Proof x = 1 with no prior determining

I'll try and prove that x = 1 without a question to solve. Here we go.

The number line is 0, 1, 2, 3, 4, you get the idea. But we will only pick 1 and 2. Let's add them in one group: [1, 2]. Now add 3 and 4: [3, 4].

[x, y] [z, w] will equal (x * y) * (z * w) in this case. Swap 2 and 1, because if we don't we will unconsciously determine, x equals 1. We need to make this a long-term effort instead of determining x = 1 in one minute and thirty-three seconds.

So, (2 * 1) * (3 * 4). Now, let's start with 2 * 1, which is 2. The rule is, x * 1 = x (no exceptions). 3 * 4 is 12. 12 * 2 is 24.

But 24 doesn't equal 1. That's where we up the ante.

Let [2, 1] [3, 4] = y. Let [2,1]^[3,4] = z. In most cases of algebra, x * y = z, but this time, y * z = x.

We know y = 24. Since [2,1] is 2 and [3,4] is 12, we get 2^12. We multiply two by itself, then the new answer twelve times.

So:

1. 2
2. 4
3. 8
4. 16
5. 32
6. 64
7. 128
8. 256
9. 512
10. 1024
11. 2048
12. 4096

So z = 4096. In which case, y * z, or 24 * 4096 does not equal 1.

24 * 4096 = 93804. Let's just betray the original purpose. So x = 93804. Don't worry, we'll jump back to 1.

In 93804, we have 90,000, 3,000, 800 and 4. Divide all values by 93804. Now sandwich them together, for a value of 0.95944 + 0.03198 + 0.00852 + 0.0004, or 1.000034. Broadly, 1.

But x = 93804, not 1! Uhh... that was just to come back to 1. I take it back, x = 1, and that is final. I know in reality, x = 1.000034, but we need five decimal points for any going off the rails. At least x can be rounded to 1.

I know I'm no math genius, and I was abusing the definition of x in this case, but who cares?