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Practice Problems
Problems from Chapter 3.1.
20. Prove that the sum of two rational numbers is rational.
Proof:
Using a direct proof. Given two rational numbers x=a/b and y =c/d where a, b, c and d are
integers, b0 and d 0, then x+y = a/b + c/d= (ad+cb)/bd (i.e. integer/integer and bd 0).
Thus x+y is rational.
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24. Prove or disprove that the product of a nonzero rational and an irrational number is
irrational.
The statement is true. We prove it using an indirect proof. Assume that x*y= z, where x is a
non-zero rational (i.e. x= a/b where a and b are integers and b0, a 0 follows from x being
non-zero) and y is irrational. Now if z is rational (negation of conclusion, assume z = c/d
where c and d are integers and d0) then y= z/x (no division by zero here!) = (bc)/(ad),
where a, b, c and d are integers and also (ad)  0 since a0 and d d0. Thus y is rational,
which contradicts the fact that y is irrational.
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26. Prove or disprove that 2n +1 is prime for all nonnegative integers n.
The statement is false. We can give a counter-example.
Consider n=6, then 26 +1=65 is not prime since it is divisible by 5.
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34. Prove that if n is a positive integer, then n is even if and only if 7n+4 is even.
Proof:
The statement given is of the form p  q, where p: n is even, q: 7n+4 is even.
Thus we have to show p  q and q  p.
p  q (by direct proof):
n is even  n=2i for some integer i.
Thus 7n+4 = 7(2i)+4= 2(7i+2) = even integer.
q  p (by indirect proof, show not p  not q):
not p  n is odd  n=2i+1 for some integer i.
Thus 7n+4 = 7(2i+1)+4= 14i+11= 14i+10+1= 2(7i+5) + 1= odd = not q.
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58. Find a counter example to the proposition:
"For every prime number n, n+2 is prime."
Consider n= 23. Here 23 is prime yet 23+2=25 is not prime.
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60. Prove or disprove that if p1, p2, ..., pn are the n smallest primes, then
k=p1p2 ... pn+1 is prime.
The statement is false. The number k is clearly not divisible by any of p1, p2, ..., pn; however,
it may be a prime or divisible by a prime number > pn. For example,
2*3*5*7*11*13+1=30031 =59*509.
Problems from Chapter 3.2.
20. Use mathematical induction to show that 3 divides n3 + 2n whenever n is a nonnegative
integer.
Proof :
Base Step:
For n=0, 3 divides 03 + 2*0=0 (3 divides 0 is true).
Also (extra check) for n=1, 3 divides 13 + 2*1=3 (3 divides 3 is true).
Induction Step:
Assume P(n) and show P(n+1).
Thus we assume that 3 divide n3 + 2n and show that 3 divides (n+1)3 + 2(n+1).
We have to show (n+1)3 + 2(n+1) is divisible by 3.
(n+1)3 + 2(n+1) = (n+1)(n2+2n+1) + 2(n+1)
= n3 + 2n2 + n + n2+ 2n+1 + 2(n+1) = n3+3n2 + 3n + 2n +3
= (n3 + 2n) + 3(n2 + n + 1)
This consists of a sum of two parts; the first part is divisible by 3 from the induction
hypothesis; the second part is clearly divisible by 3 (it is a multiple of 3).
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34. Prove by induction that 1* 21 + 2 * 22 + ... + n*2n = (n-1) 2n+1 + 2.
Proof:
Base Step:
For n=1, LHS = 1* 21 = 2. RHS = (1-1) 21+1 + 2 = 2 = LHS.
Induction Step:
Assume P(n) and show P(n+1).
Thus we assume that 1* 21 + 2 * 22 + ... + n*2n = (n-1) 2n+1 + 2,
and show that 1* 21 + 2 * 22 + ... + n*2n + (n+1)*2n+1 = n 2n+2 + 2.
LHS (of what is to be shown) = 1* 21 + 2 * 22 + ... + n*2n + (n+1)*2n+1
= (n-1) 2n+1 + 2 + (n+1)*2n+1 by the induction hypothesis
= [n-1+n+1] 2n+1 + 2
= (2n) 2n+1+ 2
= n 2n+2+ 2
= RHS.