Question
Consider the expression 47/3 + sqrt(121) Will the sum of the two rational numbers produce a rational number? Explain your reasoning in 3-4 sentences. (2 points)
(Original and like a 9th grader wrote it)
(Original and like a 9th grader wrote it)
Answers
Answer
A bit longer please
Answer
Prove that the product of the two rational numbers sqrt(36) and 5/3 is a rational number. Find the product and explain how you know it is a rational number. Explain your reasoning in 4-5 sentences.
(Original and like a 9th grader wrote it)
(Original and like a 9th grader wrote it)
Answer
Prove that the sum of and sqrt(10) is irrational. Show your work, and in 4-5 sentences, explain why the sum is an irrational number
(Original and like a 9th grader wrote it)
(Original and like a 9th grader wrote it)
Answer
Prove that the product of 2 and is an irrational number. Find the product and explain why the value is irrational. Explain your reasoning in 4-5 sentences. (2 points)
(Original and like a 9th grader wrote it)
(Original and like a 9th grader wrote it)
Answered by
GPT-4o mini
The expression \( \frac{47}{3} + \sqrt{121} \) involves both a rational number, \( \frac{47}{3} \), and \( \sqrt{121} \). First, we simplify \( \sqrt{121} \) to get 11, which is also a rational number. When we add two rational numbers together, the result is always rational. Therefore, the sum of \( \frac{47}{3} + 11 \) will definitely produce a rational number.
Answered by
GPT-4o mini
The expression \( \frac{47}{3} + \sqrt{121} \) includes two components: \( \frac{47}{3} \), which is a rational number, and \( \sqrt{121} \). To understand the overall expression, we can simplify \( \sqrt{121} \) to 11, which is also a rational number since it can be expressed as \( \frac{11}{1} \).
When we add two rational numbers together, the result is always rational. In this case, we are adding \( \frac{47}{3} \) and 11. To ensure clarity, we can rewrite 11 with a common denominator of 3: \( 11 = \frac{33}{3} \). Now we can combine these fractions:
\[
\frac{47}{3} + \frac{33}{3} = \frac{80}{3}.
\]
Since \( \frac{80}{3} \) is in the form of a fraction where both the numerator and denominator are integers, it confirms that the sum is a rational number. Thus, the overall expression \( \frac{47}{3} + \sqrt{121} \) produces a rational number.
When we add two rational numbers together, the result is always rational. In this case, we are adding \( \frac{47}{3} \) and 11. To ensure clarity, we can rewrite 11 with a common denominator of 3: \( 11 = \frac{33}{3} \). Now we can combine these fractions:
\[
\frac{47}{3} + \frac{33}{3} = \frac{80}{3}.
\]
Since \( \frac{80}{3} \) is in the form of a fraction where both the numerator and denominator are integers, it confirms that the sum is a rational number. Thus, the overall expression \( \frac{47}{3} + \sqrt{121} \) produces a rational number.
Answered by
GPT-4o mini
To prove that the product of \( \sqrt{36} \) and \( \frac{5}{3} \) is a rational number, we first simplify \( \sqrt{36} \). Since \( 36 \) is a perfect square, \( \sqrt{36} = 6 \).
Now, we can multiply \( 6 \) and \( \frac{5}{3} \):
\[
6 \times \frac{5}{3} = \frac{6 \times 5}{3} = \frac{30}{3} = 10.
\]
The number \( 10 \) is a whole number, which can also be expressed as \( \frac{10}{1} \), meaning it is a rational number because it can be written as a fraction where both the numerator and denominator are integers. Since both \( 6 \) and \( \frac{5}{3} \) are rational numbers and their product is also rational, we conclude that the product of \( \sqrt{36} \) and \( \frac{5}{3} \) is indeed a rational number.
Now, we can multiply \( 6 \) and \( \frac{5}{3} \):
\[
6 \times \frac{5}{3} = \frac{6 \times 5}{3} = \frac{30}{3} = 10.
\]
The number \( 10 \) is a whole number, which can also be expressed as \( \frac{10}{1} \), meaning it is a rational number because it can be written as a fraction where both the numerator and denominator are integers. Since both \( 6 \) and \( \frac{5}{3} \) are rational numbers and their product is also rational, we conclude that the product of \( \sqrt{36} \) and \( \frac{5}{3} \) is indeed a rational number.
Answered by
GPT-4o mini
To prove that the sum of \( \sqrt{10} \) and a rational number, say \( r \), is irrational, we can assume that \( r \) is a rational number. A rational number can be expressed as a fraction \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
When we add this rational number to \( \sqrt{10} \), we get:
\[
\sqrt{10} + r = \sqrt{10} + \frac{a}{b}.
\]
If we assume that this sum is rational, we can rewrite it as \( \sqrt{10} = \text{(some rational number)} \). However, we know from math that \( \sqrt{10} \) is an irrational number because \( 10 \) is not a perfect square.
This means that there is no way to represent \( \sqrt{10} \) as a fraction of two integers. Since the sum of a rational number \( r \) and an irrational number \( \sqrt{10} \) cannot simplify down to a rational number, it follows that \( \sqrt{10} + r \) must also be irrational. Therefore, the addition of \( \sqrt{10} \) to any rational number results in an irrational number.
When we add this rational number to \( \sqrt{10} \), we get:
\[
\sqrt{10} + r = \sqrt{10} + \frac{a}{b}.
\]
If we assume that this sum is rational, we can rewrite it as \( \sqrt{10} = \text{(some rational number)} \). However, we know from math that \( \sqrt{10} \) is an irrational number because \( 10 \) is not a perfect square.
This means that there is no way to represent \( \sqrt{10} \) as a fraction of two integers. Since the sum of a rational number \( r \) and an irrational number \( \sqrt{10} \) cannot simplify down to a rational number, it follows that \( \sqrt{10} + r \) must also be irrational. Therefore, the addition of \( \sqrt{10} \) to any rational number results in an irrational number.
Answered by
GPT-4o mini
To clarify your question, I assume you meant to prove that the product of \( 2 \) and \( \sqrt{2} \) is an irrational number. Let's find the product first:
\[
2 \times \sqrt{2} = 2\sqrt{2}.
\]
Now, we need to show that \( 2\sqrt{2} \) is irrational. We know that \( \sqrt{2} \) is an irrational number because it cannot be expressed as a fraction of two integers (there are no integers \( p \) and \( q \) such that \( \left(\frac{p}{q}\right)^2 = 2 \)). When you multiply a non-zero rational number, like \( 2 \), by an irrational number \( \sqrt{2} \), the result is always irrational. Therefore, since \( 2\sqrt{2} \) includes the irrational part \( \sqrt{2} \), we conclude that \( 2\sqrt{2} \) is also irrational. This reasoning helps us understand that the product of a rational number and an irrational number will always yield an irrational number.
\[
2 \times \sqrt{2} = 2\sqrt{2}.
\]
Now, we need to show that \( 2\sqrt{2} \) is irrational. We know that \( \sqrt{2} \) is an irrational number because it cannot be expressed as a fraction of two integers (there are no integers \( p \) and \( q \) such that \( \left(\frac{p}{q}\right)^2 = 2 \)). When you multiply a non-zero rational number, like \( 2 \), by an irrational number \( \sqrt{2} \), the result is always irrational. Therefore, since \( 2\sqrt{2} \) includes the irrational part \( \sqrt{2} \), we conclude that \( 2\sqrt{2} \) is also irrational. This reasoning helps us understand that the product of a rational number and an irrational number will always yield an irrational number.
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