Problems

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?

During the ball every young man danced the waltz with a girl, who was either more beautiful than the one he danced with during the previous dance, or more intelligent, but most of the men (at least 80%) – with a girl who was at the same time more beautiful and more intelligent. Could this happen? (There was an equal number of boys and girls at the ball.)

On a plane there is a square, and invisible ink is dotted at a point $P$. A person with special glasses can see the spot. If we draw a straight line, then the person will answer the question of on which side of the line does $P$ lie (if $P$ lies on the line, then he says that $P$ lies on the line).

What is the smallest number of such questions you need to ask to find out if the point $P$ is inside the square?

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 $$\begin{array}{rl}a-b+(a+b)& =b-a+(a+b)\\ 2a& =2b\\ a& =b.\end{array}$$ 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 $a\ne b$. 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 $${(b-\frac{a+b}{2})}^2={(a-\frac{a+b}{2})}^2$$ 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 $a\ne b$ 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 $A{B}^2-B{C}^2$. This expression can be represented in the form of a product $$A{B}^2-B{C}^2=(AB-BC)(AB+BC)$$ or $$A{B}^2-B{C}^2=-(BC-AB)(AB+BC)$$ 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 $c$?