Is the sum of an irrational number and a rational?
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Ruby Flores
Studied at University of Sydney, Lives in Sydney, Australia
As a field expert in mathematics, I specialize in the study of numbers and their properties. In particular, the distinction between rational and irrational numbers is a fundamental concept in the realm of real numbers. Rational numbers are those that can be expressed as the quotient or fraction \( \frac{p}{q} \) where \( p \) and \( q \) are integers, and \( q \neq 0 \). On the other hand, irrational numbers are numbers that cannot be expressed as a simple fraction; they have non-repeating, non-terminating decimal expansions, such as \( \sqrt{2} \), \( \pi \), and \( e \) (Euler's number).
Now, let's delve into the question at hand: Is the sum of an irrational number and a rational number rational or irrational?
To address this, we can consider a proof by contradiction. Suppose we have an irrational number \( x \) and a rational number \( y \). If we were to assume that their sum \( x + y \) is rational, we would be contradicting the established properties of irrational numbers.
Proof by Contradiction:
1. Assumption: Let \( x \) be an irrational number and \( y \) be a rational number. Assume that their sum \( x + y \) is rational.
2. Contradiction: Since \( y \) is rational, it can be expressed as a fraction \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \). If \( x + y \) is also rational, it can be expressed as a fraction \( \frac{c}{d} \), where \( c \) and \( d \) are integers and \( d \neq 0 \).
3. Rearrangement: Adding \( y \) to both sides of the equation \( x + y = \frac{c}{d} \) gives us \( x = \frac{c}{d} - y \). Substituting \( y \) with \( \frac{a}{b} \), we get \( x = \frac{c}{d} - \frac{a}{b} \).
4. Simplification: To combine the fractions on the right side, we find a common denominator, which would be \( bd \), and we get \( x = \frac{bc - ad}{bd} \). Since \( a \), \( b \), \( c \), and \( d \) are all integers, \( bc - ad \) is also an integer, say \( e \), making \( x = \frac{e}{bd} \).
5. Conclusion: This implies that \( x \) can be expressed as a fraction \( \frac{e}{bd} \), which contradicts our initial assumption that \( x \) is an irrational number. If \( x \) were rational, it would have a repeating or terminating decimal expansion, which is not the case for irrational numbers.
From this contradiction, we conclude that our initial assumption was false. Therefore, the sum of an irrational number and a rational number cannot be rational; it must be irrational.
This conclusion is consistent with the properties of real numbers and the nature of irrational numbers, which are inherently non-divisible into simpler fractions without resulting in a contradiction.
Now, let's delve into the question at hand: Is the sum of an irrational number and a rational number rational or irrational?
To address this, we can consider a proof by contradiction. Suppose we have an irrational number \( x \) and a rational number \( y \). If we were to assume that their sum \( x + y \) is rational, we would be contradicting the established properties of irrational numbers.
Proof by Contradiction:
1. Assumption: Let \( x \) be an irrational number and \( y \) be a rational number. Assume that their sum \( x + y \) is rational.
2. Contradiction: Since \( y \) is rational, it can be expressed as a fraction \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \). If \( x + y \) is also rational, it can be expressed as a fraction \( \frac{c}{d} \), where \( c \) and \( d \) are integers and \( d \neq 0 \).
3. Rearrangement: Adding \( y \) to both sides of the equation \( x + y = \frac{c}{d} \) gives us \( x = \frac{c}{d} - y \). Substituting \( y \) with \( \frac{a}{b} \), we get \( x = \frac{c}{d} - \frac{a}{b} \).
4. Simplification: To combine the fractions on the right side, we find a common denominator, which would be \( bd \), and we get \( x = \frac{bc - ad}{bd} \). Since \( a \), \( b \), \( c \), and \( d \) are all integers, \( bc - ad \) is also an integer, say \( e \), making \( x = \frac{e}{bd} \).
5. Conclusion: This implies that \( x \) can be expressed as a fraction \( \frac{e}{bd} \), which contradicts our initial assumption that \( x \) is an irrational number. If \( x \) were rational, it would have a repeating or terminating decimal expansion, which is not the case for irrational numbers.
From this contradiction, we conclude that our initial assumption was false. Therefore, the sum of an irrational number and a rational number cannot be rational; it must be irrational.
This conclusion is consistent with the properties of real numbers and the nature of irrational numbers, which are inherently non-divisible into simpler fractions without resulting in a contradiction.
2024-05-13 15:06:31
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Helpful(2)
Studied at Stanford University, Lives in Palo Alto, CA
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. 1.
2023-06-10 22:31:56
Harper Cooper
QuesHub.com delivers expert answers and knowledge to you.
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. 1.
