This problem is from Ancient Rome.
A rich senator died, leaving his wife pregnant. After the senator’s death it was found out that he left a property of 210 talents (an Ancient Roman currency) in his will as follows: “In the case of the birth of a son, give the boy two thirds of my property (i.e. 140 talents) and the other third (i.e. 70 talents) to the mother. In the case of the birth of a daughter, give the girl one third of my property (i.e. 70 talents) and the other two thirds (i.e. 140 talents) to the mother.”
The senator’s widow gave birth to twins: one boy and one girl. This possibility was not foreseen by the late senator. How can the property be divided between three inheritors so that it is as close as possible to the instructions of the will?
Take any two non-equal numbers \(a\) and \(b\), then we can write \[a^2 - 2ab + b^2 = b^2 - 2ab + a^2\] Using the formula \((x-y)^2 = x^2 - 2xy + y^2\), we complete the squares and rewrite the equality above as \[(a-b)^2 = (b-a)^2.\] As we take a square root from the both sides of the equality, we get \[a-b = b-a.\] Finally, adding to both sides \(a+b\) we get \[\begin{aligned} a-b + (a+b) &= b-a + (a+ b)\\ 2a&= 2b\\ a&=b. \end{aligned}\] Therefore, All NON-EQUAL NUMBERS ARE EQUAL! (This is gibberish, isn’t it?)
Consider equation \[x-a=0\] Dividing both sides of this equation by \(x-a\), we get \[\frac{x-a}{x-a} = \frac{0}{x-a}.\] But \(\frac{x-a}{x-a}=1\) and \(\frac{0}{x-a}=0\). Therefore, we get \[1=0.\]
Let \(n\) be some positive number. It is obvious that \[2n-1<2n.\] Take another positive number \(a\), and multiply both sides of the inequality by \((-a)\) \[-2na +a< -2na.\] Now, subtracting \((-2na)\) from both sides of the inequality we get \(a<0\).
Thus, ALL POSITIVE NUMBERS ARE NEGATIVE!
Suppose \(a \neq b\). We can write \[-a = b - (a+b)\] and \[-b = a - (a+b)\] Since \((-a)b = a(-b)\), then \[( b - (a+b))b = a(a - (a+b))\] Removing the brackets, we have \[b^2 - (a+b)b = a^2 - a(a+b)\] Adding \(\left(\frac{a+b}{2}\right)^2\) to each member of the equality we may complete the square of the differences of two numbers \[\left(b - \frac{a+b}{2}\right)^2 = \left(a - \frac{a+b}{2}\right)^2\] From the equality of the squares we conclude the equality of the bases \[b - \frac{a+b}{2} = a - \frac{a+b}{2}.\] Adding \(\frac{a+b}{2}\) to both sides of equality we get \(a=b\). Therefore, WE HAVE SHOWED THAT FROM \(a\neq b\) IT FOLLOWS \(a=b\).
Let \(x\) be equal to 1. Then we can write \(x^2=1\), or putting it differently \(x^2 -1 =0\). By using the difference of two squares formula we get \[(x+1)(x-1)=0\] Dividing both sides of the equality by \(x-1\) we obtain \[x+1=0,\] in other words \(x=-1\). But earlier we assumed \(x=1\). THUS \[-1=1\ !\]
In every right-angled triangle the arm is greater than the hypotenuse. Consider a triangle \(ABC\) with right angle at \(C\).
The difference of the squares of the hypothenuse and one of the arms is \(AB^2 -BC^2\). This expression can be represented in the form of a product \[AB^2 -BC^2 = (AB - BC)(AB+BC)\] or \[AB^2 -BC^2 = -(BC - AB)(AB+BC)\] Dividing the right hand sides by the product \(-(AB-BC)(AB+BC)\), we obtain the proportion \[\frac{AB+BC}{-(AB+BC)} = \frac{BC-AB}{AB-BC}.\] Since the positive quantity is greater than the negative one we have \(AB+BC>-(AB+BC)\). But then also \(BC-AB>AB-BC\), and therefore \(2BC>2AB\), or \(BC>AB\), i.e. THE ARM IS GREATER THAN THE HYPOTENUSE!
Having had experience with some faulty proofs above, can you now answer the following questions
(a) From the equality \((a-b)^2=(m-n)^2\) may one draw the conclusion that \(a-b=m-n\)?
(b) For what value of \(x\) do the following expressions lose their meaning?
(1) \(\frac{x^3-1}{x-1}\), (2) \(\frac{1}{x^2-1}\), (3) \(\frac{x+1}{1-2^x}\).
(c) If \(a>b\), can one conclude that \(ac>bc\) for any number \(c\)?
We prove by mathematical induction that all horses in the world are of the same colour.
Base case: There is a single horse. It has some coat colour. Because there are no other horses, all the horses have the same coat colour.
Induction step: We have \(n\) horses. We assume all of them have the same coat colour. Now we add an additional \((n+1)\)st horse. We don’t know what colour it has, but if we for now get rid of one horse from the group we had before, we suddenly have a group of \(n\) horses which includes the new one. Since we have our claim proven for \(n\), all of these horses have the same coat colour and therefore the new horse has the same coat colour as all the other ones. So every group of \(n+1\) horses has the same colour.
The third step: due to mathematical induction rule, all the horses in the world have the same coat colour. THUS WE HAVE PROVED THAT ALL HORSES IN THE WORLD ARE OF THE SAME COLOUR!
Alice finally decided to do some arithmetic. She took four different integer numbers, calculated their pairwise sums and products, and the results ( the pairwise sums and products) wrote down in her wonderful book. What could be the smallest number of different numbers Alice wrote in her book?