- #1

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I'd like to know if it's possible to solve an equation with a modulus like this one :

$$ ( \frac{200}{15x}) mod 2 = 0 $$

Thanks in advance.

Regards!

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- Thread starter hugo_faurand
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- #1

- 57

- 9

I'd like to know if it's possible to solve an equation with a modulus like this one :

$$ ( \frac{200}{15x}) mod 2 = 0 $$

Thanks in advance.

Regards!

- #2

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- #3

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Mmmmmh. I don't really understand. What is k? Does it mean that the solution is a kind of pair like in a 2 unknown equation? https://www.physicsforums.com/threads/equation-with-modulus.952619/reply?quote=6035365

- #4

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Mmmmmh. I don't really understand. What is k? Does it mean that the solution is a kind of pair like in a 2 unknown equation?

Is ##x## supposed to be an integer? Or any real number?

- #5

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Normally it's an integer. Does it change something to solve the equation?Is ##x## supposed to be an integer? Or any real number?

- #6

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Normally it's an integer. Does it change something to solve the equation?

Have you looked for integer solutions?

Hint: 15 has a factor of 3.

- #7

mfb

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If x is an integer, then 200/(15x) won't be one.

- #8

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The solution is [itex]x=1\mod 2[/itex]. If [itex]x=0\mod 2[/itex] then [itex]\frac{200}{15x}[/itex] is undefined[itex]\mod 2[/itex] (so not a solution according to Russell's theory of definite descriptions).

This assumes you are looking for solutions in the field of integers [itex]\mod 2[/itex], so there are but two possible solutions. The equation reduces to [tex]\frac{0}{x}=0\mod 2[/tex]As in any field any nonzero element is a solution of this equation.

This assumes you are looking for solutions in the field of integers [itex]\mod 2[/itex], so there are but two possible solutions. The equation reduces to [tex]\frac{0}{x}=0\mod 2[/tex]As in any field any nonzero element is a solution of this equation.

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- #9

mfb

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As you can easily verify, x=1 for example is not a solution, but you claim it would be.

200/(15x) is well-defined for x=2 and you can consider 200/(15x) mod 2 while x=0 mod 2 is satisfied.

Why would you assume that?This assumes you are looking for solutions in the field of integers mod 2

You can't reduce equations like that.The equation reduces to

- #10

Stephen Tashi

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@Martin Rattiqan: None of that is right.

It's not right according to the interpretation that @hugo_faurand probaby has in mind, where mod 2 arithmetic is considered an algorithm performed on the integers.

However, from the more advanced viewpoint of talking about ##\mathbb{Z}/(2 \mathbb{Z}) ## we have that the notation "200" represents the coset ## [200]= [0] = \{...-4,-2,0,2,4,...\}## and it is indeed true that ##[200]/([15][x] )= [0]/([1] [x])##. Whether to call that "reduction" could be debated. It's also true that letting ##[x] = [1]## gives us ##[0]/([1][1]) = [0]/[1] = [0][1]^{-1} = [0][1] = [0]##.

So if @Martin Rattigan errs ,he errs only in taking a more advanced viewpoint that the original poster.

- #11

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$$ \frac{a}{bx} \mod 2 = 0 $$

In case of I'm not clear enough, I want an equation which looks like that :

$$ x =

\text{something with b and a }

$$

- #12

Stephen Tashi

Science Advisor

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The only subject for debate is the meaning of your question.Mmmmh it's a quite interesting debate.

Notation, by itself, does not specify a context.But to give a context I'd like to know if I can have an expression of x with this equation :

$$ \frac{a}{bx} \mod 2 = 0 $$

Instead of "have an expression of x", you probably mean to ask whether a certain equation "has a solution for x".

Defining the meaning of your notation "##\frac{a}{bx} \mod 2 = 0##" requires some words.

For example, what is your interpretation of "##\mod 2##"? By some interpretations of "##\mod 2##" it is correct to write "##4 = 0 \mod 2##" because, in that context, the relation "=" is not the ordinary equivalence relation for numbers.

A different interpretation is that "## \frac{c}{d} \mod 2##" denotes an operation on two numbers ##c,d## that is performed by taking the positive remainder after c is divided by d. That interpretation is not specific until you define the set of numbers that may be operated upon. For example, must c and d be integers?

If you don't know the meaning of your notation "##\frac{a}{bx} \mod 2##", perhaps you can tell us where you saw this notation. What book? What course?

- #13

mfb

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- #14

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The only subject for debate is the meaning of your question.

Notation, by itself, does not specify a context.

Instead of "have an expression of x", you probably mean to ask whether a certain equation "has a solution for x".

Defining the meaning of your notation "##\frac{a}{bx} \mod 2 = 0##" requires some words.

For example, what is your interpretation of "##\mod 2##"? By some interpretations of "##\mod 2##" it is correct to write "##4 = 0 \mod 2##" because, in that context, the relation "=" is not the ordinary equivalence relation for numbers.

A different interpretation is that "## \frac{c}{d} \mod 2##" denotes an operation on two numbers ##c,d## that is performed by taking the positive remainder after c is divided by d. That interpretation is not specific until you define the set of numbers that may be operated upon. For example, must c and d be integers?

If you don't know the meaning of your notation "##\frac{a}{bx} \mod 2##", perhaps you can tell us where you saw this notation. What book? What course?

For me "x mod 2" is like dividing x by 2 and take the rest of the division.

Why a value from 0 inclusive to 2 exclusive ?

- #15

mfb

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That's what computers will usually do. Here is python, for example:Why a value from 0 inclusive to 2 exclusive ?

>>> 2%2

0

>>> 3.141%2

1.141

>>> -1%2

1

>>> -5.5%2

0.5

- #16

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That's what computers will usually do. Here is python, for example:

>>> 2%2

0

>>> 3.141%2

1.141

>>> -1%2

1

>>> -5.5%2

0.5

OK but now I have two unknowns k and x .SO How can I solve it ?

- #17

mfb

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See above. There are many values of x that satisfy the equation, one for every k.

- #18

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But if you runThat's what computers will usually do. Here is python, for example:

>>> 2%2

0

>>> 3.141%2

1.141

>>> -1%2

1

>>> -5.5%2

0.5

<p id="a">

<script>

function m2(exp){document.getElementById("a").innerText += (exp)%2+"\n"}

m2(2)

m2(3.141)

m2(-1)

m2(-5.5)

m2(200/(15*1))

</script>

in e.g. Chrome, javascript gives:

0

1.141

-1

-1.5

1.333333333333334

You can't rely on different computer programs to give the same answers.

Given that this is a mathematics forum rather than a programming forum I expect OP would prefer an answer conforming with generally accepted mathematics.

In this case only the odd numbered elements in my list are

Javascript gets the wrong answer for my fifth - it should be 0 (or rather

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- #19

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See above. There are many values of x that satisfy the equation, one for every k.

But really the whole usefulness of arithmetic modulo [itex]n[/itex] is that the canonical isomorphism from [itex]\mathbb{Z}[/itex] to [itex]\mathbb{Z}/n\mathbb{Z}[/itex] allows you to ignore the [itex]k[/itex]s. There is only one value[itex]\mod 2[/itex] that satisfies OP's equation as stated in my original post.

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- #20

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The problem with the operator you describe is that its not very useful mathematically if it's applied to the real numbers, because it doesn't necessarily respect arithmetic in the reals. E.g.

<p id="a">

<script>

function m2(exp){document.getElementById("a").innerText += (exp)%2+"\n"}

m2(Math.sqrt(5)*Math.sqrt(5))

m2((Math.sqrt(5)%2)*(Math.sqrt(5)%2))

</script>

when run in Chrome gives

1.0000000000000009

0.055728090000841266

For any ring [itex]R[/itex] there is a canonical homomorphism from [itex]R[/itex] to [itex]R/nR[/itex] (where [itex]nR[/itex] is defined as [itex]\{r+r+\ldots+r\ (n\text{ times}):r \in R\}[/itex], which might be described as a[itex]\mod n[/itex] operator (and, of course, respects the arithmetic in [itex]R[/itex]), but if [itex]R[/itex] is a field then the quotient is either the single element ring or the field you started with depending on the field's characteristic, ergo not much use. The operator would not be the one you described in the case [itex]R=\mathbb{R}[/itex] because in that case the ring of even integers is not the ideal [itex]2\mathbb{R}[/itex].

The [itex]\mod n[/itex] notation in mathematics is usually used when [itex]R[/itex] is [itex]\mathbb{Z}[/itex] or possibly [itex]\{i/j:i,j\in \mathbb{Z}\text{ and }(j,n)=1\}.[/itex]

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- #21

mfb

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JS gets many things wrong, but not that.Javascript gets the wrong answer for my fifth - it should be 0 (or ratherIask it to get the wrong answer because the "/" is interpreted as floating point division - there is no native javascript division ##\mod n##)

Check the order of operation. The mod function is evaluated on the

What you seem to want is ##\frac{200 \mod 2}{15 \mod 2}## but that is different from OP's problem.

- #22

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JS gets many things wrong, but not that.

Check the order of operation. The mod function is evaluated on thenumber200/(15*1) = 13.333... and 13.333... mod 2 = 1.333...

What you seem to want is ##\frac{200 \mod 2}{15 \mod 2}## but that is different from OP's problem.

As I said in parentheses.

And yes, I want the division performed in the integers mod 2.

I assumed from OP's inclusion of "mod 2" in his equation that that was exactly the problem he wanted solved. The expression ##\frac{200}{15x}## is undefined in the ring of integers for any integral ##x##, so to obtain a solution the division must be carried out mod 2.

The result of ##\frac{200}{15x}\% 2## using the % operator in python or javascript is, apart from being inconsistent, just not useful in mathematical terms. (See my post #20.)

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- #23

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I'm totally lost...

- #24

mfb

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Why would you do that?I assumed from OP's inclusion of "mod 2" in his equation that that was exactly the problem he wanted solved.

If you see the expression ##\frac{a}{b}+1##, you don't assume that the operation "add 1" has to be done in numerator and denominator separately either.

It is consistent and it leads to a non-trivial result. Your interpretation is inconsistent and leads to a trivial result only.The result of ##\frac{200}{15x}\% 2## using the % operator in python or javascript is, apart from being inconsistent, just not useful in mathematical terms. (See my post #20.)

You could help us by clarifying how the question was meant. What exactly does the "mod 2" there mean? How was it introduced, how was it used in previous problems?I'm totally lost...

- #25

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Why would you do that?

If you see the expression ##\frac{a}{b}+1##, you don't assume that the operation "add 1" has to be done in numerator and denominator separately either.It is consistent and it leads to a non-trivial result. Your interpretation is inconsistent and leads to a trivial result only.You could help us by clarifying how the question was meant. What exactly does the "mod 2" there mean? How was it introduced, how was it used in previous problems?

I'd like to solve a problem that you can find here https://www.codingame.com/training/medium/aneo . To sum up it, you have to find the max speed which a car can have to cross a road whithout stop at the traffic lights. Each traffic lights have only 2 "state" : red and green. And all traffic lights keep their state for x seconds (x is an integer). At the start all the traffic lights are green.

I wrote my equation to know if when the car cross the traffic light the traffic light is green or red.

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