Mathematical Proofs: Foundations of Mathematics Assignment 3 Analysis

This document presents solutions for Assignment 3, focusing on mathematical proofs. The first proof demonstrates that among any three consecutive integers, one must be divisible by 3. The second proof, using proof by contradiction, establishes that if a, b are rational (b≠0) and r is irrational, then a + b*r is also irrational. The third solution selects statement (b) and provides a proof by mathematical induction to show that 8^n – 3^n is divisible by 5 for any integer n >= 0. The final proof, also using proof by contradiction, demonstrates that at least one of n consecutive integers must be divisible by n. The solutions are carefully constructed and thoroughly explained, making them a valuable resource for students studying foundations of mathematics.

Running head: ASSIGNMENT 3
ASSIGNMENT 3
Name of the Student
Name of the University
Author Note

1ASSIGNMENT 3
1)
Let, the consecutive integers are n, n+1 and n+2.
Where n is any real integer.
Now, when divided by three then a number is not divisible if remainder is either 1 or 2.
Now, as three numbers are in sequence then one number must have remainder 1 or 2.
Hence, there exist one number which is divisible by 3 when a set of three consecutive
integers are divided by 3.
2) Let us assume as contradiction that a+br is a rational number.
Thus a + b*r can be expressed by, a+b*r = p1/q1 (where p1,q1 are any integers with q1≠0).
Hence, a = p2/q2 (q2≠0) and b = p3/q3 (q3≠0).
Now, from given condition
a + b*r = p1/q1
 p2/q2 + (p3/q3)*r = p1/q1
 (p3/q3)*r = (p1q2 – p2q1)/(q1q2)
 r = (p1q2 – p2q1)q3/(q1q2p3)
As, q1q2p3 ≠ 0 and r is written in a form as a numerator over a non-zero integer which makes
r is rational. This is a contradiction as given r is irrational. Hence, a+b*r is only rational if r is
also rational.
Thus it is proved that when a, b are rational (b≠0) and r is irrational then a + b*r is also
irrational.
3) The statement (b) is chosen.

2ASSIGNMENT 3
b) 8^n – 3^n is divisible by 5 for any integer n >=0
P(0): 8^0 – 3^0 = 0 (divisible by 5)
P(1): 8^1 – 3^1 = 5 (divisible by 5)
Let P(m) is true where m <n
P(m): 8^m – 3^m is divisible by 5 for any integer n >=0
Or, P(m) =8^m-3^m= 5*k (where k is any integer)
Thus P(m+1) = 8^(m+1) – 3^(m+1) = 8*8^m – 3*3^m
= 8*(5k+3^m) – 3*3^m = 5*8k + (8-3)*3^m = 5*(8k+3^m) (divisible by 5)
Hence, P(m+1) is true when P(m) is true.
Thus by method of mathematical induction it can be shown that P(n): 8^n – 3^n is divisible
by 5 is true for any integer n>=0.
4)
Let by contradiction assume that none of the n consecutive integers are divisible by n. Now,
when divided by n then there are n-1 distinct remainders where n remainders are needed.
Thus by using the pigeonhole principle two of the divisions must have same remainders as n
numbers are assigned to n-1 remainders. However, this would imply that the difference of the
two dividends is divisible by n, but the largest difference between the n numbers is n-1. This
is contradiction and thus one of the n consecutive integers must be divisible by n.