A little over a month ago, we proved that 64 = 65. Now, let's take that down a notch, and prove that 1 = 2. We will prove it two ways, one will be a little easier to understand, and one will involve complex numbers, which we worked with around four weeks ago.

Easy Proof: Let's say that a = b. Pretty simple. Now, we'll multiply both sides by a. This will be step one.

a = b

a(a) = a(b)

a^2 = ab

How about we add a^2 to both sides. This is step two.

a^2 = ab

a^2 + a^2 = a^2 + ab

2a^2 = a^2 + ab

For step three, we will subtract 2ab from both sides.

2a^2 = a^2 + ab

2a^2 - 2ab = a^2 + ab - 2ab

2a^2 - 2ab = a^2 - ab

For step four, we will factor out a two from the left hand side.

2a^2 - 2ab = a^2 - ab

2(a^2 - ab) = a^2 - ab

For step five, we will divide both sides of the equation by a^2 - ab to give us 2 = 1.

2(a^2 - ab) = a^2 - ab

(2(a^2 - ab))/(a^2 - ab) = (a^2 - ab)/(a^2 - ab)

2 = 1

Complex Proof (literally!): This one does involve complex numbers, so it might become a little bit challenging. However, it is pretty easy to understand, as long as you realize that √(-1) = i.

Let's remind ourselves that -1/1 = 1/-1. For step one, let's take the square root of both sides.

-1/1 = 1/-1

√(-1/1) = √(1/-1)

We can now simplify that to give us:

√(-1)/√(1) = √(-1)/√(1)

For step three, we can eliminate all of the square roots and replace them with 1s and is.

i/1 = 1/i

For step four, let's divide each side by two.

i/2 = 1/2i

For step five, how about we add 3/2i to both sides. Strange attempt, but we can go ahead and do it.

i/2 + 3/2i = 1/2i + 3/2i

Let's multiply through by i. That will be our step six.

i(i/2 + 3/2i = 1/2i + 3/2i)

i^2/2 + 3i/2i = i/2i + 3i/2i

For step seven, we can simplify this whole mess and see what we get.

i^2/2 + 3i/2i = i/2i + 3i/2i

-1/2 + 3/2 = 1/2 + 3/2

2/2 = 4/2

1 = 2

Again, we are ending up with the strange solution of 1 = 2.

Why on earth could this be? Maybe arithmetic and geometry can make mistakes, but algebra! Turns out, algebra is fine. These proofs are fallacies. See if you can figure out which step is incorrect, and then read the below.

Easy Proof's Fallacy: Turns out, step five was a fallacy. Where were we there?

2(a^2 - ab) = a^2 - ab

We divided both sides by a^2 - ab to get 2 = 1. The problem lies in the a^2 - ab division. Let's look at it. We know that a = b, so let's plug a in for the b.

a^2 - ab

a^2 - a(a)

a^2 - a^2

0

We have ended up dividing by zero. Since this is not allowed in mathematics, we cannot do this step. That means that 2 ≠ 1, or at least this does not prove it.

Complex Proof's Fallacy: Again, our simplification was where the fallacy lied. In this problem, it was in step two, when we separated the square roots. Let's look at it:

√(-1/1) = √(1/-1)

√(-1)/√(1) = √(-1)/√(1)

We have jumped one step too far. This property we just used is only true for positive square roots. Remember, a square root is a number when multiplied itself gives you the radicand, or number inside the square root symbol. If the square root is negative, then it does not hold true all the time.

If that was hard to understand, let me lay it out in more simple terms. Say we have:

√((-1)(-1))

One way we could do it is multiply together the -1s and get 1.

√((-1)(-1))

√(1)

1

But if you used this property, you would have:

√(-1)√(-1)

i • i

i^2

-1

This gave us two different answers, so this property just doesn't hold in these circumstances. There are tons and tons of proofs like this, even some that involve calculus. I think that these false statements are really cool to look at and try to figure out what the mistake is. So fortunately, algebra has not went under yet.

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