@mathprofbrony  
She probably assigned it to see how many kids figure out the answer is ∅ instead of just searching and searching

@Prometheus labs CEO
 
They wouldn’t.
 
The easiest way to make this equation be true is in a group of order 26. 0 = 26 if you’re considering, say, modular addition on the 26 letters of the English alphabet (so that, say, 11 + 20 = 31 = 5). But in that case any x solves this equation.
 
Honestly though this is all just sort of navel-gazing, as it is clearly just a nonsensical item on the chalkboard and Cheerilee is not teaching her students modular arithmetic.

@mathprofbrony  
From what I can tell complex numbers wouldn’t work either

@mathprofbrony  
The only example of non-commutative addition I can find is ordinal arithmetic, and even that doesn’t work here. Right subtraction isn’t always defined, but where it is, the equation never holds.

@Prometheus labs CEO
 
Just to be clear, the equation on the board does not have a real-number solution. For it to have a solution one must have strange things going on like 26 being equivalent to 0, or addition being noncommutative. None of these happen in the usual numbers you normally use, including with negative numbers.

@HDPlayer  
Oh right, I forget about negative numbers

@Prometheus labs CEO  
wrong
 
x - 5 = x + -5

Since when have 0 and 26 being equal?

A number minus a number cannot equal itself plus another number

@Ereiam  
Yeah, I should have been more specific.

@Genevamode  
You should’ve included that in your reasoning in the first place then, because that’s not how I understood it.  
You gotta be rigorous when you do math.

@Ereiam  
The reasoning is solid in the instance of that particular equation because it points out the fact that two equal numbers added or subtracted from by two unequal numbers cannot be equal. The numbers being used to subtract from x would have to be equal to work. Substituting a value works in this case purely by allowing the result to be evaluated, any value will work.
 
If the equation was different like you mentioned there would no longer be a contradiction, and thus the entire point of a proof would be moot. You could simply subtract x from 2x normally.
 
tl;dr: substituting a value just makes it easier to prove the contradiction.

@Background Pony #15D8  
In the first paragraph, I meant to say, “in the same way that on a clock, 7+8 = 3 o’clock”, not 7·8.

@Ninji  
The notation ℤ[i]/(3+2i)ℤ[i] is notation from abstract algebra constructing a particular algebraic structure that is different from the natural numbers, real numbers, or complex numbers in which the equation on Cherilee’s chalkboard is true. In this algebraic structure ℤ[i]/(3+2i)ℤ[i], it is true (though an abuse of notation) to say “13 = 0” in the same way that on a clock, 7·8 = 3 o’clock. Algebraic structures are types of structures, and structures as a general phenomenon are the domain of model theory.
 
Background Human’s original observation was that the equation on Cherilee’s chalkboard is true mod 13, which is saying it’s true in the field ℤ/13ℤ which is like a clock in which the numbers go up to 13 so that 13 = 0 and 7·8 = 56 = 4·13+4 = 4 (in the algebraic structure ℤ/13ℤ, like winding around 13 o’clock 4 times and winding up at 4 o’clock). The point of my comment was mocking the utility of this observation by saying how it’s also true in infinitely many other structures and that moving beyond the standard algebraic structures blows apart the interpretational epistemic hinges.

@Genevamode  
Poor reasoning. You can’t just substitute the unknown with a random value and say that the equation fails when it gives you unequal results between the two sides.
 
Let’s say, if the equation was 2x - 7 = 19 + x instead, then your method (picking x = 2) would give us:
 
2*2 - 7 = 19 + 2  
4 - 7 = 19 + 2  
-3 = 21
 
However, we can easily figure out that:
 
2x - x = 19 + 7  
x = 26
 
Back to our x - 7 = 19 + x however, the issue (as previous commenters already pointed it out several times) is that the x on each side of the equation would cancel each other out, leaving -7 = 19 (or 0 = 26) as a result, which is clearly absurd (with real numbers, in any case), thus making that equation a failed one, as you put it.

Total trick question, the two sides of the equation aren’t equal.
 
Let x = 2
 
2-7 = 19+2  
-5 = 21  
This is clearly false, therefore the equation fails.

So what you’re saying is  
x always equals ‘z/13z’ It couldn’t possibly be x=1/13.
 
“13 = (3+2i)(3-2i)”  
13=9-6i+6i-4i² so after the conjugates cancel out, 9 minus 4 times imagination squared?
 
Then we got “x ∈ ℤ[i]/(3+2i)ℤ[i].”
 
x ∈ ℤ[i]/3ℤ[i]+2ℤ[i²] ?
 
I only got up to factoring polynomials and quadratic equations.
 
From what I get from it this thing ‘∈’ is used to define stipulations to create a mathematical system, or model. ℤ is a non zero integer and [i] is any imaginary factor for ℤ that must exists to uphold the stipulation?
 
By ‘solving’ do you mean you created a parameter that describes the contraindication?

@Background Pony #15D8
 
So you turned the contradiction into something more meaningful to the nature of mathematics?

ofc.
 

 

 
A contradiction, or I’ll just take this as ‘no value to quantity’.  
‘x’ becomes eliminated while transposition to the left (or right).

@Ninji  
Mathematics has nothing to do with physical reality. Physical processes can often be modeled with mathematical structures, but mathematics is about mathematics and nothing else. This has been the case since Euclid.

@Background Pony #15D8  
‘Most mathematics is just understanding mathematics for the sake of itself.’  
Isn’t the best way to understand math through the application of math?

@Ninji  
Mathematics. Most mathematics is just understanding mathematics for the sake of itself; mathematical utility is measured by how mathematically useful it is to understanding mathematics.  
 
In terms of what you might have heard of, model theory is necessary in the formulation and interpretation of the likes of Gödel’s incompleteness theorems and Tarski’s theorem on the undefinability of truth. Arithmetic propositions that are logically independent of whatever axiomatic foundation one chooses (e.g. Peano arithmetic) correspond to different models of arithmetic, possibly nonstandard ones.

@Background Human  
What is model theory even for?

@Background Pony #D24B  
@Background Pony #3C49  
@Alkie  
It bothers me that you guys all go into the trouble of arranging to one side before removing the unknown. What, you think having 0 on one side will save you? That 0 is closer to an unknown than any other number?