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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
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
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
He’s correct; it is impossible to prove in any way.
He’s correct; it is impossible to prove in any way.
by using a proof by contradiction to show multiplication is a closed operation for rational numbers
Responses
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
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
He’s correct; it is impossible to prove in any way.
He’s correct; it is impossible to prove in any way.
by using a proof by contradiction to show multiplication is a closed operation for rational numbers
Answers
Answered by
https://askanewquestion.com/
so what's the answer
Answered by
GPT-4o mini
The best response to Ryiaz's claim is:
**by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.**
To elaborate, we can present a proof demonstrating that the product of a non-zero rational number and an irrational number is always irrational. Here's a simple outline of such a proof:
1. **Definition of Rational and Irrational Numbers**: A rational number can be expressed as the fraction \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \). An irrational number cannot be expressed in this form.
2. **Let’s Assume**: Let \( r = \frac{a}{b} \) be a rational number (where \( a \) and \( b \) are integers and \( b \neq 0 \)) and let \( x \) be an irrational number. We want to prove that the product \( rx \) is irrational.
3. **Proof by Contradiction**: Suppose that \( rx \) is rational. Then, we can express \( rx \) as \( \frac{m}{n} \) for some integers \( m \) and \( n \) (where \( n \neq 0 \)).
4. **Rearranging the Equation**: From our assumption, we have:
\[
rx = \frac{m}{n} \implies x = \frac{m}{rn}
\]
Here, \( rn \) is also rational since the product of two rational numbers (r and n) is rational.
5. **Conclusion**: Since \( x \) can be expressed as \( \frac{m}{rn} \), this would mean that \( x \) is rational, which contradicts our assumption that \( x \) is irrational.
Therefore, \( rx \) must be irrational. This direct proof shows that multiplying a rational number by a non-zero irrational number results in an irrational number.
**by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.**
To elaborate, we can present a proof demonstrating that the product of a non-zero rational number and an irrational number is always irrational. Here's a simple outline of such a proof:
1. **Definition of Rational and Irrational Numbers**: A rational number can be expressed as the fraction \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \). An irrational number cannot be expressed in this form.
2. **Let’s Assume**: Let \( r = \frac{a}{b} \) be a rational number (where \( a \) and \( b \) are integers and \( b \neq 0 \)) and let \( x \) be an irrational number. We want to prove that the product \( rx \) is irrational.
3. **Proof by Contradiction**: Suppose that \( rx \) is rational. Then, we can express \( rx \) as \( \frac{m}{n} \) for some integers \( m \) and \( n \) (where \( n \neq 0 \)).
4. **Rearranging the Equation**: From our assumption, we have:
\[
rx = \frac{m}{n} \implies x = \frac{m}{rn}
\]
Here, \( rn \) is also rational since the product of two rational numbers (r and n) is rational.
5. **Conclusion**: Since \( x \) can be expressed as \( \frac{m}{rn} \), this would mean that \( x \) is rational, which contradicts our assumption that \( x \) is irrational.
Therefore, \( rx \) must be irrational. This direct proof shows that multiplying a rational number by a non-zero irrational number results in an irrational number.