Prove Non-Square Integer: Ceiling Function

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In summary: We can rewrite this as:$(a+1)^2+(2a)^2 \geq d^2$Applying the Pythagorean theorem, we can see that this is a contradiction because it implies that there exists a right triangle with sides of length $a+1$, $2a$, and $d$, which violates the fundamental theorem of arithmetic.Therefore, our assumption that $a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil$ is a perfect square is false. In summary, we have proven that $a^2+\Bigl\lceil
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Let $a$ and $b$ be two positive integers. Prove that the integer $ a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil$ is not a square.

(Here $\lceil z \rceil$ denotes the least integer greater than or equal to $z$.)
 
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Thank you for bringing up this interesting problem. I would like to provide a proof for the statement that $a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil$ is not a perfect square for any positive integers $a$ and $b$.

First, let us assume that $a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil$ is a perfect square, denoted by $c^2$, where $c$ is also a positive integer. This means that we can write the following equation:

$a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil = c^2$

We can rewrite the ceiling function as follows:

$a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil = a^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil + 1-1 = a^2+1+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil -1$

Since $a^2$ and $1$ are perfect squares, their sum is also a perfect square. Therefore, we can rewrite the equation as:

$a^2+1+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil -1 = d^2$, where $d$ is a positive integer.

We can further simplify this equation by combining the terms $a^2$ and $1$:

$(a+1)^2+\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil -1 = d^2$

Since $b$ is a positive integer, we can write $b \geq 1$. Therefore, $\dfrac{4a^2}{b} \geq 4a^2$. This means that the ceiling function will always round up to at least $4a^2$, making the term $\Bigl\lceil \dfrac{4a^2}{b}\Bigr\rceil -1$ always greater than or equal to $4a^2-1$.

Substituting this into the equation, we get:

 

FAQ: Prove Non-Square Integer: Ceiling Function

1. What is the definition of the ceiling function?

The ceiling function, denoted as ⌈x⌉, takes a real number x as input and returns the smallest integer greater than or equal to x. In other words, it rounds up the input to the nearest integer.

2. How do you prove that a number is not a perfect square using the ceiling function?

To prove that a number is not a perfect square, we can use the fact that the square root of a perfect square is always an integer. Therefore, if we take the square root of a number and apply the ceiling function, and the result is not equal to the original number, then we can conclude that the number is not a perfect square.

3. Can the ceiling function be applied to negative numbers?

Yes, the ceiling function can be applied to negative numbers. For example, ⌈-2.5⌉ = -2, as -2 is the smallest integer greater than or equal to -2.5.

4. Are there any other methods to prove that a number is not a perfect square?

Yes, there are other methods to prove that a number is not a perfect square. Some common methods include checking the last digit of the number (if it is not 0, 1, 4, 5, 6, or 9, then it is not a perfect square) and using modular arithmetic.

5. What is the significance of proving a number is not a perfect square using the ceiling function?

Proving a number is not a perfect square using the ceiling function is a simple and straightforward method that can be easily applied to any real number. It also helps to eliminate the need for complex calculations and can be used as a quick check for the perfect square property of a number.

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