(1 point) Let r,s
 be rational numbers.

Prove that 2r2+3rs+5s2
 is rational.

Solution.
Since r
 is rational, then
r=AB,
where A,B
 are 
integers
 ,
(for the preceding answer, see the definition of rational numbers on page 26 of the presentation, where a rational number is defined to be a quotient of these kinds of numbers)
and B≠
0
 .

Since s
 is rational, then
s=CD,
where C,D
 are 
integers
 ,
(for the preceding answer, see the definition of rational numbers on page 26 of the presentation, where a rational number is defined to be a quotient of these kinds of numbers)
and D≠
0
 .

Then, after substituting for r
 and s
 from above,
2r2+3rs+5s2=2(
A/B
 )2+3(
A/B
 )(
 
C/D
 )+5(
C/D
 )2


After separating each square of a fraction, or product of fractions, into a single numerator and single denominator,
2r2+3rs+5s2

=2(
A
 )2/(
B
 )2+3(
A/B^2
 )/(
 
C/D^2
 )+5(
C
 )2/(
D
 )2


After finding a (smallest possible) common denominator, and combining all three fractions into a single fraction,
2r2+3rs+5s2

=(2(
AB^2
 )+3(
 )
 +5(
 ))/(
 )
.

To see that 2r2+3rs+5s2
, observe that
- the numerator of the last expression, namely 
 , is an 
 ,
- the denominator of the last expression, namely 
 , is an 
 ,
- the denominator of the last expression, namely 
 ,≠
 
 (since B≠0,D≠0
).

Thus 2r2+3rs+5s2
 is rational.

Let's clean up the solution to make it clear and accurate.

We are given that \( r \) and \( s \) are rational numbers, and we are asked to prove that the expression \( 2r^2 + 3rs + 5s^2 \) is rational.

### Step-by-step solution:

1. **Represent \( r \) and \( s \) as rational numbers:**

   Since \( r \) and \( s \) are rational, we can write them as fractions of integers:
   \[
   r = \frac{A}{B} \quad \text{and} \quad s = \frac{C}{D}
   \]
   where \( A, B, C, D \) are integers and \( B \neq 0 \), \( D \neq 0 \).

2. **Substitute these values into the given expression:**

   Substitute \( r = \frac{A}{B} \) and \( s = \frac{C}{D} \) into the expression \( 2r^2 + 3rs + 5s^2 \):
   \[
   2r^2 + 3rs + 5s^2 = 2\left(\frac{A}{B}\right)^2 + 3\left(\frac{A}{B}\right)\left(\frac{C}{D}\right) + 5\left(\frac{C}{D}\right)^2
   \]

3. **Simplify each term:**
   
   We now simplify each of the terms separately:
   \[
   2\left(\frac{A}{B}\right)^2 = 2\frac{A^2}{B^2} = \frac{2A^2}{B^2}
   \]
   \[
   3\left(\frac{A}{B}\right)\left(\frac{C}{D}\right) = \frac{3AC}{BD}
   \]
   \[
   5\left(\frac{C}{D}\right)^2 = 5\frac{C^2}{D^2} = \frac{5C^2}{D^2}
   \]

4. **Find the common denominator:**

   To combine these fractions, we need a common denominator. The common denominator of \( B^2 \), \( BD \), and \( D^2 \) is \( B^2D^2 \). So we rewrite each fraction with this common denominator:
   \[
   \frac{2A^2}{B^2} = \frac{2A^2D^2}{B^2D^2}
   \]
   \[
   \frac{3AC}{BD} = \frac{3ACB}{B^2D^2}
   \]
   \[
   \frac{5C^2}{D^2} = \frac{5C^2B^2}{B^2D^2}
   \]

5. **Combine the terms:**

   Now, we can combine the terms into a single fraction:
   \[
   2r^2 + 3rs + 5s^2 = \frac{2A^2D^2 + 3ACBD + 5C^2B^2}{B^2D^2}
   \]

6. **Conclusion:**

   The numerator \( 2A^2D^2 + 3ACBD + 5C^2B^2 \) is an integer, since it is the sum of products of integers. The denominator \( B^2D^2 \) is also an integer, and it is non-zero because \( B \neq 0 \) and \( D \neq 0 \).

   Therefore, the entire expression is a rational number, because it is a fraction where both the numerator and denominator are integers, and the denominator is non-zero.

Thus, \( 2r^2 + 3rs + 5s^2 \) is rational.

### Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. Can the sum of two irrational numbers be rational? Why or why not?
2. How would you prove that the product of two rational numbers is rational?
3. What is the relationship between rational numbers and integers?
4. Can an irrational number raised to a rational power be rational?
5. What other operations (e.g., subtraction, division) preserve the rationality of numbers?

**Tip:** Rational numbers can always be written as a fraction of two integers, which is the key to proving many properties involving them!

Let r, s be rational numbers. Prove that 2r^2 + 3rs + 5s^2 is rational.

Explore a step-by-step proof showing that the expression 2r^2 + 3rs + 5s^2 is rational when r and s are rational numbers. This algebra problem covers rational numbers, polynomial expressions, and the closure property of rational numbers, ideal for students in Grades 9-12.

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