Problems from Operations: Sum and product of real numbers

Development:

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

$$0,3$$

$$0,33$$

$$0,333$$

$$0,3333$$

$$\ldots$$

For the number $$\pi$$ the approximations are:

$$3,1$$

$$3,14$$

$$3,141$$

$$3,1415$$

$$\ldots$$

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

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

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

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

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

$$\ldots$$

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

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

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

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

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

$$\ldots$$

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

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

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

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

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

$$\ldots$$

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

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

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

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

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

$$\ldots$$

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

$$0,1$$

$$0,14$$

$$0,142$$

$$0,1428$$

$$\ldots$$

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

$$1,4$$

$$1,41$$

$$1,414$$

$$1,4142$$

$$\ldots$$

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

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

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

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

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

$$\ldots$$

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

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

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

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

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

$$\ldots$$

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

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

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

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

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

$$\ldots$$

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

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

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

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

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

$$\ldots$$

Solution:

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

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

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

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

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

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

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

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

Hide solution and development

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 $$P'$$ 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 $$(\sqrt{3})^{-1}=\dfrac{1}{\sqrt{3}}$$. 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 $$P'$$ on the auxiliary straight line. Then, we only need to move the point $$P'$$ 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 $$P'$$ that, being moved to the real straight line, gives us point $$\sqrt{2}\cdot \dfrac{1}{\sqrt{3}}= \dfrac{\sqrt{2}}{\sqrt{3}}.$$

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 $$\sqrt{2}\cdot \dfrac{1}{\sqrt{3}}= \dfrac{\sqrt{2}}{\sqrt{3}}.$$

Hide solution and development