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 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 by using a proof by contradiction to show multiplication is a closed operation for rational numbers

Question

GPT-4o mini · Answer

The correct choice is: **by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer**. ### Proof Let’s denote a rational number as $ r $ and an irrational number as $ x $. By the definition of rational numbers, we can write $ r $ as $ \frac{a}{b} $, where $ a $ and $ b $ are integers and $ b \neq 0 $. We want to show that the product $ rx $ is irrational. For the sake of contradiction, assume that $ rx $ is rational. This means that we can express $ rx $ as a fraction of integers $ \frac{c}{d} $, where $ c $ and $ d $ are integers and $ d \neq 0 $. From our assumption $ rx = \frac{c}{d} $, we can rearrange this to solve for $ x $: $$ x = \frac{rx}{r} = \frac{\frac{c}{d}}{r} = \frac{c}{dr} $$ Since $ r $ is rational (as $ r = \frac{a}{b} $), $ dr $ is also rational (since $ d $ is an integer). Thus, $ x $ can be expressed as a fraction of two integers $ \frac{c}{dr} $, which means $ x $ would also be rational. However, this is a contradiction because we assumed $ x $ is irrational. Therefore, our initial assumption that $ rx $ is rational must be false. Hence, we conclude that the product $ rx $ must be irrational. This proof demonstrates that the product of a non-zero rational number and an irrational number is indeed irrational.

Answer

this does not seem right

GPT-4o mini · Answer

I apologize for any confusion in my explanation. Let's clarify the proof regarding the multiplication of a rational number by an irrational number.

Claim:

The product of a non-zero rational number and an irrational number is irrational.

Proof:

Let $ r $ be a non-zero rational number. By definition, we can express $ r $ as $ r = \frac{a}{b} $, where $ a $ and $ b $ are integers, and $ b \neq 0 $.
Let $ x $ be an irrational number.
We want to show that the product $ rx $ is irrational.
Suppose, for the sake of contradiction, that the product $ rx $ is rational. This means we can express it as $ rx = \frac{c}{d} $, where $ c $ and $ d $ are integers, and $ d \neq 0 $.
From the equation $ rx = \frac{c}{d} $, we can solve for $ x $: \[ x = \frac{rx}{r} = \frac{\frac{c}{d}}{r} = \frac{c}{d} \cdot \frac{b}{a} = \frac{bc}{ad} \]
Since $ a, b, c, $ and $ d $ are all integers (and $ ad \neq 0 $), the expression $ \frac{bc}{ad} $ is a ratio of two integers, hence a rational number.
This leads us to conclude that $ x $ is rational, which contradicts our initial assumption that $ x $ is irrational.

Conclusion:

Thus, the assumption that $ rx $ (the product of the rational number $ r $ and the irrational number $ x $) is rational must be false. Therefore, the product $ rx $ is irrational.

This proof clearly shows that the multiplication of a non-zero rational number with an irrational number results in an irrational product. Thank you for your patience, and I hope this clearer explanation is helpful!