Course MT2002 Analysis

## Solution 1

1. If Achilles catches the tortoise after t seconds, then the distance he runs is:
10 t = 100 +10t/17 16t/17 = 10 t = 170/16 or 10 5/8 seconds.

Generating the Geometric series gives:
10 + 10/17 +10/172 + 10/173 + ... and [recalling that the sum of a + ar + ar2 + ar3 + ... = a/(1 - r)] this sum is 10/(1 - 1/17) which is the same.

2. (a) Look at the Venn diagram:

Both sides of the identity are the shaded area.

(b) Take D to be the empty set and it is easy to see that the statement is then false.

(c) If (x, y) LHS then x A B and y A C and so (x A or x B) and (y A or y C). Looking at these four possibilities gives (x, y) A A or (x, y) A C or (x, y) B A or (x, y) B C and so LHS RHS.
The proof that RHS LHS is similar.

3. The first statement is true since one can take y = x.
The second statement is false since there is no choice of y which will work for all x.
In the third statement we may take y = 0 and this choice will work for all values of z and so the statement is true.
If the statement (x)P is true then (x)P is false and so there must be some x for which P is false. That is (x)(P).
Similarly, if (x)P is false then (x)P is true and so there is no x for which P is true. That is (x)(P) is false.

4. If x + r = s were rational then x = s - r would be rational and this gives a contradiction.
The proof that x - r and xr are irrational is similar.

The number x + y is not necessarily irrational. For example, x = -y = 2 gives a counterexample.
Similarly, taking x = y = 2 shows that x - y, xy and x/y may be rational. Of course, these combinations may be irrational for some values of x, y.

Take the point which divides the interval (r, s) in the ratio of 1 : 2. This is (2 r + s)/(2 + 1) and lies between r and s. It is a simple exercise to verify that if this were rational then 2 would be also.

5. If n = a/b (with a/b in lowest terms) then a2/b2 = n and so b2 divides a2. If a and b > 0 and have no common factors, this is impossible.

If (n + 4) + (n - 4) is rational, then its square is rational n + 4 +2(n + 4)(n - 4) + n - 4 is rational (n2 - 16) is rational (from the last result) n2 - 16 = m2 for some integer m. Then (n - m)(n + m) =16 and there are only a few possibilities n = 5, m = 3 and putting n = 5 in the original expression gives a solution.

Solutions of the quadratic are -b (b2 - 2) and so are rational b2 - 2 = d2 for d rational. Then (b - d)(b + d) = 2 and this has no integer solutions.

6. Suppose 3 = a/b. Then 3b2 = a2 and so a is divisible by 3. Then the RHS is divisible by 9 and so b2 is divisible by 3 and b is divisible by 3. Thus a, b have a common factor and we could have assumed that a/b was in "Lowest terms" and get a contradiction.

A similar proof (using divisibility by either 2 or 3) shows that 6 is irrational.

If 2 + 3 were rational, then its square: 2 + 26 + 3 would be also, from which one could deduce that 6 is rational.

If m + n = r is rational then r2= M + n + 2mn and so mn = s would be rational.
Then n = s2/m and r2 = m + s2/m + 2s = (s+m)2/m. Hence m = (s+m)2/r2 = [(s+m)/r]2 and so m would be rational.

7. With P on one side and Q on the other, if P were true then Q would be true P would be false.
If P were false then Q would be false P would be true. So in either case we get an inconsistent system.

If P were on both sides of the card then P can be either true or false and we get no contradiction.

With Q on both sides, Q false Q true and vice versa and we get an inconsistent system