Discrete Mathmatics Contradiction & Contraposition

Question

Consider the following statement.

For every integer m, 7m + 4

 is not divisible by 7.

Construct a proof for the statement by selecting sentences from the following scrambled list and putting them in the correct order. 

Subtracting 7m from both sides of the equation gives 4 = 7k − 7m = 7(k − m).

Suppose that there is an integer m such that 7m + 4 is not divisible by 7.

But k − m is an integer and 7/4

is not an integer.

Dividing both sides of the equation by 4 results in 7/4

= k − m.

By definition of divisibility 7m + 4 = 7k for some integer k.

1. Suppose that there is an integer m such that 7m + 4 is divisible by 7.

But k − m is an integer and 4/7

is not an integer.

Dividing both sides of the equation by 7 results in 4/7

= k − m.

By definition of divisibility 4m + 7 = 4k, for some integer k.

Subtracting 4m from both sides of the equation gives 7 = 4k − 4m = 4(k − m).

Answer 1

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Suppose there is an integer m such that 7m+4 is divisible by 7.

7m+4 = 7k for integer k by definition of divisibility

4 = 7k - 7m = 7(k-m) by subtracting 7m from both sides

divides both sides by 4

1 = 7/4(k-m)

k-m is an integer by CLOSURE PROPERTY of integer subtraction

and 7/4 is not an integer.

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Prove:

f(M)= 7M +4 is not divisible by 7 for any integer M.

M=0 --> f(M) = 4 which is not divisible by 7.

M=1 --> f(M) = 11 which is not divisible by 7.

Suppose f(M) is not divisible by 7 for some positive integer M.

(this is the GIVEN induction hypothesis)

7(m+1)+4 = 7m+7+4 = 7m +4 + 7 = (7m+4)+7

dividing by 7, the quotient is [(7m+4)+7]/7

= (7m+4)/7 + 1

which is NOT divisible by 7 per induction hypothesis...