Does infinity exist?

[Farinata]

Take.

3x=1y

Loosen out the x.

The standard way is either to devide both sides of the equal sign with 3
The other way which gives exactly the same answer is to move the 3 to the other side and changing the sign to a devision.

My way to operate is to put -2x on the both sides of the equalsign. Giving the operation 3x=1y
3x-2x=1y-2x
1x=1y-2x

lets define x and y were x is defined as 2 and y is defined as 5. giving the equation the following look.

3*2=1*5

2=5-4

This is the solution using my technique to solve an equation.
But there is a missing link as you can see...

3*2-2*2 = 1*5-2*2

6-4=5-4

What i have done here is redefined the number 5 to a 6 and 2 to a 1 using an unorthodox technique of solveing an equation. But we don't do that because that is just plain silly, we need to go further

To get it stright, i need to move 6 to the other side of the equal sign and changing the operator to a - sign and finally add a -1 to the side without numebers giving the answer 5-6=-1

I chose to operate diffrently and use other techniques other then the standards. Here i had to add a -1 for the equation to turn out right.

But atleast i knew what I was doing. 5-6=-1 is a solution to 3x=1y when the variables are defined in such manners i wished to define it and i add -1. Lets call it for a missing link were you need to imagine a number so hard the equation turns out to be alright!!

3x=1y if we define x as 2 and y as 5, which is fully possible because it is a variables, we will however get an error on the calcualtor, but that is because 5 is not a 6 and 6 is not a 5 unless we become silly. But lets push it further. And define x as 2a:s and y as 5b:s. Since we have defined x and y to other variables the equation turns out to be correct when solving it. Even though using an unorthodox or foreign way to operate the equation...

3*2a=1*5b
6a=5b

6a-2x=5b-2x
6a-4a=5b-4a

Still correct.

Fully possible, without any objections. As we have defined the variables x and y to other variables the solution of the equation turns out alright. Using this unorthodox technique to solve the equation will give those numbers. And i am aware that the answer is diffrently from the orthodox way to solve it. But that doesn't change the fact that my way to operate will give that outcome...

 

[Eldalar]

Take.

3x=1y

Loosen out the x.

The standard way is either to devide both sides of the equal sign with 3
The other way which gives exactly the same answer is to move the 3 to the other side and changing the sign to a devision.

My way to operate is to put -2x on the both sides of the equalsign. Giving the operation 3x=1y
3x-2x=1y-2x
1x=1y-2x

lets define x and y were x is defined as 2 and y is defined as 5. giving the equation the following look.

3*2=1*5

2=5-4

This is the solution using my technique to solve an equation.
But there is a missing link as you can see...

3*2-2*2 = 1*5-2*2

6-4=5-4

What i have done here is redefined the number 5 to a 6 and 2 to a 1 using an unorthodox technique of solveing an equation. But we don't do that because that is just plain silly, we need to go further

To get it stright, i need to move 6 to the other side of the equal sign and changing the operator to a - sign and finally add a -1 to the side without numebers giving the answer 5-6=-1

I chose to operate diffrently and use other techniques other then the standards. Here i had to add a -1 for the equation to turn out right.

Or in short you altered the equation in a way that isn't mathematically correct. You didn't add -1, since adding minus one would have given you:
5-6-1=-1 which would be -2=-1 and therefore is wrong, in reality you altered the values of the equation on only one side, making it a different equation altogether.

What you have proven with that is, that x=2 and y=5 are valid for 3x-1=1y and that they are not valid for 3x=1y.

But atleast i knew what I was doing. 5-6=-1 is a solution to 3x=1y when the variables are defined in such manners i wished to define it and i add -1. Lets call it for a missing link were you need to imagine a number so hard the equation turns out to be alright!!

Yes, if you alter the equation you can prove everything you want, I can prove x=3, y=5 is correct or x=4, y=5 etc. etc. in short I can prove that any number from positive to negative infinity for x or y is a possible solution as long as I add a certain amount to the equation in the way you have done it.

In fact here is the equation that can be used to calculate the alteration you need to make:
y=3x+t (t being the alteration)
t=y-3x if we now insert x= 2 and y=5 we get
t=5-6=-1
and for some other values
x,y
3,5:t=-4
4,5:t=-7
-259,135:t=912

That is not an unorthodox method but a plain and simple incorrect one for this equation, except if the question was: How do I need to alter the equation 3x=1y in order for (2,5) to be valid?

3x=1y if we define x as 2 and y as 5, which is fully possible because it is a variables, we will however get an error on the calcualtor, but that is because 5 is not a 6 and 6 is not a 5 unless we become silly.

Yes you can define the variables however you like, that doesn't mean that those variables will make any sense in the given context(equation).

But lets push it further. And define x as 2a:s and y as 5b:s. Since we have defined x and y to other variables the equation turns out to be correct when solving it. Even though using an unorthodox or foreign way to operate the equation...

3*2a=1*5b
6a=5b

6a-2x=5b-2x
6a-4a=5b-4a

Still correct.

Fully possible, without any objections. As we have defined the variables x and y to other variables the solution of the equation turns out alright. Using this unorthodox technique to solve the equation will give those numbers. And i am aware that the answer is diffrently from the orthodox way to solve it. But that doesn't change the fact that my way to operate will give that outcome...

Totally correct way of doing it, no objection except for one: What is the solution? You have just artificially added 2 new equations, namely:
x=2a and
y=5b
and put them into the given equation, but that isn't the solution, it can be a way to find the solution but it isn't one itself. The last one tried to find a solution, but this one just stops somewhere in the middle.

This equation 3x=1y defines a line, try to draw it and then draw the point (2,5) and see if it is on the line, if not then x=2 and y=5 is not a possible solution for 3x=1y.

I hope I didn't misread or misinterpreted anything in your post but that is how I see your unorthodox solution.

 

[Lillen]

whatever, I just try to shine...

What my math teacher did when I operated diffrently from the book was to give me that insight that i had operated diffrently rather then wrong and that is why the answers was diffrent from the tests. It took me 6 years to figure out what he ment by it...

 

[Lillen]

If we define one of my letters, i.e, dale, to the solution for every mathematical equation, it will be per definition be the solution for every equation. All you would do is answer dale and your right on that equation, it do however requires the mathematical community to establish that definition...

It is God who has defined everlasting, or eternal. He himself has no beggining and no end. Without the bible no such thing would be knowned by man...

 

[Eldalar]

A number or rather a set of numbers that contains every possible solution for an equation would as far as I know be C (set of complex numbers). But that doesn't mean that C is a possible answer for every equation. Since equations can include or exclude possible values for certain variables (at least values that form a possible variable pair, triplet etc.). It just means that every variable is included in C.

What you seem to want defined is the answer: "The solution to that equation is some number that is a solution to that equation.", which is correct but already given by the fact that well ... the solution to an equation is in the set of numbers that are possible solutions to that equation. So it would be a valid solution, that is simply a repetition of the equation as the definition of the answer.

 

[Wiccan_Child]

If we define one of my letters, i.e, dale, to the solution for every mathematical equation, it will be per definition be the solution for every equation. All you would do is answer dale and your right on that equation, it do however requires the mathematical community to establish that definition...

You could define 'dale' to be 'the solution to every mathematical equation', but it is trivial to show that this is meaningless. Observe:

Equation 1: x + 4 = 5, solution, x = dale, i.e., dale + 4 = 5, i.e., dale = 1
Equation 2: x + 4 = 6, solution, x = dale, i.e., dale + 4 = 6, i.e., dale = 2

So 1 equals 2, which is wrong. Therefore, the premise (that 'dale' is the solution to every equation) is itself wrong.

In other words, no, you can't just arbitrarily decide new rules to mathematics.

What you can do is define 'dale' to be a set containing all solutions to all mathematical equations, which is fine (barring some strange paradox), but that's not what you're saying. Again, mathematics isn't arbitrary.

It is God who has defined everlasting, or eternal. He himself has no beggining and no end. Without the bible no such thing would be knowned by man...

What relevance does this have to the discussion?

 

[GrowingSmaller]

My argument for infinity is that objects have length. Number lines map onto lenght. Number lines are infinite. Therefore (by analogy) lengths are infinite.

 

[JoeyArnold]

An infinity takes an infinity to develop, physically speaking.

 

[Wiccan_Child]

An infinity takes an infinity to develop, physically speaking.

What is, 'an infinity'?