The following is a “proof” that 1 = 2:
Let a and b be any numbers but a must equal b. For example 3 = 3 is a true maths sentence but let’s just keep things in terms of a and b. Given this as a starting point, we can do the following:
Step 1: a = b
Step 2: Multiply the left side by a and the right side by b, which is valid since a = b. This gives a2 = ab
Step 3: Add a2 to both sides: a2 + a2 = a2 + ab
Step 4: Add terms on left side: 2 a2 = a2 + ab
Step 5: Subtract 2ab from both sides: 2 a2 – 2ab = a2 + ab – 2ab
Step 6: Add the like terms on the right side together: 2 a2 – 2ab = a2 – ab
Step 7: Factor out the 2 on the left side: 2(a2 – ab) = a2 – ab
Step 8: Noting that (a2 – ab)/ (a2 – ab) = 1, divide both sides by a2 – ab: 2 = 1
Is this true? Has all our maths training been one big pile of fertiliser?
As much as my jokester half would like to say “yes it has”, there is something wrong with the above “proof”. All the steps are perfectly valid except for the last one. If a = b, then a2 – ab = 0. So the last step is effectively dividing both sides of the equation by 0.
Dividing by 0 is against maths law because allowing it would make maths inconsistent and all sorts of false equations can result. See what happens if you try to divide any number by 0, even 0 itself, by 0 on your calculator. (Do this in secret as I will not be responsible for getting you out of maths jail.)
Borrowing from an American commercial about drugs:
This is your mind: 2 = 2
This is your mind after dividing by 0: 2 = 1.