Suppose <msqrt> 2 </msqrt> is an irrational number and 0 is a rational number. Because

Kazeljkaml5n9y

Kazeljkaml5n9y

Answered question

2022-05-17

Suppose 2 is an irrational number and 0 is a rational number. Because
2 = 2 + 0
the sum of an irrational number and a rational number is an irrational number.
Is this correct?

Answer & Explanation

kazneni3tr2b

kazneni3tr2b

Beginner2022-05-18Added 17 answers

If you have two rational numbers, their difference (one minus the other) must also be rational. This can be proved easily by reduction to the definition of rationality, since a difference between two ratios of integers can always be written as a single ratio of integers. From this, it follows that it is impossible to get a rational number as the sum of a rational and an irrational; if this were possible, you could re-arrange this to get an irrational as the difference between two rationals.

Theorem: If r is rational and z + r is rational then z must be rational.
Proof: Since r is rational, this means there are integers a and b 0 such that r = a / b. Since z + r is also rational, this means there are integers c and d 0 such that z + r = c / d. Hence, we have:
z = z + r r = a b c d = a d b c b d .
Since a , b , c , d are all integers (and b 0, d 0), the numerator a d b c is an integer, and the denominator b d 0 is a non-zero integer. Hence, z can be written as a ratio of integers, and so z is rational.

Corollary: If z is irrational and r is rational then z + r is irrational.
Proof: Follows trivially from the above theorem using a proof-by-contradiction.

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