Problems

Using mathematical induction prove that $$1+3+5+\cdots +(2n-1)=n^2.$$

Circles and lines are drawn on the plane. They divide the plane into non-intersecting regions, see the picture below.

Show that it is possible to colour the regions with two colours in such a way that no two regions sharing some length of border are the same colour.

Numbers $1,2,\dots ,n$ are written on a whiteboard. In one go Louise is allowed to wipe out any two numbers $a$ and $b$, and write their sum $a+b$ instead. Louise enjoys erasing the numbers, and continues the procedure until only one number is left on the whiteboard.

What number is it? What if instead of $a+b$ she writes $a+b-1$?

Prove that

(a) $$1^2+2^2+3^2+\cdots +n^2=\frac{1}{6}n(n+1)(2n+1)$$

(b) $$1^2+3^2+5^2+\cdots +(2n-1{)}^2=\frac{1}{3}n(2n-1)(2n+1).$$

Using mathematical induction prove that $2^n>n$ for all natural numbers.

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 $$-1=1!$$