A cryptographer is analyzing a security protocol that relies on the properties of prime numbers. They are required to prove the following theorem for any positive integer nnn:
“If n2n^2n2 is a multiple of 7, then nnn must be a multiple of 7.”
The start of the proof by contradiction is shown in the box below.
Assumption: There exists an integer n∈Z+n \in \mathbb{Z}^+n∈Z+, such that n2n^2n2 is a multiple of 7, and nnn is NOT a multiple of 7.
Case 1: Let n=7k+1n = 7k + 1n=7k+1 for some integer kkk. n2=(7k+1)2=49k2+14k+1=7(7k2+2k)+1n^2 = (7k + 1)^2 = 49k^2 + 14k + 1 = 7(7k^2 + 2k) + 1n2=(7k+1)2=49k2+14k+1=7(7k2+2k)+1 which is not a multiple of 7.
Case 2: Let n=7k+2n = 7k + 2n=7k+2 for some integer kkk. n2=(7k+2)2=49k2+28k+4=7(7k2+4k)+4n^2 = (7k + 2)^2 = 49k^2 + 28k + 4 = 7(7k^2 + 4k) + 4n2=(7k+2)2=49k2+28k+4=7(7k2+4k)+4 which is not a multiple of 7.
Case 3: Let n=7k+3n = 7k + 3n=7k+3 for some integer kkk. n2=(7k+3)2=49k2+42k+9=7(7k2+6k+1)+2n^2 = (7k + 3)^2 = 49k^2 + 42k + 9 = 7(7k^2 + 6k + 1) + 2n2=(7k+3)2=49k2+42k+9=7(7k2+6k+1)+2 which is not a multiple of 7.
Case 4: Let n=7k+4n = 7k + 4n=7k+4 for some integer kkk. n2=(7k+4)2=49k2+56k+16=7(7k2+8k+2)+2n^2 = (7k + 4)^2 = 49k^2 + 56k + 16 = 7(7k^2 + 8k + 2) + 2n2=(7k+4)2=49k2+56k+16=7(7k2+8k+2)+2 which is not a multiple of 7.
Show the calculations and statements required to complete this part of the proof.
Hence prove, by contradiction, that 7\sqrt{7}7 is an irrational number.
Practise Edexcel A Level Maths 1.1 Proof by Contradiction with exam-style questions for A Level Maths. 100 questions, matched to the Edexcel A Level Maths (9MA0) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.