Test Flight Problem Set for Introduction to Mathematical Thinking on Coursera Guo Yading 2017/2/13 1. Say whether the following is true or false and support your answer by a proof. (∃m ∈ ℕ) (∃n ∈ ℕ) (3m + 5n = 12) The statement is false. Proof: By list all the possible pairs of m and n . For pairs contains m ≥ 4 or n ≥ 3 , there is no possible way to produce the equation 3m + 5n = 12 . The reason is that: If m ≥ 4 , then 3m + 5n > 12 , because n is a natural number. Similarly, If n ≥ 3 , then 3m + 5n > 12 , because m is a natural number.
m must be one of 1, 2, 3 and n must be one of 1, 2 6 pairs of m and n can be formed, they are: m = 1 and n = 1 , 3m + 5n = 8 m = 1 and n = 2 , 3m + 5n = 13 m = 2 and n = 1 , 3m + 5n = 11 m = 2 and n = 2 , 3m + 5n = 16 m = 3 and n = 1 , 3m + 5n = 14 m = 3 and n = 2 , 3m + 5n = 19 Thus, no possible pairs of m and n in natural numbers can lead to the equation 3m + 5n = 12 . (∃m ∈ ℕ) (∃n ∈ ℕ) (3m + 5n = 12) is false.
2. Say whether the following is true or false and support your answer by a proof: The sum of any five consecutive integers is divisible by 5 (without remainder).
The statement is false. Proof: By given arbitrary five consecutive integers. List these five numbers for smallest one to biggest one. Suppose the third of the list is a . Then the list can be represented as following: a−2, a−1 , a , a+1, a+2 As they are consective integers. Their sum is (a − 2 ) + (a − 1) + a + (a + 1) + (a + 2) = 5a, by algeba. 5a 5 = a , since a is a integers. The sum is divisible by 5. As required.
3. Say whether the following is true or false and support your answer by a proof: For any integer n , the number n2 + n + 1 is odd. The statement is true. Proof: By assuming n is odd or even. Suppose n is odd. n2 is odd. Because the product of two odd numbers is still odd. n2 + n is even. Because the sum of two odd numbers is even. n2 + n + 1 is odd. Because the sum of one odd number, 1 in this formula, and one even number is odd. Suppose n is even. n2 is even. Because the product of two even numbers is still even. n2 + n is even. Because the sum of two even numbers is even. n2 + n + 1 is odd. Because the sum of one odd number, 1 in this formula, and one even number is odd. That is either n is even or odd, n2 + n + 1 is odd. Because integers are only composited by even numbers and odd numbers. Thus for any integer n , the number n2 + n + 1 is odd.
4. Prove that every odd natural number is of one of the forms 4n + 1 or 4n + 3 , where n is an integer. Proof: By division theorem:
∀a, b ∈ ℤ, b =/ 0 : ∃!q, r ∈ ℤ : a = qb + r, 0 ≤ r < |b| For any natural number a and b = 4 . There always exist n, 0 ≤ r < 4 which is 0, 1, 2, 3 . Thus, any natural number is one of following form: 4n , divided by 2 is 2n , even number. 4n + 1 , an even number plus 1, odd number. 4n + 2 , divided by 2 is 2n + 1 , even number. 4n + 3 , and even number plus 3, odd number. Only 4n + 1 and 4n + 3 are odd number in the above representations. And the above representations contains all the natural numbers. So every odd natural number is of one of the forms 4n + 1 or 4n + 3 .
5. Prove that for any integer n , at least one of the integers n , n + 2 , n + 4 is divisible by 3. Proof: First I will prove ∀a ∈ ℤ , at least one of the integers a , a + 1 , a + 2 is divisible by 3. By division theorem:
∀a, b ∈ ℤ, b =/ 0 : ∃!q, r ∈ ℤ : a = qb + r, 0 ≤ r < |b| For any integer a and b = 3 . There always exist n, 0 ≤ r < 3 which is 0, 1, 2 . That means a can only be in one of following three forms: 3n , which is clearly divisible by 3. So in this situation, at least one of the integers a , a + 1 , a + 2 is divisible by 3.
3n + 1 , a + 2 = 3n + 3 , which is also clearly divisible by 3. So in this situation, at least one of the integers a , a + 1 , a + 2 is divisible by 3. 3n + 2 , a + 1 = 3n + 3 , which is also clearly divisible by 3. So in this situation, at least one of the integers a , a + 1 , a + 2 is divisible by 3. In summary, in all the situations, at least one of the integers a , a + 1 , a + 2 is divisible by 3. Second I will prove n , n + 2 , n + 4 has the same property as a , a + 1 , a + 2 . If n is even, I can write n = 2k . The n , n + 2 , n + 4 become 2k , 2k + 2 , 2k + 4 which by algebra can be transformed to 2 (k) , 2 (k + 1) , 2 (k + 2) . Clearly, 2 (k) , 2 (k + 1) , 2 (k + 2) is obtained by multiplying 2 into a , a + 1 , a + 2 . Because at least
one of the integers a , a + 1 , a + 2 is divisible by 3. At least one of the integers 2 (k) , 2 (k + 1) , 2 (k + 2) is divisible by 3.
If n is odd, I can write n = 2k + 1 . The n , n + 2 , n + 4 become 2k + 1 , 2k + 3 , 2k + 5 . As I have already proved at least one of the integers k , k + 1 , k + 2 is divisible by 3. If k is divisible by 3. Obviously, 2k + 3 is divisible by 3. In this situation, at least one of the integers 2k + 1 , 2k + 3 , 2k + 5 is divisible by 3. If k + 1 is divisible by 3. 2k + 5 = 2(k + 1) + 3 is clearly divisible by 3. In this situation, at least one of the integers 2k + 1 , 2k + 3 , 2k + 5 is divisible by 3. If k + 2 is divisible by 3. 2k + 1 = 2(k + 2) − 3 is clearly divisible by 3. In this situation, at least one of the integers 2k + 1 , 2k + 3 , 2k + 5 is divisible by 3. In conclusion, at least one of the integers n , n + 2 , n + 4 is divisible by 3 when n is even or odd. Proof completes.
6. A classic unsolved problem in number theory asks if there are infinitely many pairs of ‘twin primes’, pairs of primes separated by 2, such as 3 and 5, 11 and 13, or 71 and 73. Prove that the only prime triple (i.e. three primes, each 2 from the next) is 3, 5, 7. Proof: By using the proved conclusion in last problem, which is: For any integer n , at least one of the integers n , n + 2 , n + 4 is divisible by 3. Thus For any natural number n , at least one of the integers n , n + 2 , n + 4 is divisible by 3. Take n as a natural number. When n > 3 , n + 2 > 3 and n + 4 > 3 . According to the truth proved above at least one of the integers n , n + 2 , n + 4 is divisible by 3. So, at least one of the integers n , n + 2 , n + 4 is not prime. Because 3 is neither 1 or one of n , n + 2 , n + 4 in this circumstance. The only choices left for n is 1, 2, 3. 1 is not a prime. Only 2 and 3 is possible. For n = 2 , n , n + 2 , n + 4 is 2, 4, 6 which is clearly not a “prime triple”. Because both 4 and 6 are not prime. The only choice left for n is 3 . And the sequence is 3, 5, 7. They are all primes. Proof done.
7. Prove that for any natural number n , 2
3
n
n+1
2 + 2 + 2 + ... + 2 = 2
−2
Proof: by induction. 1 1+1 For n = 1 . Left side is 2 = 2 . Right side is 2 − 2 = 2 . Equation checked. Assuming it is true for a natural number n , that is: 2
3
n
n+1
2 + 2 + 2 + ... + 2 = 2 While n become n + 1 , left side of equation is: 2
3
n
n+1
2 + 2 + 2 + ... + 2 + 2
n+1
=2
n+1
−2+2 n+1
= 2•2
n+1+1
=2
−2
by using assumption of induction
− 2 by algebra
− 2 by algebra
The equation hold for n + 1 if it hold for n . Here I complete my proof.
∞
8. Prove (from the definition of a limit of a sequence) that if the sequence {an}n=1 tends to limit L as ∞ n → ∞ , then for any fixed number M > 0 , the sequence {Man}n=1 tends to the limit M L . Proof: Take an arbitrary real number ε . ∞
Because the sequence {an}n=1 tends to limit L as n → ∞ . For the number that when m > n0 , |am − L| < Mε which, by algebra, can be transformed to: |Mam − M L| < ε .
ε M
, there always exist an n0
Combining above two points, that is for an arbitrary real number ε , there always exist an n0 that when ∞ m > n0 , |Mam − M L| < ε . This is necessary and sufficient for the sequence {Man}n=1 tends to the limit M L . As required.
9. Given an infinite collection An , n = 1, 2, ... of intervals of the real line, their intersection is defined to be ∞
An = {x| (∀n) (x ∈ An)} ∩ n=1 Give an example of a family of intervals An , n = 1, 2, ... such that An+1 ⊂ An for all n and
∞
An = ∅ . ∩ n=1
Prove that your example has the stated property. An example is An is (1, 1 + an ] for n = 1, 2, ... , a is a positive real number. For this example Proof: By contradiction. Suppose
∞
An =/ ∅ ∩ n=1
Then x ∈
∞
, there are at least one element in
An , because ∩ n=1
Because limit of
∞
An . The element is x . ∩ n=1
An is (1, 1 + an ] , x > 1 that is x − 1 > 0. Say x − 1 = ε.
{ an } is 0. For
ε , there always exists a n0 that when m > n0 , || ma − 0|| < ε. a m < ε. Because a is a positive real number
∞
An = ∅ . ∩ n=1
a m
< x − 1. Because I assume x − 1 = ε 1 + ma < x. By algebra That clearly show that x is out of boundary of interval (1, 1 + ma ] when m > n0. x don’t belong to interval (1, 1 +
a m].
This result contradicts to the assumpt x ∈
∞
∞
An . So x must not exists. Not element exist in ∩ n=1
An . The statement is proved. ∩ n=1 10. Give an example of a family of intervals An , n = 1, 2, ... such that An+1 ⊂ An for all n and
∞
An ∩ n=1
consists of a single real number. Prove that your example has the stated property. An example is An is (− an , an ) for n = 1, 2, ... , a is a positive real number. For this example
∞
An only ∩ n=1
contains a real number, 0. Proof: By elimination the real number other than 0. Suppose There are at least one element in
∞
An . The element is x and x is not zero. ∩ n=1
is 0. For x , there always exists a n0 that when m > n0 , || ma − 0|| < x. Because a is a positive real number. Get ma < x. x don’t belong to interval (− ma , ma ) . If x < 0 , because limit of {− an } is 0. For x , there always exists a n0 that when m > n0 , ||− ma − 0|| < |x| . If x > 0 , because limit of
{ an }
Because a is a positive real number and x is negative. belong to interval (− ma , ma ) .
a m
<− x. By algebra I get x <− ma . Clearly, x don’t
For real number 0. I can simply prove that for any m and positive real number a . − ma < 0 and 0 < ma . And 0 is belong to any (− ma , ma ) . Thus I proved 0 is the only real number in
∞
An . ∩ n=1