Assume sqrt(2) is a rational number.

To prove that sqrt(2) is an irrational number, we use a proof by contradiction. Step 1: Assume sqrt(2) is a rational number. If sqrt(2) is rational, it can be expressed as a fraction (p)/(q), where p and q are integers, q ≠ 0, and p and q have no common factors other than 1 (i.e., the fraction is in its simplest form). sqrt(2) = (p)/(q) Step 2: Square both sides of the equation. (sqrt(2))^2 = ((p)/(q))^2 2 = (p^2)/(q^2) 2q^2 = p^2 This equation implies that p^2 is an even number, because it is equal to 2 times an integer (q^2). Step 3: Deduce that p must be an even number. If p^2 is even, then p itself must be even. (If p were odd, p^2 would also be odd, e.g., (2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1). Since p is even, we can write p as 2k for some integer k. p = 2k Step 4: Substitute p=2k back into the equation 2q^2 = p^2. 2q^2 = (2k)^2 2q^2 = 4k^2 Divide both sides by 2: q^2 = 2k^2 This equation implies that q^2 is an even number, because it is equal to 2 times an integer (k^2). Step 5: Deduce that q must be an even number. If q^2 is even, then q itself must be even. Step 6: Reach a contradiction. We have shown that p is even and q is even. This means that both p and q have a common factor of 2. However, in Step 1, we assumed that p and q have no common factors other than 1. This is a contradiction. Step 7: Conclude that the initial assumption was false. Since our initial assumption that sqrt(2) is rational leads to a contradiction, the assumption must be false. Therefore, sqrt(2) must be an irrational number. Thus, under root 2 is an irrational number. 3 done, 2 left today. You're making progress.