In the previous section we completed the discussion on the addition and subtraction of irrational numbers. We also saw some solved examples. In this section, we will see their multiplication.
Consider ⊿ABC and the attached square CBDE in fig.16.23(a) below:
We have seen such combination before. But we will write the calculation steps again:
• Applying Pythagoras theorem to ⊿ABC, we get: BC2 = AB2 + AC2
⇒ BC2 = 12 + 12 ⇒ BC2 = 1 + 1 ⇒ BC2 = 2. ∴ BC = √2
• The square CBDE has the side √2. So it's perimeter = √2 + √2 + √2 + √2
⇒ 4 times √2 ⇒ 4 × √2
• Usually, we avoid the '×' sign, and write this as 4√2
• Let us put the value of √2. We know that √2 = 1.4142135....
• So 4√2 = 4 × 1.4142135... = 5.65685... We can write:
■ Perimeter of square = 5.65 m (correct to 1 cm)
■ Perimeter of square = 5.656 m (correct to 1 mm)
■ So we find that, an irrational number can be multiplied by a natural number
• Now consider the ⊿ABC separately, as in fig.b. The length of the hypotenuse BC is √2
• We know how to draw the 'perpendicular bisector' of any line. Such a bisector for BC is drawn in fig.c
• BC is bisected at F. What is the length of CF and BF?
• Obviously, they are both equal to (1⁄2 × √2). This is usually written as √2⁄2
• So half of the hypotenuse = √2⁄2 = 1.4142135...⁄2 = 0.7071.... m. We can write:
■ Half of hypotenuse = 0.70 m (correct to 1 cm)
■ Half of hypotenuse = 0.701 m (correct to 1 mm)
Let us consider another situation: Fig.16.24(a) shows an equilateral triangle of side 2 m. In fig.b, it is cut into two equal halves.
1. Applying Pythagoras theorem to the green half, we will get the altitude as √3.
2. In fig.c, the two halves are rearranged to form a rectangle. The base 1 m of the triangle becomes the width of the rectangle. The altitude √3 of the triangle becomes the length of the rectangle
3. Now, what is the perimeter of the rectangle?
4. Let us add the sides: 1 + 1 + √3 + √3 = 2 + 2√3
5. Let us put the values of √3. We get: 2√3 = 2 × 1.73205... = 3.4641...
6. So 2 + 2√3 = 2 + 3.4641... = 5.4641... We can write:
■ Perimeter of rectangle = 5.46 m (correct to 1 cm)
■ Perimeter of rectangle = 5.464 m (correct to 1 mm)
In the rectangle that we considered above, the length was an irrational number. But the width was a natural number. What if both are irrational numbers?
So we have completed the discussion on multiplication. In the next section, we will see some solved examples.
Consider ⊿ABC and the attached square CBDE in fig.16.23(a) below:
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| Fig.16.23 |
• Applying Pythagoras theorem to ⊿ABC, we get: BC2 = AB2 + AC2
⇒ BC2 = 12 + 12 ⇒ BC2 = 1 + 1 ⇒ BC2 = 2. ∴ BC = √2
• The square CBDE has the side √2. So it's perimeter = √2 + √2 + √2 + √2
⇒ 4 times √2 ⇒ 4 × √2
• Usually, we avoid the '×' sign, and write this as 4√2
• Let us put the value of √2. We know that √2 = 1.4142135....
• So 4√2 = 4 × 1.4142135... = 5.65685... We can write:
■ Perimeter of square = 5.65 m (correct to 1 cm)
■ Perimeter of square = 5.656 m (correct to 1 mm)
■ So we find that, an irrational number can be multiplied by a natural number
• Now consider the ⊿ABC separately, as in fig.b. The length of the hypotenuse BC is √2
• We know how to draw the 'perpendicular bisector' of any line. Such a bisector for BC is drawn in fig.c
• BC is bisected at F. What is the length of CF and BF?
• Obviously, they are both equal to (1⁄2 × √2). This is usually written as √2⁄2
• So half of the hypotenuse = √2⁄2 = 1.4142135...⁄2 = 0.7071.... m. We can write:
■ Half of hypotenuse = 0.70 m (correct to 1 cm)
■ Half of hypotenuse = 0.701 m (correct to 1 mm)
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| Fig.16.24 |
2. In fig.c, the two halves are rearranged to form a rectangle. The base 1 m of the triangle becomes the width of the rectangle. The altitude √3 of the triangle becomes the length of the rectangle
3. Now, what is the perimeter of the rectangle?
4. Let us add the sides: 1 + 1 + √3 + √3 = 2 + 2√3
5. Let us put the values of √3. We get: 2√3 = 2 × 1.73205... = 3.4641...
6. So 2 + 2√3 = 2 + 3.4641... = 5.4641... We can write:
■ Perimeter of rectangle = 5.46 m (correct to 1 cm)
■ Perimeter of rectangle = 5.464 m (correct to 1 mm)
Is there an easier method to get the above result? Let us take a deviation after the step (4) above:
4. Let us add the sides: 1 + 1 + √3 + √3 = 2 + 2√3
5a. The above can be written as (2 × 1 + 2√3) ⇒ (2 × √1 + 2√3) ⇒ (2√1 + 2√3)
6a. Let us take the common factor '2' outside the brackets. We get: 2(√1 + √3) ⇒ 2(1 + √3) [∵ √1 = 1]
7a. Let us put the values of √3. We get: (1 + √3) = 1 + 1.73205... = 2.73205...
8a. So 2(√1 + √3) = 2 × 2.73205... = 5.4641... We can write:
■ Perimeter of rectangle = 5.46 m (correct to 1 cm)
■ Perimeter of rectangle = 5.464 m (correct to 1 mm)
This is the same result that we obtained in (6) above
Thus we get a general form:
k√x + k√y = k(√x + √y)
That is., we can take the common factor 'k' outside the brackets
So we have calculated the perimeter of the rectangle. Let us now calculate it's area. The rectangle in fig.16.25(a) below is the same one in fig.16.24(c) which we saw above.
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| Fig.16.25 |
1. We know that the area of a rectangle = length × width
2. So for our rectangle, the area = (1 × √3) m2
3. Let us put the approximate values of √3.
4. If we take √3 = 1.7, the length will not be equal to the actual √3.
5. This is shown in fig.b. The green rectangle will have a lesser area than the required yellow area
6. If we take √3 = 1.73, then also, the length will not be equal to the actual √3.
7. This is shown in fig.c. The green rectangle, still will have a lesser area than the required yellow area. But this time, the result is better than that in fig.b
8. If we take √3 = 1.732, then also, the length will not be equal to the actual √3.
9. This is shown in fig.d. The green rectangle, still will have a lesser area than the required yellow area. But this time, the result is better than that in fig.c
10. So, we find that, when the number of decimal places increases, the area of the green rectangle gets closer and closer to the required value
■ We can write: Area of the rectangle = 1 × 1.73205...= 1.73205... m2
The fig.16.26(a) below shows such a rectangle
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| Fig.16.26 |
1. The length of the rectangle in fig.(a) is √3, and the width is √2.
2. So for our rectangle, the area = (√3 × √2) m2
12. Squaring both sides we get: Area2 = (√3)2 × (√2)2
⇒ Area2 = 3 × 2 ⇒ Area2 = 6
2. So for our rectangle, the area = (√3 × √2) m2
3. Let us put the approximate values of √3 and √2.
4. If we take √3 = 1.7, the length will not be equal to the actual √3.
Also, if we take √2 = 1.4, the width will not be equal to the actual √2.
Also, if we take √2 = 1.4, the width will not be equal to the actual √2.
5. This is shown in fig.b. The green rectangle will have a lesser area than the required yellow area
6. If we take √3 = 1.73, the length will not be equal to the actual √3.
Also, if we take √2 = 1.41, the width will not be equal to the actual √2.
Also, if we take √2 = 1.41, the width will not be equal to the actual √2.
7. This is shown in fig.c. The green rectangle, still will have a lesser area than the required yellow area. But this time, the result is better than that in fig.b
8. If we take √3 = 1.732, the length will not be equal to the actual √3.
Also, if we take √2 = 1.414, the width will not be equal to the actual √2.
Also, if we take √2 = 1.414, the width will not be equal to the actual √2.
9. This is shown in fig.d. The green rectangle still will have a lesser area than the required yellow area. But this time, the result is better than that in fig.c
10. So, we find that, when the number of decimal places increases, the area of the green rectangle gets closer and closer to the required value
■ We can write: Area of the rectangle = 1.73205... × 1.41421...= 2.44948... m2
Based on the above discussion, we can derive a very interesting result:
11. Consider the equation: Area = √3 × √212. Squaring both sides we get: Area2 = (√3)2 × (√2)2
⇒ Area2 = 3 × 2 ⇒ Area2 = 6
13. In (10) above, we obtained Area = 2.44948... m2 .
14. So what is Area2 ? It is (2.44948...)2
15. Using a calculator, we will get (2.44948...)2 = 5.999952
16.This is very close to '6' that we obtained in (12)
17. Combining the results in (12), (14) and (15), we can write:
Area2 = │(√3)2 × (√2)2│ = │3 × 2│ = │6│ = │(2.44948...)2│ = │5.999952│
18. From (17) we can write: (√3)2 × (√2)2 = 6
19. We know that (a2 × b2)= (ab)2. Some examples:
• 32 × 42 = 9 × 16 = 144
♦ (3×4)2 = 122 = 144
• 52 × 82 = 25 × 64 = 1600
♦ (5×8)2 = 402 = 1600
So we can write: (√3)2 × (√2)2 = (√3 × √2)2
20. So (18) becomes: (√3 × √2)2 = 6
21. Taking square root on both sides we get: (√3 × √2) = √6
Thus we get an equation (related to irrational numbers) which is frequently used in science and engineering:
√x × √y = √xy
If we use a calculator, we can write numerous examples to prove the above. Some examples are:
• √2 × √7 = 1.4142... × 2.6457... = 3.74154894
♦ √(2 × 7) = √14 = 3.74165...
• √3 × √5 = 1.732050... × 2.236067... = 3.872979
♦ √(3 × 5) = √15 = 3.87298
• √2 × √7 = 1.4142... × 2.6457... = 3.74154894
♦ √(2 × 7) = √14 = 3.74165...
• √3 × √5 = 1.732050... × 2.236067... = 3.872979
♦ √(3 × 5) = √15 = 3.87298
It may be noted that, We had proved the above equation for rational numbers in earlier classes. Some examples:
• √25 × √16 = 5 × 4 = 20
♦ √(25 × 16) = √400 = 20
• √9 × √64 = 3 × 8 = 24
♦ √(9 × 64) = √576= 24
Here, √25, √16, √9 etc., are rational numbers. For such numbers, we had proved the validity of the equation in earlier classes. Now we proved that, it is valid for irrational numbers (like √2, √3, √5, √6 etc.,) also.
Now we will see a practical application of the above equation:
1. Consider ⊿ABC in fig.16.27(a) below. Both the legs AB and AC have a length of 3 cm.
Applying Pythagoras theorem, we get: Hypotenuse BC = √(32 + 32) = √(9 + 9) = √18
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| Fig.16.27 |
2. But 18 can be written as 9 × 2. So √18 can be written as √(9 × 2)
3. Now, applying the equation √x × √y = √xy, we get: √(9 × 2) = √9 × √2
4. That is., √18 = √9 × √2. But √9 = 3. So we can write:
5. √18 = 3 × √2 = 3√2
That means: √18 is 3 times √2. This can be proved geometrically as shown in fig.(b)
6. The 3 cm long leg AC is divided into 3 equal parts. Each part is 1 cm long. 'p' and 'q' are the points of division.
7. Similarly, the 3 cm long leg AB is also divided into 3 equal parts. Each part is 1 cm long. 'r' and 's' are the points of division.
8. Horizontal lines are drawn through p and q. Also vertical lines are drawn through r and s.
9. These horizontal and vertical lines divide the interior of the ⊿ABC into 3 squares and 3 right triangles
10. For our present discussion, we need only the three right triangles. They are: ⊿Cpt, ⊿tuv and ⊿vsB
11. Look at them carefully. We see that, all 3 of them have the same 1 cm long legs. So the 3 of them will have the same hypotenuse
12. Length of each hypotenuse = √(12 + 12) = √(1 + 1) = √2
13. There are three such hypotenuse. Total length of all three = 3 × √2 = 3√2
14. But these three hypotenuses make up the hypotenuse √18 of the original triangle. Thus we prove that 18 = 3√2
6. The 3 cm long leg AC is divided into 3 equal parts. Each part is 1 cm long. 'p' and 'q' are the points of division.
7. Similarly, the 3 cm long leg AB is also divided into 3 equal parts. Each part is 1 cm long. 'r' and 's' are the points of division.
8. Horizontal lines are drawn through p and q. Also vertical lines are drawn through r and s.
9. These horizontal and vertical lines divide the interior of the ⊿ABC into 3 squares and 3 right triangles
10. For our present discussion, we need only the three right triangles. They are: ⊿Cpt, ⊿tuv and ⊿vsB
11. Look at them carefully. We see that, all 3 of them have the same 1 cm long legs. So the 3 of them will have the same hypotenuse
12. Length of each hypotenuse = √(12 + 12) = √(1 + 1) = √2
13. There are three such hypotenuse. Total length of all three = 3 × √2 = 3√2
14. But these three hypotenuses make up the hypotenuse √18 of the original triangle. Thus we prove that 18 = 3√2
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