NCERT Solution Class 9th Maths Chapter – 1 Number System Examples

Solution – Recall that to find a rational number between r and s, you can add r and
s and divide the sum by 2, that is 2/r+s lies between r and s. So, 3/2 is a number
between 1 and 2. You can proceed in this manner to find four more rational numbers between 1 and 2. These four numbers are 5/4, 11/8, 13/8, and 7/4

Solution – It is easy to see how the Greeks might have discovered 2 . Consider a square OABC, with each side 1 unit in length (see Fig. 1.6). Then you can see by the Pythagoras theorem that OB = 1²+1² = √2 How do we represent 2 on the number line?
This is easy. Transfer Fig. 1.6 onto the number line making sure that the vertex O
coincides with zero (see Fig. 1.7)

We have just seen that OB = √2 . Using a compass with Centre O and radius OB,
draw an arc intersecting the number line at the point P. Then P corresponds to √2 on the number line.

Solution – Let us return to Fig. 1.7.
Construct BD of unit length perpendicular to OB (as in Fig. 1.8). Then using the
Pythagoras theorem, we see that OD = (√2)² + 1² = √3. Using a compass, with
centre O and radius OD, draw an arc which intersects the number line at the point Q.
Then Q corresponds to √3 .
In the same way, you can locate √n for any positive integer n, after √n − 1 has been
located.

Solution –

What have you noticed? You should have noticed at least three things:
(i) The remainders either become 0 after a certain stage, or start repeating themselves.
(ii) The number of entries in the repeating string of remainders is less than the divisor
(in 10/3 one number repeats itself and the divisor is 3, in 1/7 there are six entries 326451 in the repeating string of remainders and 7 is the divisor).
(iii) If the remainders repeat, then we get a repeating block of digits in the quotient
(for 10/3, 3 repeats in the quotient and for 1/7 , we get the repeating block 142857 in the quotient).

Solution – We have 3.142678 =3142678/1000000, and hence is a rational number

Solution – Since we do not know what 0 3. is , let us call it ‘x’ and so
x = 0.3333…
Now here is where the trick comes in. Look at
10 x = 10 × (0.333…) = 3.333…
Now, 3.3333… = 3 + x, since x = 0.3333…
Therefore, 10 x = 3 + x
Solving for x, we get
9x = 3, i.e., x = 1/3

Solution – Let x = 1.272727… Since two digits are repeating, we multiply x by 100 to get
100 x = 127.2727…
So, 100 x = 126 + 1.272727… = 126 + x
Therefore, 100 x – x = 126, i.e., 99 x = 126
i.e., x =126/99 = 14/11
You can check the reverse that 14/11 = 1 27 . .

Example 9 : Show that 0.2353535… = 0 235 . can be expressed in the form p/q, where p and q are integers and q ≠ 0.

Solution – Let x = 0 235 . . Over here, note that 2 does not repeat, but the block 35
repeats. Since two digits are repeating, we multiply x by 100 to get
100 x = 23.53535…
So, 100 x = 23.3 + 0.23535… = 23.3 + x
Therefore, 99 x = 23.3
i.e., 99 x = 233/10, which gives x = 233/990
You can also check the reverse that 233/990 = 0 235

Solution – We saw that 1/7 = 0142857 . . So, you can easily calculate 2/7 = 0 285714.
To find an irrational number between 1/7 and 2/7, we find a number which is non-terminating non-recurring lying between them. Of course, you can find infinitely many such numbers.
An example of such a number is 0.150150015000150000…