For all numbers n, the symbol [n] denotes the greatest integer less than or equal to n. What is the...

1. Translate the problem requirements: The symbol \(\lfloor n \rfloor\) means the "floor function" - find the largest integer that is less than or equal to n. We need to find \(\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{5} \rfloor + \lfloor \sqrt{7} \rfloor\), which means finding the greatest integer ≤ \(\sqrt{2}\), plus the greatest integer ≤ \(\sqrt{5}\), plus the greatest integer ≤ \(\sqrt{7}\).
2. Identify reference points using perfect squares: Use perfect squares as benchmarks to determine which consecutive integers each square root falls between, since we know \(\sqrt{1} = 1\), \(\sqrt{4} = 2\), \(\sqrt{9} = 3\), etc.
3. Determine the floor value for each square root: For each of \(\sqrt{2}\), \(\sqrt{5}\), and \(\sqrt{7}\), identify which two consecutive integers it falls between, then select the smaller integer as the floor value.
4. Sum the individual floor values: Add the three floor values together to get the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is asking us to do in everyday language.

The symbol \(\lfloor n \rfloor\) is called the "floor function" - it simply means we need to find the largest whole number that doesn't exceed our given number. Think of it like this: if you have 3.7 cookies, the largest whole number of cookies you can count is 3. So \(\lfloor 3.7 \rfloor = 3\).

Our job is to find \(\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{5} \rfloor + \lfloor \sqrt{7} \rfloor\). This means we need to:
• Find the largest whole number ≤ \(\sqrt{2}\)
• Find the largest whole number ≤ \(\sqrt{5}\)
• Find the largest whole number ≤ \(\sqrt{7}\)
• Add these three whole numbers together

Process Skill: TRANSLATE - Converting the floor function notation into clear mathematical understanding

2. Identify reference points using perfect squares

To figure out where our square roots fall, let's use perfect squares as our guideposts. Perfect squares give us exact whole number results that we can easily remember:

\(\sqrt{1} = 1\) exactly
\(\sqrt{4} = 2\) exactly
\(\sqrt{9} = 3\) exactly
\(\sqrt{16} = 4\) exactly

These act like mile markers on a number line. Any square root between two perfect squares will fall between their corresponding whole numbers. For example, any number between \(\sqrt{4}\) and \(\sqrt{9}\) will have a square root between 2 and 3.

3. Determine the floor value for each square root

Now let's figure out where each of our square roots falls:

For \(\sqrt{2}\):
Since \(1 < 2 < 4\), we know that \(\sqrt{1} < \sqrt{2} < \sqrt{4}\)
This means: \(1 < \sqrt{2} < 2\)
So \(\sqrt{2}\) is between 1 and 2, which means \(\lfloor \sqrt{2} \rfloor = 1\)

For \(\sqrt{5}\):
Since \(4 < 5 < 9\), we know that \(\sqrt{4} < \sqrt{5} < \sqrt{9}\)
This means: \(2 < \sqrt{5} < 3\)
So \(\sqrt{5}\) is between 2 and 3, which means \(\lfloor \sqrt{5} \rfloor = 2\)

For \(\sqrt{7}\):
Since \(4 < 7 < 9\), we know that \(\sqrt{4} < \sqrt{7} < \sqrt{9}\)
This means: \(2 < \sqrt{7} < 3\)
So \(\sqrt{7}\) is between 2 and 3, which means \(\lfloor \sqrt{7} \rfloor = 2\)

Process Skill: INFER - Drawing the non-obvious conclusion that we can use perfect square inequalities to determine floor values

4. Sum the individual floor values

Now we simply add up our three floor values:

\(\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{5} \rfloor + \lfloor \sqrt{7} \rfloor = 1 + 2 + 2 = 5\)

Final Answer

The value of \(\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{5} \rfloor + \lfloor \sqrt{7} \rfloor\) is 5.

Looking at our answer choices, this corresponds to choice C.

To verify: We found that \(\sqrt{2}\) is between 1 and 2 (so floor = 1), \(\sqrt{5}\) is between 2 and 3 (so floor = 2), and \(\sqrt{7}\) is between 2 and 3 (so floor = 2). Adding these gives us \(1 + 2 + 2 = 5\).

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the floor function notation
Students may confuse \(\lfloor n \rfloor\) with other mathematical notations like absolute value \(|n|\) or rounding to nearest integer. The floor function specifically means "greatest integer less than or equal to n," not "round to nearest" or "make positive." For example, they might think \(\lfloor 2.7 \rfloor = 3\) (rounding up) instead of \(\lfloor 2.7 \rfloor = 2\) (floor down).

2. Attempting to calculate exact decimal values
Students might waste time trying to calculate \(\sqrt{2} = 1.414...\), \(\sqrt{5} = 2.236...\), \(\sqrt{7} = 2.645...\) to multiple decimal places instead of recognizing that they only need to determine which consecutive integers each square root falls between. This approach is time-consuming and unnecessary for floor function problems.

Errors while executing the approach

1. Incorrectly ordering the perfect square inequalities
Students may set up wrong inequalities like \(4 < 2 < 9\) instead of \(1 < 2 < 4\), leading to incorrect conclusions about where \(\sqrt{2}\) falls. They need to be careful to place each number (2, 5, 7) between the correct consecutive perfect squares (1&4, 4&9, 4&9 respectively).

2. Confusing which perfect squares to use as reference points
Students might use non-consecutive perfect squares or forget common perfect squares. For instance, they might not immediately recall that \(4 = 2^2\) and \(9 = 3^2\), or they might incorrectly think 8 or 6 are perfect squares, leading to wrong interval determinations.

Errors while selecting the answer

1. Adding the square root values instead of their floor values
After correctly finding \(\lfloor \sqrt{2} \rfloor = 1\), \(\lfloor \sqrt{5} \rfloor = 2\), \(\lfloor \sqrt{7} \rfloor = 2\), students might accidentally add the original numbers (\(2 + 5 + 7 = 14\)) or approximate decimal values instead of adding the floor function results (\(1 + 2 + 2 = 5\)). This type of error occurs when students lose track of what the question is actually asking for in the final step.