Step 1. subtract, writing a as a fraction
step 2. for a contradiction, assume that t is rational, and write r and t as fractions.
step 3 realize a contradiction.
step 4. find a common denominator
A B C or D
A. step 1, step 3, step 2, step 4
B.step 3, step 4, step 2, step 1
C. step 4, step 2, step 1, step 3
D. step 2, step 4, step 1, step 3
9 answers
To prove that the sum \( t = r + s \) is irrational if \( r \) is rational and \( s \) is irrational, the correct sequence of steps is as follows:
Thus, the correct sequence is:
D. step 2, step 4, step 1, step 3
Let's clarify the proof that \( t = r + s \) is irrational if \( r \) is rational and \( s \) is irrational.
Given this breakdown, the correct order of steps is indeed:
D. step 2, step 4, step 1, step 3.
I apologize for any confusion in my previous response, but my answer about the correct order is accurate. Thank you for your patience!
The product of \(\pi\) (which is an irrational number) and \(\frac{3}{4}\) (which is a rational number) will also be an irrational number.
When you multiply an irrational number by a non-zero rational number, the result is always irrational. Therefore, the product \(\frac{3}{4} \times \pi\) is irrational.
The expression \( \frac{47}{3} + \sqrt{121} \) will produce a rational number because \( \frac{47}{3} \) is rational and \( \sqrt{121} = 11 \) is also rational. The sum of two rational numbers is always rational, so their total will be rational.
To prove that the product of the two rational numbers \( \sqrt{36} \) and \( \frac{5}{3} \) is rational, we first calculate the product:
\[ \sqrt{36} = 6 \quad (\text{which is rational}) \] Now, multiplying by \( \frac{5}{3} \):
\[ 6 \times \frac{5}{3} = \frac{30}{3} = 10 \]
The product is \( 10 \), which is a rational number. A number is rational if it can be expressed as a fraction of two integers, and here \( 10 \) can be expressed as \( \frac{10}{1} \). Thus, the product is confirmed to be rational.
