Mathematics
Mathematics, 25.11.2019 00:31, anzeller6363

The rational number 2 2/9

is equal to

answer
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Mathematics, 24.06.2019 04:00, itsme77
Try to prove the hypothesis from part b that the sum of a rational and an irrational number is an irrational number. let’s say is x an irrational number and y is a rational number. the rational number y can be written as y=a/b , where a and b are integers and b does not equal 0. leave the irrational number x as x since it can’t be written as the ratio of two integers. prove the hypothesis using a proof by contradiction. in other words, try to show that x+y equals a rational number instead of an irrational number. let the sum equal m/n , where m and n are integers and n does not equal 0. plug in the values to get this equation: x+a/b=m/n now, solve the equation for x .
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Mathematics, 13.07.2019 03:30, Weser17
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.
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Mathematics, 26.08.2019 07:30, yeehaw777
50) let m be a rational number and n be an irrational number. (m)(n) = c, assume c is rational mn m = c m n = c m , so n is rational, contradicting the initial assumption. the proof shows that the product of an irrational and rational number is explain. a) rational. since an irrational number cannot equal a rational number. b) irrational. since an irrational number cannot equal a rational number. c) rational. since you can write it as the division of two rational numbers. d) irrational. since you can write it as the division of two rational numbers.
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The rational number 2 2/9

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