743/10000 (base 10 precise notation), or is it "some indeterminate number between -397425/100000 inclusive and -397435/100000 exclusive (scientific notation, significant digits), or is it -3.974299999999999943867123874952085316181182861328125 exactly? Because the last of those is what MATLAB is going to use.