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 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 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 choosing random numbers to plug into multiplication problems until he’s convinced
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The correct response is: by using a proof by contradiction to show multiplication is a closed operation for rational numbers.
The correct answer is: by using a proof by contradiction to show multiplication is a closed operation for rational numbers.
To prove that a rational number multiplied by an irrational number gives an irrational product, we can assume the opposite (which is a rational product), and then show that it leads to a contradiction. This is known as a proof by contradiction.
Here's how the proof would go:
1. Assume that there exists a rational number (let's call it "a") and an irrational number (let's call it "b") such that their product is a rational number (let's call it "c").
2. Now, since a is a rational number, it can be expressed as a ratio of two integers, say a = p/q, where p and q are integers and q is not equal to zero.
3. Similarly, since b is an irrational number, it cannot be expressed as a ratio of two integers.
4. Multiply a by b: c = a * b = (p/q) * b.
5. Since c is a rational number, it can also be expressed as a ratio of two integers, say c = r/s, where r and s are integers and s is not equal to zero.
6. Now, let's equate the two expressions for c: (p/q) * b = r/s.
7. Rearrange the equation: b = (r/s) * (q/p) = (r * q) / (s * p).
8. Since r, q, s, and p are all integers, (r * q) and (s * p) are also integers.
9. Therefore, this implies that b (which is an irrational number) can also be expressed as a ratio of two integers (r * q) and (s * p), which contradicts the definition of an irrational number.
10. Thus, our assumption that the product of a rational number and an irrational number is rational leads to a contradiction.
Therefore, we can conclude that a rational number multiplied by an irrational number always results in an irrational product.