Prove that there exist irrational real numbers x and y where xy is rational. Numbers x and y are not required to be different. (Hint: consider z z where z is the square root of 2. It is not clear whether that is rational or irrational. Consider both cases.)
Let z be the square root of 2, which is known to be irrational. Note that z2 = 2. There are two cases.
If z z is rational, choose x = z and y = z. Then x and y are irrational and xy is rational.
If z z is irrational, choose x = z z and y = z. Then x and y are both irrational and xy = (z z) z = z z⋅z = z2 = 2 since z⋅z = z2 = 2.
Prove or disprove that if you have an 8 gallon jug filled with water and you have two empty 5 gallon and 3 gallon jugs, then you can measure exactly 4 gallons of water by successively pouring some or all of the water from one jug to another. You cannot make eyeball estimates.
Here is one solution. There are many solutions, and students came up with some innovative ones.
Each row shows how many gallons each container holds. The jugs are shown in decreasing order by capacity: 8, 5, 5, 3, 3. From one row to the next, water is poured from just one jug to just one other jug, filling the second jug. The jugs that had their contents changed from the previous row are in red.
| 8 | 0 | 0 | 0 | 0 |
| 3 | 5 | 0 | 0 | 0 |
| 3 | 2 | 0 | 3 | 0 |
| 0 | 2 | 3 | 3 | 0 |
| 0 | 2 | 5 | 1 | 0 |
| 0 | 2 | 2 | 1 | 3 |
| 0 | 4 | 0 | 1 | 3 |
Prove that, for all real numbers x and y, min(x,y) + max(x,y) = x + y.
If x = y, the claim is clearly true. That leaves two cases: (1) x < y and (2) y < x. Without loss of generality, assume that x < y.
Then min(x, y) = x and max(x, y) = y. So min(x,y) + max(x,y) = x + y.