Fill in the blanks of the following proof by contradicition that 7 + 42 is an ir

Question

Fill in the blanks of the following proof by contradicition that 7 + 42 is an irrational number.

(You may use the fact that 2 is irrational.)

Proof: Suppose not. That is suppose that 7 + 42 is (i) ___.

By definition of rational, 7 + 42 = a/b, where (ii) ___. Multiplying both sides by b gives 7b + 4b2 = a

so if we subtract 7b from both sides we have 4b2 = (iii) ____

Dividing both sides by 4b gives 2 = (iv) ____.

But then 2 would be rational number because (v) ___. This contradicts our knowlegde that 2 is irrational. Hence the supposition is false and the given statement is true.

A. (i). rational
(ii). A and B are integers and b cannot equal 0
(iii). a+7b
(iv). (a+7b)/4b
(v). both a-7b and 4b are integers (since products and differences of interger are integers) and so 2 would be a rational number.

B. (i). irrational
(ii). A and B are integers and b cannot equal 0
(iii). a-7b
(iv). (a-7b)/4b
(v). both a-7b and 4b are integers (since products and differences of interger are integers) and so 2 would be a irrational number.

C. (i). irrational
(ii). A and B are integers and b cannot equal 0
(iii). a-7b
(iv).( a-7b)/4b
(v). both a-7b and 4b are integers (since products and differences of interger are integers) and so 2 would be a rational number.
D. B. (i). rational
(ii). A and B are integers and b cannot equal 0
(iii). a-7b
(iv).( a-7b)/4b
(v). both a-7b and 4b are integers (since products and differences of interger are integers) and so 2 would be a rational number.

Explanation / Answer

(i) Rational

(ii) A and B are integers and b cannot equal 0

(iii) a - 7b

(iv) (a - 7b)/4b

(v) both a-7b and 4b are integers (since products and differences of interger are integers) and so 2 would be a rational number

Thus, D option

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