Is e rational ?

We all know that ‘e’ is an irrational number and is given by:

e = 1 + (1/2!) + (1/3!) | + (1/4!) + …

Here is a proof that ‘e’ is rational. Find what is wrong with it.

We know that sum of two rational numbers is rational. Using mathematical induction, we can say that sum of any number of rational numbers is also rational. From the equation above, we can see that ‘e’ is indeed a sum of rational numbers. Hence ‘e’ must be rational as well.

What is wrong with the above proof ?

Sum of any two rational numbers is rational. Using induction, we can only say that sum of any finite number of rational numbers is rational. But ‘e’ is a summation of an infinite number of rational numbers.

Tiling a floor with Polygons

We know that square tiles can be put together to cover a floor without gaps in between. This coverability holds for triangular and hexagonal tiles as well.

Are there any other symmetric tiles that can cover a floor?. Let us approach this question mathematically.

Triangles, squares and hexagons have 3, 4 and 6 sides respectively. Consider the figure below:

Some observations, we can make:

• Triangle: Has 3 sides. All internal angles are 60 degrees. Six triangles meet at a corner-point
• Square: Has 4 sides. All internal angles are 90 degrees. Four squares meet at a corner-point
• Hexagon: Has 6 sides. All internal angles are 120 degrees. Three hexagons meet a corner-point

Note that, for a tile of k sides, the sum of internal angles is (k–2)*180. (This can be proved easily and is left as an exercise to the reader.)

In the general case, let k be the number of sides of a symmetric shape that can be tiled to cover a floor. The following must hold true:

1. The sum of internal angles in the tile is (k–2)*180. Since the tiles are symmetric, every internal angle is the same. Hence the internal angle is (k–2)*180/k
2. Adjacent tiles that share a corner point, must subtend a total angle of 360 degrees at the corner point. This holds for all corner points.

 

If m tiles share a corner point, then the conditions above imply: m*(k–2)*180/k = 360, which simplifies to: k = 2m/(m–2) and m = 2k/(k–2)

Let us try various values of k and see which result in m being in integer. Obviously k > 2, and m > 2

• k=3 implies that m=6 (Six triangular tiles meet at a corner-point)
• k=4 implies that m=4 (Four square tiles meet at corner-point)
• k=5 implies that m=10/3, which is not possible, since 10/3 is not an integer
• k=6 implies that m=3 (Three hexagonal tiles meet at a corner-point)

We can keep trying for further values of k=7, 8 ,9 and so on. But let us simplify the analysis. Since we want to try for k=7, 8, 9, etc., let k = 6 + k’, where k’ >= 1. Then we get:

We observe the following:

(a) Since k’>1, the second term in the above equation: (4/(4+k’))  < 1. This implies that, m < 3

(b) Since k = 2m/(m-2), we have m > 2

No integer value of m can simultaneously satisfy the above two conditions. So, there is no valid value of m, for k=7, 8, 9 and so on. The only valid values of m are 3, 4 and 6, which correspond to k=6, 4 and 3 respectively.

Thus there are only three symmetric shapes that can cover a floor: triangles, squares and hexagons.

Turning Radius of a Car

Suppose a car salesman is trying to convince you about a new model car. The car is 2m long (i.e. the distance between the centres of the front wheel and real wheel). He claims that it can make a very sharp turn of radius 0.8m. Will you believe him?

Let us see how maths can help here. All of us know that, turning radius depends on how far you can rotate the steering wheel. Intuitively, we also feel that a shorter car can make a sharper turn, than a longer car.

Let us draw a car which is making a turn. The situation is show in the figure below. The wheels are shown shaded in the figure.

The front wheel’s center is P and the rear wheel’s center is Q. Let O be the center of the circle, along whose circumference the car moves. Let L=PQ be the length of the car and let R be the radius of the circle. Sharpest turn corresponds to smallest possible value of R. Note that PQ is a chord of the circle. Let us say that the front wheel can rotate at most by an angle w relative to the ‘straight’ position.

If we observe closely, the front wheel is always travelling along the line that is tangent to the circle at point P. Note that P itself is moving. The rear wheel is always aligned with the body. The picture shows the trigonometry worked out.

The smaller the value of L, the smaller the turning radius. So our calculation confirms the intuition that a shorter car can move along a smaller circle, and hence can make a sharper turn.

Also we note that maximum value of sin is 1, no matter what w is. Hence the smallest turning radius is (L/2), which happens when PQ is a diameter of the circle. Thus car of length 2m, can have a smallest turning radius of 1m. So the salesman was lying.

Chemistry: Balancing an Equation

Suppose a chemistry student comes to us, having failed at balancing the equation below.

H2O + PO2 = H3PO4

i.e. water and phosphorus dioxide reacting to produce phosphoric acid. But what if we know nothing about chemistry of phosphorus ?

 We can try the usual method taught at school: progressively adjusting the number of molecules of the two reactants and the product to see if the equation becomes balanced. (We urge the reader to try balancing the equation, before reading further.)

After a few attempts we will see that we are not able to balance the equation, no matter how many combinations we try. Let us change our approach – look at it mathematically. We have to find positive integers a, b and c such that:

a H2O + b PO2 = c H3PO4.

In other words, a molecules of water and b molecules of phosphorus dioxide react to product c molecules of phosphoric acid. For the equation to be balanced the following three conditions must be met:

1. Number of oxygen atoms is the same on both sides of the equation: a + 2b = 4c
2. Number of hydrogem atoms is the same on both sides of the equation: 2a = 3c
3. Number of phosphorus atoms is the same on both sides of the equation: b = c

So, we have got three linear equations. Using equations (2) and (3) in equation (1), we get: 7c = 8c. Thus, we get: c=0

Using c=0 in equations (2) and (3) we get a=0 and b=0. In other words, the only solution is: a=0, b=0, c=0. So, the equation can not be balanced no matter how many times we try.

Without really know anything about the chemistry of phosphorus, we can tell the student that the reaction is not possible