Problems from Situation of the irrational numbers on the straight line

Development:

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 $$e=2,71828182846\ldots$$:

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

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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