# Proofs: Problem 1

### Model Solution

Let x and y be real numbers, x non-zero. Suppose that x and $$\frac{y+1}{3}$$ are rational.

Since x is rational, $$x = \frac{m}{n}$$ where m and n are integers (n non-zero). Similarly, since $$\frac{y+1}{3}$$ is rational, $$\frac{y+1}{3} = \frac{p}{q}$$ where p and q are integers (q non-zero).

Since $$x = \frac{m}{n}$$ and x is non-zero, $$\frac{1}{x} = \frac{n}{m}$$.

Since $$\frac{y+1}{3} = \frac{p}{q}$$, $$y+1 = \frac{3p}{q}$$. So $$y = \frac{3p}{q} -1 = \frac{3p-q}{q}$$.

Then $$\frac{1}{x} + y = \frac{n}{m} + \frac{3p-q}{q} = \frac{nq + m(3p-q)}{mq}$$.

Let s = nq + m(3p-q) and t = mq. s is an integer, because n, q, m, and p are integers. Similarly, t is a non-zero integer, because m and q are non-zero integers.

Then $$\frac{1}{x} + y = \frac{s}{t}$$ where s and t are integers and t is not zero. So $$\frac{1}{x} + y$$ is rational, which is what we needed to prove.

### Self-check

Did you declare your variables (x and y above) at the start of the proof?

Does your proof start by supposing the hypothesis of the claim (x and $$\frac{y+1}{3}$$ are rational) and end with the conclusion of the claim ($$\frac{1}{x} + y$$ is rational)?

When the definition of rational is first used near the start of the proof, it introduces two new variables. Did you pick a fresh pair of variable names the second time you used it?

At the end of the proof, did you get $$\frac{1}{x} + y$$ into the form of a single fraction? Did you check that the top and bottom were both integers?