In the first part of the paradox series, I listed four categories of paradoxes. Let’s look at an old favourite today, which falls in one of the categories I haven’t given an example of yet.

We have a proof that 1 = 0:

- Let
`x`

and`y`

be positive numbers such that`x = y`

- Square both sides:
`x2 = y2`

- Subtract
`y2`

from both sides:`x2 - y2 = 0`

- Factor
`x2 - y2`

:`(x + y)(x - y) = 0`

- Divide both sides by
`x - y`

:`x + y = 0`

- Since
`x = y`

, substitute`x`

for`y`

:`2x = 0`

- Divide both sides by
`2x`

:`1 = 0`

*— Quod Erat Demonstrandum.*

The flaw, of course, is in step 5: if `x = y`

then `x - y = 0`

, and we’re dividing by zero in step 5. (But note that because I made `x`

a positive number, there’s no division by zero in step 7.) Since division by zero is not well defined, everything after that point is invalid. That puts this into category two:

It actually is not true, because it’s based on a false premise (that seems true, or that’s been lost in the weeds).

There are many variations on this, “proving” slightly different things, but they all involve dividing by zero. When we use an invalid premise — here, the premise that division by zero is legitimate — we can claim to deduce anything from it. That’s because of how conditional statements work.

Put mathematically, if `P`

and `Q`

are statements, and `P`

is false, the statement `if P then Q`

is always true. But that doesn’t mean that `Q`

is true — if `P`

is false, we can say nothing at all about `Q`

, either way — and that’s the part that people often miss.

We’ll come back to that point later in this series.

## 2 comments:

Barry, I can't help but comment on a frustrating, recurring result of some of my pathetic attempts at math in high school, when I would be trying to solve a problem and would end up with the useless (but very true) equation 0=0. Not a paradox, but it might as well have been, for all the good it did me!

haha, M.W. :)

That came as a result of you replacing a solution of an equation back into that very same equation you manipulated.

It's akin to finding a calf in a flock of sheep, putting the calf back into the flock, and then looking for the odd one out... calf = calf... still :P

True, but you've discovered nothing you didn't already know!

It's a common, and frustrating, mistake high school students make in simultaneous equation problems, with a simple resolve... substitute your solution into the unused equation (ie. put the calf with the other calfs):)

Pity this weight couldn't have been relieved from your shoulders back in high school!

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