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Fill in the Blanks of the Following Proof by Contradiction 7+427 + 4 \sqrt { 2 }

Question 17

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Fill in the blanks of the following proof by contradiction that 7+427 + 4 \sqrt { 2 } is an irrational number. (You may use the fact that 2\sqrt { 2 } is irrational.)
Proof: Suppose not. That is, suppose that 7+427 + 4 \sqrt { 2 } is (i). By definition of rational, 7+42=7 + 4 \sqrt { 2 } = ab\frac { a } { b } , where (ii). Multiplying both sides by bb gives
7b+4b2=a,7 b + 4 b \sqrt { 2 } = a ,
so if we subtract 7b7 b from both sides we have
4b2= (iii) .4 b \sqrt { 2 } = \underline { \text { (iii) } } .
Dividing both sides by 4b4 b gives
2= (iv). \sqrt { 2 } = \text { (iv). }
But then 2\sqrt { 2 } would be a rational number because (v). This contradicts our knowledge that 2\sqrt { 2 } is irrational. Hence (vi).

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(i). rational
(ii). blured image and blured image are integers a...

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