Question

Use the method of proof by contradiction to prove the following statements. (In each case, you should also think about how a direct or contrapositive proof would work. You will find in most cases that proof by contradiction is easier.) Prove that ✓6 is irrational.

Answer 1

To prove that sqrt(6) is irrational, we assume that sqrt(6) is rational and show that it leads to a contradiction. We start by assuming sqrt(6) can be written as p/q, where p and q are integers with no common factors. By squaring both sides and simplifying, we find that p and q are both divisible by 2, which contradicts our assumption. Therefore, sqrt(6) is irrational.

Step-by-step explanation:

To prove that √6 is irrational using the method of proof by contradiction, we assume that √6 is rational. This means that we can write √6 as √6 = p/q, where p and q are integers with no common factors other than 1 and q ≠ 0.

Squaring both sides of the equation, we get 6 = p2/q2, which can be rewritten as 6q2 = p2.

From this equation, we can see that p2 is divisible by 6, which means that p and p2 are also divisible by 2 and 3. Therefore, p must be divisible by 2 and 3. Let's write p as p = 2k, where k is an integer. Substituting this into the equation, we get 6q2 = (2k)2 = 4k2, which simplifies to 3q2 = 2k2.

Now we can see that q2 is divisible by 2, which means that q and q2 are also divisible by 2. Therefore, q must be divisible by 2. Let's write q as q = 2m, where m is an integer. Substituting this into the equation, we get 3(2m)2 = 2k2, which simplifies to 12m2 = 2k2. Dividing both sides by 2, we get 6m2 = k2.

Now we have shown that both p and q are divisible by 2, which means that p/q is not in simplest form. This contradicts our assumption that √6 can be written as a fraction, so our initial assumption is false and √6 is irrational.