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How is the sum of a rational and irrational number irrational? - Answers

This can easily be proved by contradiction. Without loss of generality, I will take specific numbers as an example. The proof can easily be extended to any rational + irrational number. Assumption: 1 plus the square root of 2 is rational. (It is a well-known fact that the square root of 2 is irrational. No need to prove it here; you can use any other irrational number will do.) This rational sum can be written as p / q, where "p" and "q" are whole numbers (this is basically the definition of a "rational number"). Then, the square root of 2, which is equal to the sum minus 1, is: p / q - 1 = p / q - q / q = (p - q) / q Since the difference of two whole numbers is a whole number, this makes the square root of 2 rational, which doesn't make sense.



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How is the sum of a rational and irrational number irrational? - Answers

https://math.answers.com/math-and-arithmetic/How_is_the_sum_of_a_rational_and_irrational_number_irrational

This can easily be proved by contradiction. Without loss of generality, I will take specific numbers as an example. The proof can easily be extended to any rational + irrational number. Assumption: 1 plus the square root of 2 is rational. (It is a well-known fact that the square root of 2 is irrational. No need to prove it here; you can use any other irrational number will do.) This rational sum can be written as p / q, where "p" and "q" are whole numbers (this is basically the definition of a "rational number"). Then, the square root of 2, which is equal to the sum minus 1, is: p / q - 1 = p / q - q / q = (p - q) / q Since the difference of two whole numbers is a whole number, this makes the square root of 2 rational, which doesn't make sense.



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https://math.answers.com/math-and-arithmetic/How_is_the_sum_of_a_rational_and_irrational_number_irrational

How is the sum of a rational and irrational number irrational? - Answers

This can easily be proved by contradiction. Without loss of generality, I will take specific numbers as an example. The proof can easily be extended to any rational + irrational number. Assumption: 1 plus the square root of 2 is rational. (It is a well-known fact that the square root of 2 is irrational. No need to prove it here; you can use any other irrational number will do.) This rational sum can be written as p / q, where "p" and "q" are whole numbers (this is basically the definition of a "rational number"). Then, the square root of 2, which is equal to the sum minus 1, is: p / q - 1 = p / q - q / q = (p - q) / q Since the difference of two whole numbers is a whole number, this makes the square root of 2 rational, which doesn't make sense.

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      This can easily be proved by contradiction. Without loss of generality, I will take specific numbers as an example. The proof can easily be extended to any rational + irrational number. Assumption: 1 plus the square root of 2 is rational. (It is a well-known fact that the square root of 2 is irrational. No need to prove it here; you can use any other irrational number will do.) This rational sum can be written as p / q, where "p" and "q" are whole numbers (this is basically the definition of a "rational number"). Then, the square root of 2, which is equal to the sum minus 1, is: p / q - 1 = p / q - q / q = (p - q) / q Since the difference of two whole numbers is a whole number, this makes the square root of 2 rational, which doesn't make sense.
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