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)

Question

Responses

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.

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.

Answer 1

Mia's statement is incorrect. The correct response is:

No, some radicals simplify to rational numbers.

For example, the square root of 4 is 2, which is a rational number. Therefore, not all sums involving radicals are necessarily irrational.

Answer 2

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

a whole number
a whole number

a rational number
a rational number

an irrational number
an irrational number

an integer

Answer 3

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

1. \( \sqrt{19} \) is an irrational number because 19 is not a perfect square.
2. Subtracting 32, which is an integer, from an irrational number will still result in an irrational number.

Therefore, the type of number that will result from the expression \( \sqrt{19} - 32 \) is:

an irrational number.

Answer 4

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

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

Answer 5

To address Ryiaz's claim, the correct response is:

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

Here's a brief outline of how this can be proven:

1. Let \( r \) be a rational number (which can be written as \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \)) and let \( s \) be an irrational number.
2. Assume for the sake of contradiction that the product \( r \times s \) is rational. This would mean that there exists a rational number \( q \) such that \( r \times s = q \).
3. Then, rearranging gives us \( s = \frac{q}{r} \). Since \( r \) is rational and \( q \) is also rational (as both are expressed as fractions of integers), \( \frac{q}{r} \) would also be a rational number.
4. This contradicts the assumption that \( s \) is irrational.

Thus, this contradiction shows that the product of a rational number and an irrational number must be irrational.

Answer 6

Find the product of 12−−√ and 5/6. What type of number is it?(1 point)
Responses

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

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

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

2.6307 . . . ; an irrational number

Answer 7

To find the product of \( \sqrt{12} \) and \( \frac{5}{6} \), we can calculate as follows:

1. First simplify \( \sqrt{12} \): \[ \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3} \]

2. Now, multiply this by \( \frac{5}{6} \): \[ \frac{5}{6} \times \sqrt{12} = \frac{5}{6} \times 2\sqrt{3} = \frac{10\sqrt{3}}{6} = \frac{5\sqrt{3}}{3} \]

Next, we need to determine the type of number. Since \( \sqrt{3} \) is an irrational number, any non-zero rational number (like \( \frac{5}{3} \)) multiplied by an irrational number results in an irrational number.

Thus, the product \( \frac{5\sqrt{3}}{3} \) is an irrational number.

To find a decimal approximation, we can calculate: \[ \sqrt{3} \approx 1.732 \] \[ \frac{5\sqrt{3}}{3} \approx \frac{5 \times 1.732}{3} \approx 2.8867 \]

Thus, the answer is:

2.8867 . . . ; an irrational number.