A researcher investigating number theory properties is formalising a proof for the theorem:
“For any positive integer mmm, if m2 m^2\,m2 is divisible by 7, then m m\,m must be divisible by 7.”
The initial steps of the proof by contradiction are provided below.
Assumption: There exists a positive integer m∈Z+m \in \mathbb{Z}^+m∈Z+ such that m2 m^2\,m2 is a multiple of 7, but m m\,m is NOT a multiple of 7.
Case 1: Let m=7k+1m = 7k + 1m=7k+1 for some integer k≥0k \ge 0k≥0.
m2=(7k+1)2=49k2+14k+1=7(7k2+2k)+1 m^2 = (7k + 1)^2 = 49k^2 + 14k + 1 = 7(7k^2 + 2k) + 1 m2=(7k+1)2=49k2+14k+1=7(7k2+2k)+1which is not a multiple of 7.
Case 2: Let m=7k+2m = 7k + 2m=7k+2 for some integer k≥0k \ge 0k≥0.
m2=(7k+2)2=49k2+28k+4=7(7k2+4k)+4 m^2 = (7k + 2)^2 = 49k^2 + 28k + 4 = 7(7k^2 + 4k) + 4 m2=(7k+2)2=49k2+28k+4=7(7k2+4k)+4which is not a multiple of 7.
Case 3: Let m=7k+3m = 7k + 3m=7k+3 for some integer k≥0k \ge 0k≥0.
m2=(7k+3)2=49k2+42k+9=7(7k2+6k+1)+2 m^2 = (7k + 3)^2 = 49k^2 + 42k + 9 = 7(7k^2 + 6k + 1) + 2 m2=(7k+3)2=49k2+42k+9=7(7k2+6k+1)+2which is not a multiple of 7.
Show the calculations and concluding logic required to complete this part of the proof by exhaustion.
Hence prove, by contradiction, that 7 \sqrt{7}\,7 is an irrational number.
107 exam-style questions on Edexcel A Level Maths 1.1 Proof by Contradiction. Each one has a worked solution and a mark scheme showing where the marks go.