`2 = 1`

, which is also false, now I am going to show where it went wrong, by doing the same steps as before, using the same forms as before.```
a = b
```

Substitute b for a in the entire rest of the article, since they are the same.

```
b = b
```

we can do whatever we want to an equation as long as we do the same thing to both sides, so first add

`b`

```
b + b = b + b
```

partially simplify

```
2b = b + b
```

subtract

`2b`

```
2b - 2b = b + b - 2b
```

partially group like terms

```
2b - 2b = b + (1-2)b
```

simplify

```
2b - 2b = b + (-1)b
```

adding a negative number is the same as subtracting

```
2b - 2b = b - b
```

factor out

`(b - b)`

```
2(b - b) = 1(b - b)
```

divide by

`(b - b)`

, which is also known as zero.```
\frac{2(b - b)}{b - b} = \frac{1(b - b)}{b - b}
```

re-factor to isolate the one, now we have zero over zero, anything can happen

```
2\left(\frac{b - b}{b-b}\right) = 1\left(\frac{b - b}{b-b}\right)
```

reduce the one, and we are just not allowed to get here

```
2(1) = 1(1)
```

simplify

```
2 = 1
```

Division by zero is not allowed, and zero over zero is just undefined.

## No comments:

## Post a Comment