Fill in the blanks of the following proof by contradicition that 7 + 4√2 is an irrational number.
(you may use the fact that √2 is irrational.)
proof: suppose not. that is suppose that 7 + 4√2 is (i)
by definition of rational, 7 + 4√2 = a/b, where (ii) multiplying both sides by b gives 7b + 4b√2 = a
so if we subtract 7b from both sides we have 4b√2 = (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
( 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
( 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.