In this article, we will delve into the world of irrational numbers, specifically focusing on the proof that there is no rational number whose square is 3. Additionally, we will explore a unique property of a specific type of parallelogram, where the ratio of the perimeter to the long side is equal to the ratio of the long side to the short side. This type of parallelogram will be referred to as 'Ejercito Parallelograms.'

Proving That There Is No Rational Number Whose Square Is 3

To prove that there is no rational number whose square is 3, we can use a proof by contradiction. Here's a step-by-step outline of the proof:

Assume the opposite: Suppose there exists a rational number ( r ) such that ( r^2 3 ). Since ( r ) is rational, it can be expressed as a fraction of two integers in simplest form: [ r frac{a}{b} ] where ( a ) and ( b ) are integers, ( b eq 0 ), and ( gcd(a, b) 1 ) meaning ( a ) and ( b ) have no common factors other than 1.

Square both sides: From our assumption, we have: [ r^2 3 implies left( frac{a}{b} right)^2 3 ] This simplifies to: [ frac{a^2}{b^2} 3 ] Multiplying both sides by ( b^2 ) gives: [ a^2 3b^2 ]

Analyze the equation: The equation ( a^2 3b^2 ) implies that ( a^2 ) is divisible by 3.

Conclude about ( a ): If ( a^2 ) is divisible by 3, then ( a ) must also be divisible by 3 (since if a prime divides the square of a number, it must also divide the number itself). Therefore, we can express ( a ) as: [ a 3k , text{for some integer} k. ]

Substitute back into the equation: Plugging ( a 3k ) back into the equation ( a^2 3b^2 ) gives: [ 3k^2 3b^2 implies 9k^2 3b^2 implies 3k^2 b^2. ]

Analyze the new equation: This equation ( b^2 3k^2 ) implies that ( b^2 ) is also divisible by 3, and by the same reasoning, ( b ) must also be divisible by 3.

Reach a contradiction: Since both ( a ) and ( b ) are divisible by 3, this contradicts our initial assumption that ( frac{a}{b} ) is in simplest form, i.e., ( gcd(a, b) 1 ).

Conclusion: Therefore, our assumption that there exists a rational number ( r ) such that ( r^2 3 ) must be false. Thus, there is no rational number whose square is 3, and thus ( sqrt{3} ) is irrational.

Properties of the Ejercito Parallelogram

Now, let us consider a parallelogram on a flat plane. A parallelogram is a quadrilateral whose opposite sides are parallel. One of their properties is that the opposite sides are congruent. If ( P ) is the perimeter, ( L ) is the length of the long side, and ( S ) is the length of the short side with ( L geq S ), then: [ P 2L 2S ]

Ratio of lengths: ( S ) can be arbitrarily small as it only needs to be a positive real number. The upper limit for ( S ) is ( L ), in which case the parallelogram is also a rhombus. So, the exclusive lower bound for the perimeter ( P ) is ( 2L ). The inclusive upper bound for ( P ) is ( 4S ) or ( 4L ).

Ratios of perimeters and lengths: Given these respective lower and upper bounds, it is possible for a parallelogram to have ( P ), ( L ), and ( S ) such that: [ frac{P}{L} frac{L}{S} ].

Is this ratio rational?

Let us assume it is. Then we should be able to pick a pair of coprime positive integer values for ( P ) and ( L ). Now, let us take a second look at the equation ( P 2L 2S ).

Subtract ( 2L ) from both sides: [ P - 2L 2S ] Divide by 2: [ S frac{P - 2L}{2} ]. So: [ frac{P}{L} frac{L}{S} frac{L}{frac{P - 2L}{2}} ]

We can multiply the right side by ( frac{2}{2} ): [ frac{P}{L} frac{2L}{P - 2L} ]

2L and P - 2L are clearly integers due to the closure of the integers over multiplication and subtraction. Recall that ( P 2L 2S ). As ( S ) is strictly positive, this means that ( 2L

But we had concluded there existed coprime integer values for ( P ) and ( L ), which would lead to the conclusion that ( frac{P}{L} ) is in lowest terms and can not be reduced to lower terms. We have a contradiction.

And thus we have proved that if the ratio of the perimeter and the long side of a parallelogram is equal to the ratio of the long side to the short side, this ratio is irrational.

Calculating the Ratio

If ( L 1 ), then ( P frac{2}{P - 2} ). Multiply by ( P - 2 ): [ PP - 2 P^2 - 2P 2 ] Subtract 2: [ P^2 - 2P - 2 0 ]

We have a quadratic equation in terms of ( P ). As ( P ) is irrational because we set ( L 1 ), we must use the quadratic formula: [ P frac{-b pm sqrt{b^2 - 4ac}}{2a} ] where ( a 1 ), ( b -2 ), and ( c -2 ).

Substitute the values into the formula: [ P frac{2 pm sqrt{4 8}}{2} frac{2 pm sqrt{12}}{2} 1 pm sqrt{3} ]

The ratio is about ( 2.7321 ).

If ( S 1 ), the perimeter of this parallelogram is ( 2(1 sqrt{3}) 2 2sqrt{3} approx 4.4641 ).

The difference of an irrational number and a rational number is irrational. So, ( 1 sqrt{3} - 1 sqrt{3} ) is about ( 1.7321 ) and is irrational.

There is no literature on these types of parallelograms where the ratio of the perimeter and the long side of a parallelogram is equal to the ratio of the long side to the short side. I propose that they be called 'Ejercito Parallelograms.'