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/sci/ - Science & Math

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11724668 No.11724668 [Reply] [Original]

Is there a formal proof that x=0 is the only solution to x=2x?
In fact, IS it the only solution if we include nonreal numbers?

>> No.11724710

Assume x not 0
Then we can divide by x and get 1=2 which is a contradiction
Now assume x = 0
Then 0 = 0 which is consistent
Therefore x = 0

>> No.11724720

>>11724710
couldn't you just use a uniqueness theorem or something to show that all x are defined as x so the only answer is the trivial one ie. 0

>> No.11724764

>>11724710
Why is 1=2 a contradiction in arbitrary algebras?

>> No.11724766

>>11724764

Unicorn = leprechaun

Qed

>> No.11724803

>>11724710
But the only reason that x=0 works despite dividing by x producing 1=2 is that division by 0 is impossible, so in order for your proof to hold we would need to prove that 0 is the only number that we cannot divide by. How can we prove that? Especially if considering nonreal numbers?

>> No.11724805

Since the real part of ai (where a is any real number and i is √−1) is 0, does that mean that division by ai is impossible? Does that mean that ai could be another solution to x=2x? Or what if we let a be a nonreal number?

>> No.11724823

>>11724764
>in arbitrary algebras
because it implies 0 = 1 by subtracting 1, and this implies x = 1 * x = 0 * x = 0 for any x in the algebra/ring
thus this is the 0 ring

>> No.11724832

>>11724823
What is a ring? what is ring algebra? can x=0 in some ring, while also x is not 0 in the same ring? and can 1x=2x for x not 0, or 1x=2x be impossible for x=0 because 1(0) not equal 2(0)?
what if 1×0 is not 2×0 in some math?

>> No.11724848

Why don't you graph that for us sonnyboy?

>> No.11724853

>>11724832
Rings are integers but fancy.

>> No.11724857

>>11724764
One of the axioms that constructs the naturals states that Successor{x} only equals Successor{y} if x=y.
1 and 2 have different successors (2 and 3), which implies 1[math]\neq[/math]2.
So you have 1=2 and 1[math]\neq[/math]2 at the same time. You might think this is ok, but consider the OR operator. It answers "TRUE" for two statements a,b if any of the following apply:
a is true AND b is not true
a is not true AND b is true
a is true AND b is true
If you have a AND not a both be true you do all kinds of crazy shit.
Consider the following statement:
Either 1[math]\neq[/math]2 OR I am Based Tooker, blessed with the powers of flight, telepathy and solving Millenium prize problems.
1[math]\neq[/math]2, therefore this statement is true.
But 1=2, and therefore the first part of this statement isn't true. But we've already proven that the statement is true, so therefore the second part must be true.
You can't resolve this without abandoning OR, XOR, XNOR and NAND as tools in your logic system, which leaves you with a logic system that can't define addition.
It's interesting to study paraconsistent logics like that but they are useless at modeling anything.

>> No.11724865

>>11724668
>Is there a formal proof that x=0 is the only solution to x=2x?

nigga, just put it into the calculator lmfaoooo

twice x will always be double therefore not the same

easy, you dumb bitch

>> No.11724867

>>11724865
Based

>> No.11724868

>>11724832
A ring has all them good properties

>> No.11724874

>>11724668
>subtract both sides by x
there you go

>> No.11724916 [DELETED] 

>>11724868
>A ring has all them good properties
so a ring has 2(0) is not 1(0)? what ring has that? your wedding ring since it doesn't exist? ;) at least you don't get fat and diabetic from a wedding cake if you don't get married

>> No.11724921

>>11724874
oh right i forgot about subtraction and was only trying division, thanks!

>> No.11724927

>>11724865
double 0 is not 0 you tard
double 0 is literally ∞
so if 2x=x then x=∞
I mean no 2x=∞ and 1x=0
so 2(0)=∞

>> No.11724937

>>11724668
Given what axioms?

>> No.11725117

>>11724874
>it took 15 posts for someone to say this
The absolute state of \sci\

>> No.11725798

Fucking idiots just do -2x on each side. Complex numbers are closed, so a polynomial of degree 1 has exactly one solution in the complex numbers. If you found a solution, it is unique.

>> No.11725898

>>11725798
x=1+x2 has two solutions retard

>> No.11726428

>>11725898
Hahaha, okay, ridicule yourself and tell us which ones. Also:
*x = 1 + 2x has two solutions, retard.

>> No.11726449

>>11724668
x = 2x
x = x+x |-x
0 = x
It's not hard Anon

>> No.11726490

>>11724927
You aren't serious. Dumbest post on this board if serious.

>> No.11726554

Whatever algebra "x = 2*x" is in has a multiplication operation. "2" is probably defined as 1+1, so there is also a multiplicative identity "1" and an addition operation. If multiplicative and addition are distributive, such as in any ring, then "x = 2*x" implies "x=(1+1)*x" implies "x=1*x+1*x" implies "x=x+x"
If, in addition, as in any ring, every element has an additive inverse, we then have "x+(-x) = (x+x)+(-x)". Associativity would give "x+((-x) = x + (x + (-x))", which evaluates to '0 = x + 0", hence "0 = x".
Without all these assumptions you can get other solutions. For example, if \infinity is an element of your algebra, perhaps \infinity = 2*\infinity. But \infinity either doesn't have an additive inverse, or it isn't additively associative.

>> No.11727337

Tooookeeeer, where are you? You won't believe what someone just posted!

>>11726554