How does the Closure Property prove that the sum of a rational and irrational number is irrational?(1 point)

Question

Responses

The sum of an irrational and rational number can’t be predicted.
The sum of an irrational and rational number can’t be predicted.

Irrational numbers can be added or subtracted.
Irrational numbers can be added or subtracted.

The sum or difference of two rational numbers is a rational number.
The sum or difference of two rational numbers is a rational number.

The sum or difference of a rational and irrational number is a rational number.
The sum or difference of a rational and irrational number is a rational number.

Answer 1

The correct response is:

The sum or difference of two rational numbers is a rational number.

This statement supports the understanding of the closure property in rational numbers. To prove that the sum of a rational number and an irrational number is irrational, consider the following:

1. Let \( r \) be a rational number and \( i \) be an irrational number.

2. Assume that their sum \( s = r + i \) is rational.

3. If \( s \) is rational, then we can rearrange this equation to find \( i = s - r \).

4. However, since \( r \) is rational, and the sum or difference of two rational numbers (in this case, \( s \) and \( r \)) is also rational, it would imply that \( i \) is rational as well.

5. This is a contradiction because \( i \) is defined as irrational.

Thus, the assumption that \( s \) (the sum of a rational and irrational number) is rational leads to a contradiction, meaning the sum must be irrational. This aligns with the closure property that the sum or difference of a rational and irrational number is NOT a rational number.

Answer 2

Mia says that anytime there is a radical in an addition expression, it automatically has an irrational number as the sum. Is she correct?(1 point)
Responses

Yes, radicals can’t be simplified, so they must be irrational.
Yes, radicals can’t be simplified, so they must be irrational.

No, some radicals simplify to rational numbers.
No, some radicals simplify to rational numbers.

No, radicals always simplify to integers, so they are rational.
No, radicals always simplify to integers, so they are rational.

Yes, radicals are irrational.
Yes, radicals are irrational.

Answer 3

The correct response is:

No, some radicals simplify to rational numbers.

Mia's statement is not accurate because while many radicals (like \(\sqrt{2}\) or \(\sqrt{3}\)) are indeed irrational, there are also radicals that simplify to rational numbers. For example, \(\sqrt{4} = 2\) and \(\sqrt{16} = 4\) are both rational numbers. Therefore, a radical in an addition expression does not automatically imply that the sum will be irrational.

Answer 4

What type of number will result from the expression 19−−√−32 ?(1 point)
Responses

an irrational number
an irrational number

a whole number
a whole number

an integer
an integer

a rational number

Answer 5

To evaluate the expression \(\sqrt{19} - 32\):

1. \(\sqrt{19}\) is an irrational number because it cannot be expressed as a fraction of two integers.
2. Subtracting the integer 32 from an irrational number (i.e., \(\sqrt{19} - 32\)) results in another irrational number.

Therefore, the correct response is:

an irrational number.

Answer 6

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 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
by choosing random numbers to plug into multiplication problems until he’s convinced

He’s correct; it is impossible to prove in any way.

Answer 7

The correct response is:

by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.

To demonstrate this, consider the following proof:

1. Let \( r \) be a rational number and \( i \) be an irrational number.
2. Assume for contradiction that the product \( p = r \cdot i \) is a rational number.
3. Since \( r \) is rational, we can express \( r \) as \( \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0\).
4. From our assumption that \( p \) is rational, we can rearrange to get \( i = \frac{p}{r} \).
5. Since \( p \) is rational and \( r \) is rational (and non-zero), the fraction \( \frac{p}{r} \) would also be rational.

This creates a contradiction because \( i \) was assumed to be irrational. Therefore, the assumption that the product \( p = r \cdot i \) is rational must be false, implying that the product of a rational number and an irrational number is indeed irrational.

Answer 8

Find the product of 12−−√ and 56 . What type of number is it?(1 point)
Responses

2.8867 . . . ; an irrational number
2.8867 . . . ; an irrational number

4.1569 . . . ; an irrational number
4.1569 . . . ; an irrational number

4.2974 . . . ; an irrational number
4.2974 . . . ; an irrational number

2.6307 . . . ; an irrational number
2.6307 . . . ; an irrational number

Answer 9

To find the product of \(\sqrt{12}\) and \(56\), we can calculate it as follows:

\[ \sqrt{12} \cdot 56 \]

First, we simplify \(\sqrt{12}\):

\[ \sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3} \]

Now, we can calculate the product:

\[ 2\sqrt{3} \cdot 56 = 112\sqrt{3} \]

Since \(\sqrt{3}\) is an irrational number, the product \(112\sqrt{3}\) is also irrational.

Now, if we want to calculate an approximate numerical value:

\[ \sqrt{3} \approx 1.732 \]

Thus,

\[ 112\sqrt{3} \approx 112 \cdot 1.732 \approx 193.184 \]

So, we find that the product is approximately \(193.184\).

However, we need to check the provided responses. None of them directly match this result or provide a clear number that we calculated from \(\sqrt{12} \cdot 56\).

Since the exact decimal approximation isn't one of the options given, we conclude:

1. The type of number is an irrational number, since it includes the term \(\sqrt{3}\).

The exact value is irrational, but based on the context of your question, if you're looking for the type alone, the final answer is:

112\sqrt{3}; an irrational number.

Unfortunately, the provided response options do not align with this answer.