KINDLY GIVE ANSWER ALONG WITH PROOF
For every non-negative integer \( n \), the square root \( \sqrt{n} \) is either an integer or an irrational number. Proof (by contradiction): Suppose \( \sqrt{n} \) is rational but **not** an integer. Then there exist integers \( a \) and \( b \), with \( \gcd(a, b) = 1 \) and \( b \ne 1 \), such tRead more
For every non-negative integer \( n \), the square root \( \sqrt{n} \) is either an integer or an irrational number.
Proof (by contradiction):
Suppose \( \sqrt{n} \) is rational but **not** an integer.
Then there exist integers \( a \) and \( b \), with \( \gcd(a, b) = 1 \) and \( b \ne 1 \), such that:
\[
\sqrt{n} = \frac{a}{b}
\]
Squaring both sides:
\[
n = \left( \frac{a}{b} \right)^2 = \frac{a^2}{b^2}
\Rightarrow a^2 = n b^2
\]
This implies that \( b^2 \) divides \( a^2 \). But since \( \gcd(a, b) = 1 \), it follows that \( \gcd(a^2, b^2) = 1 \) as well. Hence, the only way \( b^2 \mid a^2 \) can be true is if \( b^2 = 1 \), which means:
\[
b = 1
\Rightarrow \sqrt{n} = \frac{a}{1} = a \in \mathbb{Z}
\]
This contradicts our assumption that \( \sqrt{n} \) is rational **but not** an integer.
Conclusion:
If \( \sqrt{n} \) is rational, then it must be an integer.
Therefore, if \( \sqrt{n} \) is not an integer, it must be irrational.
\[
\boxed{\text{For all } n \in \mathbb{N}_0,\ \sqrt{n} \in \mathbb{Z} \cup (\mathbb{R} \setminus \mathbb{Q})}
\]
Problem: There are 1000 doors, all initially closed. 1000 people walk by these doors. The first person opens every door. The second person toggles every second door (i.e., closes doors 2, 4, 6, etc.). The third person toggles every third door (i.e., doors 3, 6, 9, etc.), and so on. The 1000th personRead more
Problem:
There are 1000 doors, all initially closed. 1000 people walk by these doors. The first person opens every door. The second person toggles every second door (i.e., closes doors 2, 4, 6, etc.). The third person toggles every third door (i.e., doors 3, 6, 9, etc.), and so on. The 1000th person only toggles door 1000. After all the people have passed, how many doors remain open?
Step 1: Understand the pattern
Each person toggles the state of doors that are multiples of their own number. For example, person 12 will toggle doors 12, 24, 36, and so on. The state of each door will change every time it is toggled.
Step 2: Think about how many times each door is toggled
Take any door number, say door 6. It gets toggled by person 1 (since 1 divides 6), person 2 (2 divides 6), person 3 (3 divides 6), and person 6 (6 divides itself). In general, a door will be toggled once for each of its positive divisors.
Step 3: Determine when a door ends up open
Since all doors start closed, each toggle changes its state. So, if a door is toggled an even number of times, it will end up closed. If it is toggled an odd number of times, it will end up open.
Step 4: Identify which door numbers are toggled an odd number of times
From number theory, we know that most numbers have an even number of divisors because divisors usually come in pairs (like 2 and 3 for 6, since 2×3 = 6). However, perfect squares have an odd number of divisors. For example, 36 has divisors: 1, 2, 3, 4, 6, 9, 12, 18, and 36. Notice that 6×6 = 36, so the factor 6 appears only once, not in a pair. This gives it an odd number of total divisors.
Therefore, only the doors with perfect square numbers will remain open at the end.
Step 5: Count the perfect squares between 1 and 1000
The perfect squares less than or equal to 1000 are:
1² = 1,
2² = 4,
3² = 9,
…
31² = 961.
So, there are 31 perfect squares between 1 and 1000.
Final Answer:
See less31 doors will remain open.