Situation of the irrational numbers on the straight line
Describe a method to draw the irrational number $\sqrt{7}$.
To draw the irrational one $\sqrt{7}$, we decompose the number $7$into an addition of another two. We have different options: $7=1+6$; $7=2+5$ or $7=3+4$. We choose the third one because $4$ is a perfect square and it is going to be easier to draw. Then, we draw a rectangular triangle whose legs have length $\sqrt{4}=2$ and $\sqrt{3}$.
To draw the leg which has length $\sqrt{3}$ we must begin the process again:
We decompose the number $3$ into an addition of another two: $3=1+2$.
We draw a rectangular triangle with legs $\sqrt{1}=1$ and $\sqrt{2}$.
It is not a problem to draw a segment of length $1$, and the segment with length $\sqrt{2}$ can be obtained by drawing a triangle with legs $1$.
Using Pythagoras' theorem, we have that the hypotenuse of the above mentioned triangle is: $$H^2=(\sqrt{2})^2+1^2$$ $$H^2=2+1=3$$ $$H=\sqrt{3}$$
Now we already have drawn the length $\sqrt{3}$ and, using a compas, we can move it to the rectangular triangle that we were drawing which has a length of legs $\sqrt{3}$ and $2$.
The hypotenuse $G$ of this triangle measures exactly $\sqrt{7}$: $$G^2=(\sqrt{3})^2+2^2=3+4=7$$
so we already have drawn a length of $\sqrt{7}$.
We decompose the number $7$ thus $7=3+4$. To draw the length of the leg $\sqrt{3}$ we decompose $3=1+2$.
Write the first five decimal approximations by defect and by excess to the number $e=2,71828182846\ldots$, as well as the first five fitted intervals of the sequence of intervals that it defines.
Given the number $e=2,71828182846\ldots$ and considering the rational number obtained by truncation in each of the decimal positions we obtain the sequence of approximation of number $e$ by defect:
$$2; 2.7; 2.71; 2.718; 2.7182; 2.71828; 2.718281; 2.7182818;$$ $$2.71828182; 2.718281828; 2.7182818284; 2.71828182846; \ldots $$
Then, we add a unit in the last digit of every term of this sequence, so we obtain the approximation by excess of the number $e$:
$$3; 2.8; 2.72; 2.719; 2.7183; 2.71829; 2.718282; 2.7182819;$$ $$2.71828183; 2.718281829; 2.7182818285; 2.71828182847; \ldots $$
And from both successions, we can construct the succession of fitted intervals that defines to number $e=2,71828182846\ldots$:
$$[2,3]; [2.7,2.8]; [2.71,2.72]; [2.718,2.719]; [2.7182,2.7183]; [2.71828,2.71829];$$ $$[2.718281,2.718282]; [2.7182818,2.7182819]; [2.71828182,2.71828183]; \ldots$$
The first five terms of the sequence of approximation by defect are: $2; 2.7; 2.71; 2.718; 2.7182$ The first five terms of the sequence of approximation by excess are: $3; 2.8; 2.72; 2.719; 2.7183$ The first five terms of the sequence of fitted intervals: $[2,3]; [2.7,2.8]; [2.71,2.72]; [2.718,2.719]; [2.7182,2.7183]$