Do all Equations have Solutions?

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jewish_scientist
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Do all Equations have Solutions?

Spoiler:
The equation x-x = y does not have a solution over the counting numbers, but does over the natural numbers.* This is basically why we made the term 'natural numbers'.

The inequality x-y > x does not have a solution over the natural numbers, but it does over the integers. This is basically why we made the term 'integers'.

The inequality 0 < x+y < 1 does not have a solution over the integers, but does over the rational numbers. This is basically why we made the term 'rational numbers'.**

The equation x/y = 21/2 does not have a solution over the rational numbers, but does over the constructible numbers. This is basically why we made the term 'constructible numbers'.

The equation (z(x+y))1/2 = 21/3 does not have a solution over the constructible numbers, but does over the algebraic numbers. This is basically why we made the term 'algebraic numbers'.

The equation (w(x+y))z = pi does not have a solution over the algebraic numbers, but does over the computable numbers. This is basically why we made the term 'computable numbers'.

The equation x+y = (-1)1/2 does not have a solution over the computable numbers, but does over the imaginary numbers. This is basically why we made the terms 'imaginary numbers'.

The equation |x+iy| = 1 does not have a solution over the imaginary xor the computable numbers, but does over the complex numbers. This is basically why we made the term 'complex numbers'.

*I am trying to do a "God made the natural numbers; all else is the work of man," thing here, so do not complain that I am counting 0 as a natural number. It is not my fault professional mathematicians did not make a term describing this set of numbers that does not reference integers.

**Again, I am trying to build up from counting numbers, so do not complain that I am not following how these discoveries happened in history.

It seems to me that whenever an equation or inequality without any boundaries on the variables is found that has no solutions, we make a name for the set of numbers that includes these solutions. However, doing this implies that solutions do exist for any equation. In other words, any operation preformed on numbers will result in numbers. I feel like that is a pretty powerful assumption to make, so I was wondering what the justification for it was.

Before anyone asks, I would not consider dividing by 0 to be a counterexample, because I can use limits to calculate what the result is. Limits are a method of calculation; it does not change what the operation in question actually is.

Bonus questions: Can the first inequality be replaced by one that has a single inequality sign? Is there an equation that can replace the inequalities? I nearly drove myself insane trying to do these two things, but it just feels like there is something really obvious that I am missing..
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gmalivuk
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Re: Do all Equations have Solutions?

Limits don't always work for dividing by zero, because limits usually distinguish between positive and negative infinity, but the limit of 1/x when x approaches 0 depends on the direction.

Limits use the extended reals, while you'd want the real projective line to give a unique solution to 1/0.

https://en.wikipedia.org/wiki/Real_projective_line
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But ignoring that and other inaccuracies in your descriptions of different number sets, the answer to the question is probably , "Yes, if we find those solutions useful."

We're not assuming any other numbers exist, rather, we're defining the characteristics of a new set that satisfies the equations in question. Often the full set comes from adding one solution to the equation and making sure it's closed under the operations we want.

To get all the complex numbers from the real numbers, add i to the set and then close it under addition and multiplication, for example.
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Eebster the Great
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Re: Do all Equations have Solutions?

It also depends on what you mean by a "number." Usually the things we think of as numbers have particular properties like total order, which would mean some equations would never have solutions. For instance, we rarely consider a matrix or a function a "number." If we allow these things too, then we could always create a solution to any equation by defining a new set according to that equation. But if we put any restrictions at all on what counts as a number, we should be able to create equations they cannot satisfy.

For example, no element x of the surreal numbers satisfies the equation -||x|| = ||x||.

Heimhenge
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Re: Do all Equations have Solutions?

I like Eebster's example. It's actually pretty easy to write an "equation" that defeats solution by choosing the right operators.

OP's question was first (?) articulated by Kronecker: https://en.wikipedia.org/wiki/Leopold_Kronecker

I don't like that proposition because it's really grounded in theology and has nothing to do with mathematics itself. Without invoking the existence of God, I suppose you could say that "physical laws" existed before physics became a field of study. But I don't think you could say "numbers" existed anymore than you could say "poetry" or "philosophy" existed ... these are constructions of a sentient mind, and before minds existed neither did these constructions.

I would LOVE to know if other sentient beings elsewhere in the cosmos came up with the same math* we did. I'm pretty sure they'd have the same physics we do (if they're equally advanced). But I'm not so sure they'd have the same mathematics. I eagerly await first contact to answer that question.

*Sure, they'd have π and zero and "equality" and other fundamental concepts, but beyond that who knows where their math might go.

gmalivuk
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Re: Do all Equations have Solutions?

jewish_scientist wrote:Bonus questions: Can the first inequality be replaced by one that has a single inequality sign? Is there an equation that can replace the inequalities? I nearly drove myself insane trying to do these two things, but it just feels like there is something really obvious that I am missing..

I assume you mean the second inequality, because the first one already has only one inequality sign.

0 < x + y < 1
For all real numbers between 0 and 1, the square of the number is less than the number itself.
(x + y)^2 < x + y

You can't in general replace inequalities with equalities, but there's no need for those inequalities in the first place.

x + 1 = 0 gets us at least one negative number.

2*x = 1 gets us a rational number.

And again, once you also decide to close your set under some basic operations, you can get all the rest once you have one. Given the natural numbers and -1, to be closed under addition you need all the other integers. And for integers x (other than 0), if I want solutions to x*y=1 *and* I want closure under multiplication, I get all the rationals. Algebraic numbers show up when you want to include solutions to polynomials. Reals when you want Cauchy sequences to have limits. Complex numbers are for when you're really serious about polynomial solutions. Et cetera.
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ucim
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Re: Do all Equations have Solutions?

Based on the way you are approaching the question, I think you're not asking the question you want the answer to.

It sounds more like "Here is a method we have used to find 'more numbers'. Can we find all the numbers?" The key point being not that
x^2= -1
"had no solution", but rather, that its solution was a different kind of number that hadn't been thought of before. It would be worth thinking about what you mean by "number", and also to check out the 3blue1brown youtube channel.

As to the literal question in the subject line, you are looking at a very limited set of equations. Solving an equation usually means simplifying the relationship so that it is more comprehensible. Sometimes the result is "just a number". But if you consider (for example) differential equations
{the rate of change of something} is equal to {some other function of that same something}
"solving it" often means expressing it as a function without derivatives, in the form of:
{something} is equal to {some function of something else}
often called "closed form". Not all equations can be reduced to closed form.

The inequality 0 < x+y < 1 does not have a solution over the integers...
consider "The solution to x/y is not always a member of the integers" (division is not closed over the integers) led to the rational numbers. x/y is closed over the rational numbers (so long as y<>0). That thinking is closer to the way other "numbers" were "discovered".

"{set of numbers} is not closed over {operation}. So, let's expand "numbers" to include some new ones. Now it's closed and we have something more fun to work with."

Jose
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cyanyoshi
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Re: Do all Equations have Solutions?

What value of "x" will solve the equation 1=0?

Eebster the Great
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Re: Do all Equations have Solutions?

I'm not sure how many X rays you need to change that byte in the various servers from = to >, but surely there is such a number.

jewish_scientist
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Re: Do all Equations have Solutions?

Eebster the Great wrote:Usually the things we think of as numbers have particular properties like total order, which would mean some equations would never have solutions.

I do not understand why total order leads to that conclusion.

gmalivuk wrote:x + 1 = 0 gets us at least one negative number.

2*x = 1 gets us a rational number.

I wanted all of the equations to have at least 2 variables, but I do not remember why. If we through out this rule that I forgot to mention, then these equations do answer my question. What equations do answer the bonus questions and follow the rule I forgot to mention?

ucim wrote:Based on the way you are approaching the question, I think you're not asking the question you want the answer to.

I tend to do that a lot.

"{set of numbers} is not closed over {operation}. So, let's expand "numbers" to include some new ones. Now it's closed and we have something more fun to work with."

What I want to know is if the underlined part always possible.
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Pfhorrest
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Re: Do all Equations have Solutions?

IIRC the surreal numbers contain basically anything that might possibly be considered a number, and AFAIK there are still some equations that do not have solutions in that set (like x=0/0), though someone else here might correct me.

ETA: actually I guess complex numbers are not included in there, but there are also surcomplex numbers, and surhypercomplex numbers, and IIRC octonions are about as hypercomplex as numbers can get without losing basically all useful properties of numbers, so suroctonions are probably the best you're going to do, and I really doubt that x=0/0 has a solution in them either.
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Eebster the Great
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Re: Do all Equations have Solutions?

x = 0/0 does have a solution in some sets of numbers in which /0 is defined, but not in any extension of the real numbers. The obscure "wheel theory," for instance, treats /x as an element of the "wheel" for every x in the wheel, with //x = x and /(xy) = /y/x. Division in this theory is a unary operator, not a binary one, but /0 is defined, and thus 0/0 = 0 * /0 is defined. You can decide for yourself whether or not this "counts."

That's sort of the point, though. If we have enough freedom with our notation, then anything can be true.

ucim
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Re: Do all Equations have Solutions?

jewish_scientist wrote:
ucim wrote:"{set of numbers} is not closed over {operation}. So, let's expand "numbers" to include some new ones. Now it's closed and we have something more fun to work with."
What I want to know is if the underlined part always possible.
I don't think it's possible to know this.

Either:
it's true (and here's how we can expand the numbers)*,
it's true (and we haven't figured out how to expand the numbers yet),
it's false (and that's why we haven't figured out how to expand the numbers yet), or
it's false (and here's the proof).

That last case requires a good definition of what a "number" is. But it's ultimately a proof that the definition of "number" can't be changed if we change the definition of "number".

* "Expanding the numbers" needs to not only close the set, but remain self consistent (wrt the operations), and consistent with the prior set. So, you can't just append "Fred" to the reals, define 0/0 to be "Fred", and call it a day.

Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

arbiteroftruth
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Re: Do all Equations have Solutions?

x=0/0 may have a solution in wheel theory, as does x=1/0, but the equation 0*x=1 doesn't have a solution without seriously breaking things.

It sort of has a 'solution' if you weaken what you mean by 'solution'. For example, one interpretation of wheel theory might identify the element 0/0 with the entirety of the underlying set of numbers. For example in the wheel generated from R, 0/0 might be interpreted as the entire projectively extended real line, essentially representing all the possible values 0/0 could take as an indeterminate form. Under such an interpretation, /0 does weakly 'solve' 0*x=1 in the sense that {1} is a subset of 0/0.

One might define a weak notion of 'solving' an equation defined by instantiating variables in such a way that the intersection of the set resulting from the left-side computation and the set resulting from the right-side computation is as small as possible without being empty. Then, if you extend the notion of a 'number' to include certain sets of numbers, it might be possible to create a system in which all expressible equations are 'solvable' in this weak sense.

But ultimately, 0*x=1 is an easy example of an equation that can't be solved without completely breaking what me mean by at least one of '0', '1', '*', or '='. You can break '0' by using a system in which 0 is not an absorbing element under multiplication, you can break '1' by making it equal 0, you can break '*' by, well, using it to denote a completely different function than multiplication, or as I've suggested above, you can break '=' by weakening it to some statement about intersections of the sets describing indeterminate forms.

The question then is, do any of those options count as 'solutions' to the original equation 0*x=1?