Problems from Operations: Sum and product of real numbers

Development:

a) For the number ${\displaystyle \frac{1}{3}}$ the approximations are:

$0,3$

$0,33$

$0,333$

$0,3333$

$\dots$

For the number $\pi$ the approximations are:

$3,1$

$3,14$

$3,141$

$3,1415$

$\dots$

For the addition of ${\displaystyle \frac{1}{3}}$ and $\pi$:

$0,3+3,1=3,4$

$0,33+3,14=3,47$

$0,333+3,141=3,474$

$0,3333+3,1415=3,4748$

$\dots$

For the subtraction of ${\displaystyle \frac{1}{3}}$ and $\pi$:

$0,3-3,1=-2,8$

$0,33-3,14=-2,81$

$0,333-3,141=-2,808$

$0,3333-3,1415=-2,8082$

$\dots$

For the multiplication of ${\displaystyle \frac{1}{3}}$ by $\pi$:

$0,3\cdot 3,1=0,93$

$0,33\cdot 3,14=1,0362$

$0,333\cdot 3,141=1,045953$

$0,3333\cdot 3,1415=1,04706195$

$\dots$

For the division of ${\displaystyle \frac{1}{3}}$ by $\pi$:

$0,3/3,1=0,096774$

$0,33/3,14=0,105095$

$0,333/3,141=0,106017$

$0,3333/3,1415=0,106095$

$\dots$

b) For the number ${\displaystyle \frac{1}{7}}$ the approximations are:

$0,1$

$0,14$

$0,142$

$0,1428$

$\dots$

For the number $\sqrt{2}$ the approximations are:

$1,4$

$1,41$

$1,414$

$1,4142$

$\dots$

For the addition of ${\displaystyle \frac{1}{7}}$ and $\sqrt{2}$:

$0,1+1,4=1,5$

$0,14+1,41=1,55$

$0,142+1,414=1,556$

$0,1428+1,4142=1,5570$

$\dots$

For the subtraction of ${\displaystyle \frac{1}{7}}$ and $\sqrt{2}$:

$0,1-1,4=-1,3$

$0,14-1,41=-1,27$

$0,142-1,414=-1,272$

$0,1428-1,4142=-1,2714$

$\dots$

For the multiplication of ${\displaystyle \frac{1}{7}}$ by $\sqrt{2}$:

$0,1\cdot 1,4=0,14$

$0,14\cdot 1,41=0,1974$

$0,142\cdot 1,414=0,20022$

$0,1428\cdot 1,4142=0,20194776$

$\dots$

For the quotient of ${\displaystyle \frac{1}{7}}$ by $\sqrt{2}$:

$0,1/1,4=0,0714285$

$0,14/1,41=0,0992907$

$0,142/1,414=0,1004243$

$0,1428/1,4142=0,1009758$

$\dots$

Solution:

a) For the addition of ${\displaystyle \frac{1}{3}}$ and $\pi$: $3,4748$

For the subtraction of ${\displaystyle \frac{1}{3}}$ and $\pi$: $-2,8082$

For the multiplication of ${\displaystyle \frac{1}{3}}$ by $\pi$: $1,04706195$

For the division of ${\displaystyle \frac{1}{3}}$ by $\pi$: $0,106095$

b) For the addition of ${\displaystyle \frac{1}{7}}$ and $\sqrt{2}$: $1,5570$

For the subtraction of ${\displaystyle \frac{1}{7}}$ and $\sqrt{2}$: $-1,2714$

For the multiplication of ${\displaystyle \frac{1}{7}}$ by $\sqrt{2}$: $0,20194776$

For the division of ${\displaystyle \frac{1}{7}}$ by $\sqrt{2}$: $0,1009758$

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Development:

1. We draw on the straight line the numbers $\sqrt{2}$ and $\sqrt{3}$ using the construction of rectangular triangles.

Now we move the segment $\overline{0\sqrt{3}}$ beginning from point $\sqrt{2}$. The point that we obtain corresponds to $\sqrt{2}+\sqrt{3}$.

1. We draw on the straight line the numbers $\sqrt{2}$ and $\sqrt{3}$. Next we move the segment $\overline{0\sqrt{3}}$ beginning from point $\sqrt{2}$, to the left (instead of to the right as we have done in the last exercice). The point that we obtain corresponds to $\sqrt{2}-\sqrt{3}$.

2. We draw on the straight line the numbers $\sqrt{2}$ and $\sqrt{3}$. This time we move towards the left the segment $\overline{0\sqrt{2}}$ from point $\sqrt{3}$, and the point that we obtain corresponds to $\sqrt{3}-\sqrt{2}$.

3. We draw on the straight line the numbers $\sqrt{2}$, $\sqrt{3}$ and the unit.

Now we move the segment $\overline{0\sqrt{3}}$ from the zero of an auxiliary straight line, finding the point $P$ .

We draw a straight line which joins the point $P$ and point $1$, and then we draw its parallel that goes through the point $\sqrt{2}$, obtaining the point $P^{\prime }$ which being moved to the real straight line, gives us the point $\sqrt{2}\cdot \sqrt{3}$.

1. We will first try to find, on the straight line, the point $(\sqrt{3}{)}^{-1}={\displaystyle \frac{1}{\sqrt{3}}}$. So, we draw on the straight line the points $\sqrt{3},1$ and $0$.

Now, we draw on an auxiliary straight line a point $P$ moving the segment $\overline{01}$. We draw the straight line that joins point $P$ with point $\sqrt{3}$, and we draw a parallel to this one which crosses point $1$. In that way, we will find point $P^{\prime }$ on the auxiliary straight line. Then, we only need to move the point $P^{\prime }$ to the real straight line, obtaining in this way the point $(\sqrt{3}{)}^{-1}$.

We place on the same straight line point $\sqrt{2}$ to be able to calculate the product between $\sqrt{2}$ and $(\sqrt{3}{)}^{-1}$.

We move the segment $\overline{0(\sqrt{3}{)}^{-1}}$ from the zero of an auxiliary straight line, finding point $P$.

We draw a straight line that joins point $P$ and point $1$, and then we draw its parallel that goes through point $\sqrt{2}$, obtaining point $P^{\prime }$ that, being moved to the real straight line, gives us point $\sqrt{2}\cdot {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{\sqrt{2}}{\sqrt{3}}}.$

Solution:

1, 2 and 3. We draw on the straight line the numbers $\sqrt{2}$ and $\sqrt{3}$. Following the established procedures we draw the corresponding numbers.

1. We draw on the straight line the numbers $\sqrt{2}$ and $\sqrt{3}$. Using the Thales' Theorem we can obtain the result of $\sqrt{2}\cdot \sqrt{3}$.

2. We draw on the straight line the numbers $\sqrt{2}$ and $\sqrt{3}$. We can then solve the operation $(\sqrt{3}{)}^{-1}$.

Using the method to multiply the number we solve the number $\sqrt{2}\cdot {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{\sqrt{2}}{\sqrt{3}}}.$

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