As a mathematician with a deep understanding of number theory, I can provide a comprehensive explanation to the question of whether the sum of a rational number and an irrational number is rational.

Rational Numbers are numbers that can be expressed as the quotient or fraction \( \frac{p}{q} \) of two integers, where the denominator \( q \) is not zero. This includes all integers, fractions, and finite or repeating decimals.

Irrational Numbers, on the other hand, are numbers that cannot be expressed as simple fractions. They have non-repeating, non-terminating decimal expansions. Examples include the mathematical constants \( \pi \), \( e \), and the square root of a non-perfect square, such as \( \sqrt{2} \).

Now, let's delve into the question at hand: Is the sum of a rational number and an irrational number rational?

To approach this, we can use a proof by contradiction, which is a common method in mathematics to establish the truth of a statement. The idea is to assume the opposite of what we want to prove and show that this assumption leads to a contradiction.

Assumption: Let's assume that the sum of a rational number \( r \) and an irrational number \( i \) is rational. This means there exists some rational number \( s \) such that \( r + i = s \), where \( r \) is rational and \( i \) is irrational.

Contradiction: If \( s \) is rational, then it can be expressed as a fraction \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \). Substituting \( s \) into our equation, we get \( r + i = \frac{a}{b} \). Since \( r \) is rational, it can also be expressed as a fraction \( \frac{c}{d} \), where \( c \) and \( d \) are integers and \( d \neq 0 \). Rearranging the terms, we get \( i = \frac{a}{b} - r = \frac{a}{b} - \frac{c}{d} \).

To combine these fractions, we find a common denominator, which is \( bd \), and we get:
\[ i = \frac{ad}{bd} - \frac{bc}{bd} = \frac{ad - bc}{bd} \]

This expression suggests that \( i \) can be written as a fraction, which contradicts our initial assumption that \( i \) is irrational. Since we have arrived at a contradiction by assuming that the sum \( s \) is rational, we must conclude that our assumption is false.

Conclusion: The sum of a rational number and an irrational number cannot be rational; it must be irrational.

This conclusion is consistent with the reference content provided, which also concludes that the assumption of a rational sum leads to a contradiction, thereby proving the sum to be irrational.

Now, let's proceed with the next step as instructed.

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Each time they assume the sum is rational; however, upon rearranging the terms of their equation, they get a contradiction (that an irrational number is equal to a rational number). Since the assumption that the sum of a rational and irrational number is rational leads to a contradiction, the sum must be irrational.read more >>