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Ryiaz claims it is impossible to prove that a rational number multiplied by an irrational number gives an irrational product. H...Question
Ryiaz claims it is impossible to prove that a rational number multiplied by an irrational number gives an irrational product. How can you prove this?(1 point)
Responses
He’s correct; it is impossible to prove in any way.
He’s correct; it is impossible to prove in any way.
by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer
by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer
by using a proof by contradiction to show multiplication is a closed operation for rational numbers
by using a proof by contradiction to show multiplication is a closed operation for rational numbers
by choosing random numbers to plug into multiplication problems until he’s convinced
by choosing random numbers to plug into multiplication problems until he’s convinced
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Responses
He’s correct; it is impossible to prove in any way.
He’s correct; it is impossible to prove in any way.
by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer
by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer
by using a proof by contradiction to show multiplication is a closed operation for rational numbers
by using a proof by contradiction to show multiplication is a closed operation for rational numbers
by choosing random numbers to plug into multiplication problems until he’s convinced
by choosing random numbers to plug into multiplication problems until he’s convinced
Skip to navigation
Answers
Answered by
Lyly
CORRECT ANSWER IS: By using a proof by contradiction to show multiplication is a closed operation for rational numbers
Answered by
GPT-4o mini
To prove that a rational number multiplied by an irrational number yields an irrational product, we can adopt a direct proof approach.
### Proof:
1. **Definition of Rational and Irrational Numbers**:
- A number \( r \) is rational if it can be expressed as the fraction of two integers \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers, and \( b \neq 0 \).
- A number \( x \) is irrational if it cannot be expressed as a fraction of two integers.
2. **Assumption**:
- Let \( r \) be a rational number, expressed as \( \frac{a}{b} \).
- Let \( x \) be an irrational number.
3. **Multiplication**:
- We want to consider the product \( r \times x \). This can be represented as:
\[
r \times x = \frac{a}{b} \times x = \frac{ax}{b}
\]
- Here, \( ax \) represents the product of the rational number \( a \) and the irrational number \( x \).
4. **Proof by Contradiction**:
- Assume, for the sake of contradiction, that \( r \times x \) is rational. Hence, we can write:
\[
r \times x = \frac{c}{d}
\]
for some integers \( c \) and \( d \) (with \( d \neq 0 \)).
- Rearranging gives:
\[
ax = \frac{bc}{d}
\]
- Since \( \frac{bc}{d} \) is rational (product of rational numbers \( b \) and \( c \), and \( d \neq 0 \)), we see that \( ax \) must also be rational.
5. **Considering \( a \) is non-zero**:
- If \( a \neq 0 \), we can divide both sides of the equation \( ax = \frac{bc}{d} \) by \( a \):
\[
x = \frac{bc}{ad}
\]
- The right side is expressed as a fraction of integers (since \( b, c, a, d \) are integers), implying that \( x \) is rational.
6. **Conclusion**:
- This contradicts our initial assumption that \( x \) is irrational. Therefore, our assumption that \( r \times x \) is rational must be false.
Thus, we can conclude that the product of a rational number and an irrational number is indeed irrational.
### Answer:
The correct response is:
- **by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer**.
### Proof:
1. **Definition of Rational and Irrational Numbers**:
- A number \( r \) is rational if it can be expressed as the fraction of two integers \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers, and \( b \neq 0 \).
- A number \( x \) is irrational if it cannot be expressed as a fraction of two integers.
2. **Assumption**:
- Let \( r \) be a rational number, expressed as \( \frac{a}{b} \).
- Let \( x \) be an irrational number.
3. **Multiplication**:
- We want to consider the product \( r \times x \). This can be represented as:
\[
r \times x = \frac{a}{b} \times x = \frac{ax}{b}
\]
- Here, \( ax \) represents the product of the rational number \( a \) and the irrational number \( x \).
4. **Proof by Contradiction**:
- Assume, for the sake of contradiction, that \( r \times x \) is rational. Hence, we can write:
\[
r \times x = \frac{c}{d}
\]
for some integers \( c \) and \( d \) (with \( d \neq 0 \)).
- Rearranging gives:
\[
ax = \frac{bc}{d}
\]
- Since \( \frac{bc}{d} \) is rational (product of rational numbers \( b \) and \( c \), and \( d \neq 0 \)), we see that \( ax \) must also be rational.
5. **Considering \( a \) is non-zero**:
- If \( a \neq 0 \), we can divide both sides of the equation \( ax = \frac{bc}{d} \) by \( a \):
\[
x = \frac{bc}{ad}
\]
- The right side is expressed as a fraction of integers (since \( b, c, a, d \) are integers), implying that \( x \) is rational.
6. **Conclusion**:
- This contradicts our initial assumption that \( x \) is irrational. Therefore, our assumption that \( r \times x \) is rational must be false.
Thus, we can conclude that the product of a rational number and an irrational number is indeed irrational.
### Answer:
The correct response is:
- **by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer**.
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