These examples may lead you to expect the following facts, which are true:
(i) The sum or difference of a rational number and an irrational number is irrational.
(ii) The product or quotient of a non-zero rational number with an irrational number is
irrational.
(iii) If we add, subtract, multiply or divide two irrationals, the result may be rational or
irrational.
We now turn our attention to the operation of taking square roots of real numbers.
Recall that, if a is a natural number, then `sqrta=b` means `b^2=a` and `b>a` he same definition can be extended for positive real numbers.
Let `a > 0` be a real number. Then `sqrta = b` menas `b^2=a b > 0.`
In Section 1.2, we saw how to represent `sqrtn` for any positive integer n on the number line. We now show how to find `sqrtx` for any given positive real number x geometrically.
For example, let us find it for x = 3.5, i.e., we find `sqrt(3.5)` geometrically.
Mark the distance 3.5 units from a fixed point A on a given line to obtain a point B such that AB = 3.5 units (see Fig. 1.15). From B, mark a distance of 1 unit and mark the new point as C. Find the mid-point of AC and mark that point as O. Draw a semicircle with centre O and radius OC.
Draw a line perpendicular to AC passing through B and intersecting the semicircle at D.
Then, `BD = sqrt(3.5)`
More generally, to find `sqrtx` for any positive real number x, we mark B so that AB = x units, and, as in Fig. 1.16, mark C so that BC = 1 unit. Then, as we have done for the case x = 3.5, we find BD `= sqrtx`
(see Fig. 1.16).
We can prove this result using the Pythagoras Theorem.
Notice that, in Fig. 1.16, `Delta OBD` is a right-angled triangle. Also, the radius of the circle is `(x+1)/2`
Therefore, `OC = OD = OA = (x+1)/2`
Now, `OB = x - ((x+1)/2) = (x-1)/2`
So, by the Pythagoras Theorem, we have
`BD^2 = OD^2 –OB^2 = ((x+1)/2)^2 - (x-1)/2 = (4x)/4 =x`
Now, `OB = x- ((x+1)/2) = (x-1)/2`
So, by the Pythagoras Theorem, we have
`BD^2 = OD^2 – OB^2 = ((x+1)/2)^2 - ((x-1)/2)^2 = (4x)/4=x`
This shows that `BD = sqrtx`
This construction gives us a visual, and geometric way of showing that `sqrtx` exists for
all real numbers `x > 0. `
If you want to know the position of `sqrtx` on the number line, then let us treat the line BC as the number line, with B as zero, C as 1, and so on. Draw an arc with centre B and radius BD, which intersects the number line in E (see Fig. 1.17). Then, E represents `sqrtx`
We would like to now extend the idea of square roots to cube roots, fourth roots, and in general nth roots, where n is a positive integer. Recall your understanding of square roots and cube roots from earlier classes.
What is `8sqrt8` ? Well, we know it has to be some positive number whose cube is 8, and you must have guessed `8sqrt8 =2` . Let us try `5sqrt(243)` Do you know some number b such that `b^2=243?` The answer is 3. Therefore, `5sqrt(243)=3`
From these examples, can you define `nsqrta` for a real number `a > 0` and a positive integer n?
Let a > 0 be a real number and n be a positive integer. Then `'sqrt'` used in `sqrt2, 3sqrt8,nsqrta` etc. is called the radical sign.
We now list some identities relating to square roots, which are useful in various ways.
You are already familiar with some of these from your earlier classes. The remaining ones follow from the distributive law of multiplication over addition of real numbers, and from the identity `(x+y)(x-y)=x^2-y^2` for any real numbers x and y. Let a and b be positive real numbers. Then
`(i) sqrt(ab) = sqrtasqrtb`
`(ii) sqrt(a/b)=sqrt(a)/sqrt(b)`
`(iii) (sqrta+sqrtb)(sqrta-sqrtb)=a-b`
`(iv)(a+sqrtb)(a-sqrtb)=a2-b`
`(v)(sqrta+sqrtb)(sqrtc+sqrtd)=sqrt(ac)+sqrt(ad)+sqrt(bd)`
`(vi) (sqrta+sqrtb)^2=a+2sqrt(Ab)+b`
These examples may lead you to expect the following facts, which are true:
(i) The sum or difference of a rational number and an irrational number is irrational.
(ii) The product or quotient of a non-zero rational number with an irrational number is
irrational.
(iii) If we add, subtract, multiply or divide two irrationals, the result may be rational or
irrational.
We now turn our attention to the operation of taking square roots of real numbers.
Recall that, if a is a natural number, then `sqrta=b` means `b^2=a` and `b>a` he same definition can be extended for positive real numbers.
Let `a > 0` be a real number. Then `sqrta = b` menas `b^2=a b > 0.`
In Section 1.2, we saw how to represent `sqrtn` for any positive integer n on the number line. We now show how to find `sqrtx` for any given positive real number x geometrically.
For example, let us find it for x = 3.5, i.e., we find `sqrt(3.5)` geometrically.
Mark the distance 3.5 units from a fixed point A on a given line to obtain a point B such that AB = 3.5 units (see Fig. 1.15). From B, mark a distance of 1 unit and mark the new point as C. Find the mid-point of AC and mark that point as O. Draw a semicircle with centre O and radius OC.
Draw a line perpendicular to AC passing through B and intersecting the semicircle at D.
Then, `BD = sqrt(3.5)`
More generally, to find `sqrtx` for any positive real number x, we mark B so that AB = x units, and, as in Fig. 1.16, mark C so that BC = 1 unit. Then, as we have done for the case x = 3.5, we find BD `= sqrtx`
(see Fig. 1.16).
We can prove this result using the Pythagoras Theorem.
Notice that, in Fig. 1.16, `Delta OBD` is a right-angled triangle. Also, the radius of the circle is `(x+1)/2`
Therefore, `OC = OD = OA = (x+1)/2`
Now, `OB = x - ((x+1)/2) = (x-1)/2`
So, by the Pythagoras Theorem, we have
`BD^2 = OD^2 –OB^2 = ((x+1)/2)^2 - (x-1)/2 = (4x)/4 =x`
Now, `OB = x- ((x+1)/2) = (x-1)/2`
So, by the Pythagoras Theorem, we have
`BD^2 = OD^2 – OB^2 = ((x+1)/2)^2 - ((x-1)/2)^2 = (4x)/4=x`
This shows that `BD = sqrtx`
This construction gives us a visual, and geometric way of showing that `sqrtx` exists for
all real numbers `x > 0. `
If you want to know the position of `sqrtx` on the number line, then let us treat the line BC as the number line, with B as zero, C as 1, and so on. Draw an arc with centre B and radius BD, which intersects the number line in E (see Fig. 1.17). Then, E represents `sqrtx`
We would like to now extend the idea of square roots to cube roots, fourth roots, and in general nth roots, where n is a positive integer. Recall your understanding of square roots and cube roots from earlier classes.
What is `8sqrt8` ? Well, we know it has to be some positive number whose cube is 8, and you must have guessed `8sqrt8 =2` . Let us try `5sqrt(243)` Do you know some number b such that `b^2=243?` The answer is 3. Therefore, `5sqrt(243)=3`
From these examples, can you define `nsqrta` for a real number `a > 0` and a positive integer n?
Let a > 0 be a real number and n be a positive integer. Then `'sqrt'` used in `sqrt2, 3sqrt8,nsqrta` etc. is called the radical sign.
We now list some identities relating to square roots, which are useful in various ways.
You are already familiar with some of these from your earlier classes. The remaining ones follow from the distributive law of multiplication over addition of real numbers, and from the identity `(x+y)(x-y)=x^2-y^2` for any real numbers x and y. Let a and b be positive real numbers. Then
`(i) sqrt(ab) = sqrtasqrtb`
`(ii) sqrt(a/b)=sqrt(a)/sqrt(b)`
`(iii) (sqrta+sqrtb)(sqrta-sqrtb)=a-b`
`(iv)(a+sqrtb)(a-sqrtb)=a2-b`
`(v)(sqrta+sqrtb)(sqrtc+sqrtd)=sqrt(ac)+sqrt(ad)+sqrt(bd)`
`(vi) (sqrta+sqrtb)^2=a+2sqrt(Ab)+b`
Check whether `7 sqrt5, 7/(sqrt5) , sqrt2 +21, pi -2` are irrational numbers or not.
