Page 18 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 1] NUMBERS 9
2
x 5x þ 6
1.3. Simplify .
2
x 2x 3
2
x 5x þ 6 ðx 3Þðx 2Þ x 2 provided that the cancelled factor ðx 3Þ is not zero, i.e., x 6¼ 3.
2
x 2x 3 ¼ ðx 3Þðx þ 1Þ ¼ x þ 1
For x ¼ 3the given fraction is undefined.

RATIONAL AND IRRATIONAL NUMBERS

1.4. Prove that the square of any odd integer is odd.
2
2
Any odd integer has the form 2m þ 1. Since ð2m þ 1Þ ¼ 4m þ 4m þ 1is1 more than the even integer
2
2
4m þ 4m ¼ 2ð2m þ 2mÞ,the result follows.
1.5. Prove that there is no rational number whose square is 2.
Let p=q be a rational number whose square is 2, where we assume that p=q is in lowest terms, i.e., p and q
have no common integer factors except 1(we sometimes call such integers relatively prime).
2
2
2
2
2
Then ðp=qÞ ¼ 2, p ¼ 2q and p is even. From Problem 1.4, p is even since if p were odd, p would be
odd. Thus p ¼ 2m:
2
2
2
2
2
Substituting p ¼ 2m in p ¼ 2q yields q ¼ 2m ,sothat q is even and q is even.
Thus p and q have the common factor 2, contradicting the original assumption that they had no
common factors other than 1. By virtue of this contradiction there can be no rational number whose
square is 2.
1.6. Show how to find rational numbers whose squares can be arbitrarily close to 2.
2
2
We restrict ourselves to positive rational numbers. Since ð1Þ ¼ 1 and ð2Þ ¼ 4, we are led to choose
rational numbers between 1 and 2, e.g., 1:1; 1:2; 1:3; .. . ; 1:9.
2
2
Since ð1:4Þ ¼ 1:96 and ð1:5Þ ¼ 2:25, we consider rational numbers between 1.4 and 1.5, e.g.,
1:41; 1:42; .. . ; 1:49:
2
Continuing in this manner we can obtain closer and closer rational approximations, e.g. ð1:414213562Þ
2
is less than 2 while ð1:414213563Þ is greater than 2.
n n 1
1.7. Given the equation a 0 x þ a 1 x þ þ a n ¼ 0, where a 0 ; a 1 ; .. . ; a n are integers and a 0 and
a n 6¼ 0. Show that if the equation is to have a rational root p=q, then p must divide a n and q
must divide a 0 exactly.
n
Since p=q is a root we have, on substituting in the given equation and multiplying by q ,the result
n
n
a 0 p þ a 1 p n 1 q þ a 2 p n 2 2 n 1 þ a n q ¼ 0 ð1Þ
q þ þ a n 1 pq
or dividing by p,
a n q n
a 0 p n 1 þ a 1 p n 2 q þ þ a n 1 q n 1
p ð2Þ
¼
Since the left side of (2)is aninteger, the right side must also be an integer. Then since p and q are relatively
n
prime, p does not divide q exactly and so must divide a n .
In a similar manner, by transposing the first term of (1) and dividing by q,wecan show that q must
divide a 0 .
p ffiffiffi p ffiffiffi
1.8. Prove that 2 þ 3 cannot be a rational number.
p ffiffiffi p ffiffiffi 2 p ffiffiffi 2 p ffiffiffi 4 2
3,then x ¼ 5 þ 2 6, x 5 ¼ 2 6 and squaring, x 10x þ 1 ¼ 0. The only possible
If x ¼ 2 þ
rational roots of this equation are 1byProblem 1.7, and these do not satisfy the equation. It follows that
p ffiffiffi p ffiffiffi
3, which satisfies the equation, cannot be a rational number.
2 þ