sequences and series - Is $sum_{n=1}^{infty} 1 = -frac{3}{12}$ true?

Here is how I derived this...

$$1+2+3+4+...=-\frac{1}{12}$$
$$2+4+6+8+...=2(1+2+3+4+...)=2(-\frac{1}{12})=-\frac{1}{6}$$
Thus $1+3+5+7+...=-\frac{1}{6}-(1+1+1+...) $ because $S_{odd}=S_{even}-(1+1+1+...)$

Also $S_{odd}=(1+2+3+4+...)-(2+4+6+8+...)=-\frac{1}{12}+\frac{1}{6}=\frac{1}{12}$

Thus $\frac{1}{12}=-\frac{1}{6}-(1+1+1+...)$
$$1+1+1+...=-\frac{3}{12}$$

What I think my mistake was (if I have one) is where I assume the sum of all numbers is half the sum of all even numbers; although, it should work since there is nothing two times infinity.

Please leave simple solution (im only 15 years old) to why this is false: Wikipedia says that $\sum_{n=1}^{\infty} 1 = \infty$.

Edit: I know now that the above notion is false. I recently watched a Numberphile video proving $\sum_{n=1}^{\infty} n = -\frac{1}{12}$. I followed the same line of reasoning to derive the untrue $\sum_{n=1}^{\infty} 1 = -\frac{3}{12}$ I'll admit; I was ignorant in believing so and I apologize for wasting your time.

Thanks for the answer kindly explaining why I was wrong. I guess I should watch this video to un-brainwash me. Sorry again.

Answer

Take two infinite sets: $A=\{1, 2, 3, \cdots\}$ and $B=\{4, 5, 6, \cdots\}.$ Then $A-B = \{1,2,3\}$ and so is of cardinality 3. So infinity minus infinity must be 3. Now by choosing a different $B$ you can make infinity minus infinity any number you like.

In fact, choose $B=\{1, 3, 5, \cdots\}$ and then $A-B =\{2, 4, 6, \cdots\},$ and so $A-B$ has cardinality equal to infinity.

The point here is that when you start juggling infinities, you can make just about anything equal to just about anything. Which means almost everything is not well-defined. Once you had a divergent series in your hand, everything else was nonsense.