Proof by Contradiction Examples

Prove the following statement by contradiction:

The sum of two even numbers is always even.

For starters, let's negate our original statement:

The sum of two even numbers is not always even.

That would mean that there are two even numbers out there in the world somewhere that'll give us an odd number when we add them. Let's try proving that.

By definition, even numbers are evenly divisible by 2. Right? Right. So we could write our new supposition as:

2a + 2b = c

Even and odd numbers are always integers (no fractions or decimals), so we know 2a and 2b are integers, which means a and b are also integers—if we divide an even number like 2a by 2, we'll always get a integer. We also know c is an odd integer, which means it's not evenly divisible by 2. Now we can factor out a 2 from the left side:

2(a + b) = c

And then divide by 2:

Hmm. The stuff on the left, a + b, must add up to an integer because the sum of two integers is always another integer. We're not gonna get a fraction or decimal when we add 5 + 16, for example. But we already decided c must be an odd number, which means we can't divide it evenly by 2. That means  is not an integer.

Wait…if a + b is an integer, but isn't an integer, there's no way our equation is true. The stuff on the left can't possibly equal the fraction on the right. Bingo, that's a contradiction.

Now we just need a nice, formal statement using our Mad Lib fill-in-the-blank from the reading:

"Since the sum of two even numbers 2a and 2b must always be an integer that's divisible by 2, this contradicts the supposition that the sum of two even numbers is not always even. Hence, our original proposition is true: the sum of two even numbers is always even."

Prove the following statement by contradiction:

The negative of any irrational number is also irrational.

First we want to negate our original statement. Careful though; the negation is not "the negative of any rational number is also rational." That's a whole different thing. We just need a new statement that makes our original statement false.

Our original proposition is saying that if n is irrational, then -n is also irrational. To negate that, we'd write:

There is at least one irrational number n such that -n is rational.

See what we did there? For the original statement to be false, we'd need one example of a number where n is irrational and -n is rational.

That's our new supposition, so now we try to prove it. By definition, a rational number can be written as a ratio (fraction) of two integers. Since -n is rational, we should be able to write it like this:

…where both a and b are integers and b ≠ 0, since we can't ever divide by 0. But what happens when we divide both sides by -1 and solve for n?

Hold up. If a is an integer, then -a is also an integer (think of 2 and -2, 37 and -37, etc.). But that would mean that n is also a rational number, since we just showed that it's the ratio of two integers: -a and b. Our supposition said that n is an irrational number, though. It can't be both rational and irrational, so there's our contradiction.

We finish up by writing a formal statement:

"Since a rational number (where a and b are integers and b ≠ 0) implies that n is a ratio of two integers -a and b, this contradicts the supposition that there is at least one irrational number n such that -n is rational. Hence, our original proposition was true: the negative of any irrational number is also irrational."

Whew, what a mouthful.

Prove the following statement by contradiction:

The sum of two positive numbers is always positive.

We've got our proposition, which means our supposition is the opposite:

The sum of two positive numbers is not always positive.

Or, in other words:

There exist two positive numbers a and b that sum to a negative number.

Proof time. Putting our supposition in math lingo, we get:

a + b < 0

Their sum needs to be negative, or less than 0. Let's rearrange our inequality a bit. How 'bout we subtract b from both sides?

a < -b

Our supposition implies that a must be less than -b. But since b is a positive number, -b must be a negative number. That would mean that a, which is positive, is less than b, which is negative. Hold the phone: anything less than a negative would also be negative. Uh, a can't be positive and negative at the same time. That would probably destroy the universe in a puff of purple smoke, and we can't have that.

We've found our contradiction: if a and b are both positive, a can't possibly be less than -b. Let's formalize this mess.

"Since a + b < 0 implies that a < -b when both a and b are positive, which means a is also negative, this contradicts the supposition that there exist two positive numbers a and b that sum to a negative number. Hence, our original proposition is true: the sum of two positive numbers is always positive."