Why x=2x doesn't have three solutions?

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In summary, the conversation discusses the concept of extending the real numbers to include positive and negative infinity, and how this can affect the number of solutions to an equation. It also touches on the idea that mathematical objects do not necessarily have to exist in a physical sense, but can be defined by their properties for the purpose of exploration and problem-solving.
  • #1
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Sorry for the slightly bizarre and definitely odd-sounding question... but why

x=2x

doesn't have three solutions? One of them is clearly x = 0, but x = +∞ and -∞ should also be solutions - not in ℕ or ℝ, of course, given that +/-∞ does not belong to ℕ or ℝ, but I heard there's something called an extended-ℝ thingie that includes +/-∞, therefore in some domain x=2x should have three solutions, should it not?

If that's true, and x=2x does have three solutions in some domain... then that doesn't make sense at all! How a first-order equation would have multiple solutions? It's incredibly bizarre!

Where did I got out of the track in this reasoning? Once again sorry if the question sounds too idiotic...
 
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  • #2
If you extend the real numbers to include plus/minus infinity you lose a lot of properties of the real numbers.

In general it shouldn't be surprising that increasing the domain can increase the number of solutions. You have this on all levels. 5+x=3 has no solutions in the positive integers. 2x=3 has no solutions in the integers. x2=2 has no solution in the rational numbers. x2=-1 has no solutions in the real numbers but has some in the complex numbers.
 
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  • #3
fbs7 said:
but why

x=2x

doesn't have three solutions?
Because the question is incomplete! You did not specify where ##x## may be taken from and use this lack of information to create a question aka confusion! A bad habit, if you ask me. The hidden assumption is ##x\in \mathbb{R}\text{ or }\mathbb{Q} \text{ or }\mathbb{Z}##. ##\{\,\pm \infty\,\}## isn't an element in either of them, so the question doesn't arise. If we are allowed to do basic arithmetic - another hidden assumption - then the equation gets: ##2x-x=(2-1)x=1x=0##. Since ##1## isn't a zero divisor in any reasonable domain, we must conclude ##x=0##.

Things change, if ##x## is regarded as a cardinal, but as this is the exception, and ordinals the standard, this would have had to be mentioned.
 
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  • #4
I see. I got me a little more reading on that extended-R thingie.

So, if we define ℝ° = ℝ ∪ {-∞, +∞}, then is it true that in ℝ° the equation x=2x has 3 solutions? Is this a valid statement or is that some hubbabah that a non-mathematician (that is, me) came up as crazy unrefined hypothesis?

I still find the whole thing flabbergasting, as for one thing this assumes that ℝ° exists, therefore it assumes that ∞ exists and is some kind of number (as we assume it can be multiplied by 2). Meanwhile I can't even imagine a definition for ∞, much less a way to prove you can it can be multiplied by 2. Really bizarre!
 
  • #5
The extended real numbers (https://en.wikipedia.org/wiki/Extended_real_number_line) as far as I have met them, are more of an abbreviation in notation than an actual object; at least not an algebraic object as a field or similar which you can calculate with. If at all, it is a topological object. One can write limits, suprema, infinma, and and values of integration more compact because theorems are often valid even if their limit is infinite, especially when it comes to higher analysis like measure theory.
 
  • #6
fbs7 said:
I see. I got me a little more reading on that extended-R thingie.

So, if we define ℝ° = ℝ ∪ {-∞, +∞}, then is it true that in ℝ° the equation x=2x has 3 solutions? Is this a valid statement or is that some hubbabah that a non-mathematician (that is, me) came up as crazy unrefined hypothesis?

I still find the whole thing flabbergasting, as for one thing this assumes that ℝ° exists, therefore it assumes that ∞ exists and is some kind of number (as we assume it can be multiplied by 2). Meanwhile I can't even imagine a definition for ∞, much less a way to prove you can it can be multiplied by 2. Really bizarre!

Mathematical objects do not exist according to some physical reality. Does a matrix exist? You don't have to assume that matrices actually exist in the real world in some sense in order to do linear algebra.

Mathematical objects are defined by their properties. So, if you extend the real numbers in some way, you simply have to define the properties that you want your new elements to have and you can take it from there. There is no concept of a proof that shows that objects can be multiplied: you simply provide a definition of the multiplication operation you want. Again, take the example of a matrix. You cannot prove that it's legitimate to multiply a matrix by a real number. Instead, you define what you mean by this multiplication. Likewise, you cannot prove that matrix multiplication is physically legitimate: that again is a definition. Albeit chosen so that matrices do what you want them to do.

You're not alone. If you look at the history of complex numbers, you will see a reluctance by the mathematicians of the past to accept them as legitimate mathematical objects. To look at this now through modern eyes seems absurd and tragic. Even when they solved problems using complex numbers, they were so concerned by issues of their actual existence, that they closed the door on them and denied themselves the opportunity to open up great avenues of mathematics.

PS There's a great series of videos here on the history of complex numbers:

 
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  • #7
Oh, wow... that's deep! Now I'm starting to get it!

It's not whether the thing exists or not, but whether it has some interesting properties to explore! What a thought -- that's a complete 90 degrees on my way of thinking! So whatever equation we're trying to solve, it depends completely on the domain we're using! wow!

So if we were to define say ℝ° = ℝ ∪ { 0/0 }, then it's not a matter of whether 0/0 is a real thingie or not, but of whether that ℝ° is useful -- and I assume it's a very useless domain with useless algebra, as opposed to ℂ, which has many interesting properties!

Now, last question, does inventing those crazy domains like ℝ ∪ {∞} or ℝ ∪ { 0/0 } go against an axiomatic basis for mathematics, even if they lead to interesting discussions?

Let me explain the question. Say that I define ℕ° = { 0, ... , 9, θ }, and then I define 0+1 = 1, 1+1 = 2, ... , 1+9 = 10, 1 * 2 = 1+1, etc... (up so far just regular ℕ), but then I define this θ as a kinky thingie such that 1 * θ = 2, 2 * θ = 3, 3 * θ = 4, etc... That will generate an algebra of some kind, which will most certainly be quite useless, where for example θ * θ doesn't have any value (although 1*θ*θ and 2*θ*θ do have values, 3 and 4 respectively).

Yet, fbs7's crazy ℕ° should (I hope) still be a valid domain for some algebra (where x=2x would have some set of solutions), but then it's up to a human to judge "yeah-yeah-yeah, but that's useless" or "wow, that's interesting, let's see how the equations of gravity look like here!". Then aren't ℝ ∪ {∞} or ℂ and its axioms up for the same a human judgement, and as human judgement is not an axiom, then using ∞ or ℂ in everyday mathematics is not due to an axiomatic basis for them but rather to human preference and aesthetics?
 
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  • #8
fbs7 said:
Oh, wow... that's deep! Now I'm starting to get it!

It's not whether the thing exists or not, but whether it has some interesting properties to explore! What a thought -- that's a complete 90 degrees on my way of thinking! So whatever equation we're trying to solve, it depends completely on the domain we're using! wow!

So if we were to define say ℝ° = ℝ ∪ { 0/0 }, then it's not a matter of whether 0/0 is a real thingie or not, but of whether that ℝ° is useful -- and I assume it's a very useless domain with useless algebra, as opposed to ℂ, which has many interesting properties!

Now, last question, does inventing those crazy domains like ℝ ∪ {∞} or ℝ ∪ { 0/0 } go against an axiomatic basis for mathematics, even if they lead to interesting discussions?

Let me explain the question. Say that I define ℕ° = { 0, ... , 9, θ }, and then I define 0+1 = 1, 1+1 = 2, ... , 1+9 = 10, 1 * 2 = 1+1, etc... (up so far just regular ℕ), but then I define this θ as a kinky thingie such that 1 * θ = 2, 2 * θ = 3, 3 * θ = 4, etc... That will generate an algebra of some kind, which will most certainly be quite useless, where for example θ * θ doesn't have any value (although 1*θ*θ and 2*θ*θ do have values, 3 and 4 respectively).

Yet, fbs7's crazy ℕ° should (I hope) still be a valid domain for some algebra (where x=2x would have some set of solutions), but then it's up to a human to judge "yeah-yeah-yeah, but that's useless" or "wow, that's interesting, let's see how the equations of gravity look like here!". Then aren't ℝ ∪ {∞} or ℂ and its axioms up for the same a human judgement, and as human judgement is not an axiom, then using ∞ or ℂ in everyday mathematics is not due to an axiomatic basis for them but rather to human preference and aesthetics?

The axiomatic basis of mathematics is essentially a logical basis. Those are deep waters.

The problem with introducing additional numbers is that you may lose some basic properties. This certainly happens if you want a number which is the multiplicative inverse of ##0##. Then you have:

##a \times 0 = b \times 0 \ \Rightarrow \ a = b## does not hold.

This means that the simple rule of ##ac = bc \ \Rightarrow \ a = b## remains only true when ##c \ne 0##. I.e. you still can't divide by ##0##.

But, in principle, if a mathematical system is useful in some way, then it's legitimate. Complex numbers and, again, matrices are examples. Matrices were initially thought of as suspicious because in general:

##AB \ne BA##

And, how can a multiplication depend on the order you multiply things?

The other obvious example is that non-Euclidean geometry may have been considered a mathematical indulgence until, would you believe it, the theory of General Relativity predicted (correctly) that our universe is spatially non-Euclidean.

It's another sad tale that some able mathematicians wasted their lives trying to prove that non-Euclidean geometry was mathematically impossible!
 
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  • #9
Wow, wonderful insights! Thank all so much for the interesting remarks in the thread!
 

1. Why does x=2x only have one solution?

When solving equations, we are looking for a value of x that makes the equation true. In the equation x=2x, there is only one value of x that satisfies this condition, which is x=0. Any other value of x would make the equation false. Therefore, there is only one solution.

2. Can an equation have more than one solution?

Yes, an equation can have more than one solution. For example, the equation x^2=4 has two solutions, x=2 and x=-2. In this case, both values of x make the equation true. However, the equation x=2x only has one solution, as explained in the previous answer.

3. Is it possible for an equation to have no solutions?

Yes, it is possible for an equation to have no solutions. This happens when there is no value of x that makes the equation true. For example, the equation x+2=0 has no solutions, as there is no number that when added to 2 equals 0.

4. Why does x=2x have no solutions?

The equation x=2x has no solutions because there is no value of x that satisfies the equation. If we divide both sides by x, we get 1=2, which is not true for any value of x. Therefore, this equation has no solutions.

5. Can an equation have an infinite number of solutions?

Yes, an equation can have an infinite number of solutions. This happens when any value of x makes the equation true. For example, the equation x=x has an infinite number of solutions, as any value of x would make the equation true. However, the equation x=2x only has one solution, as explained in the first answer.

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