STEP I 2008 question 1 solution

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A number is irrational if it not expressable as a fraction p/q where p and q are integers.

We assume that both p and q are expressible as fractions m/n and r/s say, then pq=mr/ns, the product of two integers is an integer so this is also the ratio of two integers, so pq is rational, contradicting that A is false.

Express p and q as above, then p+q = m/n + r/s = (ms+rn)/ns, since addition and multiplication are closed on the integers this is again the ratio of two integers so cannot be irrational, so pq cannot be irrational, contradicting that B is false.

Let p=x-q where x is some rational, then p+q=x which is also rational, p=sqrt(2), q=-sqrt(2) gives p+q=0 contradicting C.

If we take pi and e as arbitrary real irrationals we can prove no such thing, if e=±pi where pi^2 is irrational two of the four are zero which is rational, there are many ways to avoid this problem, the most obvious being that the five numbers listed be all distinct, or to assume some properties of e and pi such as 2<e<3<pi<4, or simply that they be positive and distinct. Suppose pi+e is rational, then e=r-pi where r is some rational, so pi-e=2pi-r which is an irrational - a rational which is irrational, pi^2-e^2 is the product of the first two so is irrational, and pi^2+e^2 is (pi+e)^2-2epi which is a raitonal- an irrational so is irrational. So assume pi+e is irraitonal, now by a similar argument pi-e is irrational, but we cannot have both the final two rational else their difference would also be rational, but this is 2e^2 which is irrational, so at most one of the four is rational. Here we have assumed that r is nonzero, if r were zero the argument fails, thus e=±pi are solutions.