Mathematics, 13.07.2019 03:30 Weser17
Fill in the blanks of the following proof by contradicition that 7 + 4β2 is an irrational number.
(you may use the fact that β2 is irrational.)
proof: suppose not. that is suppose that 7 + 4β2 is (i)
by definition of rational, 7 + 4β2 = a/b, where (ii) multiplying both sides by b gives 7b + 4bβ2 = a
so if we subtract 7b from both sides we have 4bβ2 = (iii)
dividing both sides by 4b gives β2 = (iv)
but then β2 would be rational number because (v) this contradicts our knowlegde that β2 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 β2 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 β2 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 β2 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 β2 would be a rational number.
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Fill in the blanks of the following proof by contradicition that 7 + 4β2 is an irrational number.
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