Problems from Situation of the irrational numbers on the straight line

Development:

Given the number $e=2,71828182846\dots$ 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:

$$2;2.7;2.71;2.718;2.7182;2.71828;2.718281;2.7182818;$$ $$2.71828182;2.718281828;2.7182818284;2.71828182846;\dots$$

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 $e$:

$$3;2.8;2.72;2.719;2.7183;2.71829;2.718282;2.7182819;$$ $$2.71828183;2.718281829;2.7182818285;2.71828182847;\dots$$

And from both successions, we can construct the succession of fitted intervals that defines to number $e=2,71828182846\dots$:

$$[2,3];[2.7,2.8];[2.71,2.72];[2.718,2.719];[2.7182,2.7183];[2.71828,2.71829];$$ $$[2.718281,2.718282];[2.7182818,2.7182819];[2.71828182,2.71828183];\dots$$

Solution:

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: $[2,3];[2.7,2.8];[2.71,2.72];[2.718,2.719];[2.7182,2.7183]$

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Development:

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 $\sqrt{7}$.

Solution:

We decompose the number $7$ thus $7=3+4$. To draw the length of the leg $\sqrt{3}$ we decompose $3=1+2$.

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