Oops! It appears that you have disabled your Javascript. In order for you to see this page as it is meant to appear, we ask that you please re-enable your Javascript!

Notes for real numbers chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.

**(1) Euclidâ€™s Division Lemma:
**

**(2) Euclidâ€™s division algorithm:Â ****To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below:
**Step 1: Apply Euclidâ€™s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 â‰¤ r < d.

Step 2: If r = 0, d is the HCF of c and d. If r â‰ 0, apply the division lemma to d and r.

Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

** For Example:Â **Use Euclidâ€™s division algorithm to find the HCF of 135 and 225.

Step 1: Here 225 > 135, on applying the division lemma to 225 and 135, we get 225 = 135 x 1 + 90

Step 2: Since, remainder â‰ 0, we again apply division lemma to 135 and 90, we get 135 = 90 x 1 + 45

Step 3: Again, applying division lemma to 90 and 45, we get 90 = 45 x 2 + 0

The remainder has become zero. And since the divisor at this step is 45, the HCF of 135 and 225 is 45.

** For Example:Â **Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

Let

As 0 â‰¤ r < 4, the possible remainders could be 0, 1, 2 and 3.

So, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.

Now, since

Hence, any odd integer is of the form 4q + 1 or 4q + 3.

**(3) The Fundamental Theorem of Arithmetic:
**

** For Example:Â **The prime factors of 32760 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5 Ã— 7 Ã— 13 = 2

** For Example:Â **Find the HCF and LCM of 96 and 404 by prime factorisation method.

The prime factorisation of 96 is 2

Hence, HCF of 96 and 404 will be 2

Now, LCM (96, 404) = (96 x 404)/(HCF (96, 404)) = (96 x 404)/4 = 9696.

** For Example:Â **Check whether 6

If a number ends with digit 0, then, it must be divisible by 10 or in other words, it will be divisible by 2 and 5 as 10 = 2 x 5.

Now, prime factorisation of 6

Here, 5 is not in the prime factorisation of 6

Thus, 6

**(4) Revisiting Irrational Numbers:
**

(i) a = p

(ii) On squaring both the sides, we get,

(iii) a

(iv) It is given that p divides a

(v) However, as per the uniqueness part of the Fundamental Theorem of Arithmetic, we can deduce that the only prime factors of a

Since, a = p

** Theorem 2:** âˆš2 is irrational.

(i) Let x and y have a common factor other than 1, and so we can divide by that common factor and assume that x and y are coprime. So, yâˆš2 = x.

(ii) Squaring both side, we get, 2y

(iii) Thus, 2 divides x

(iv) Hence, x = 2z for some integer z.

(v) Substituting x, we get, 2x

(vi) Now, from theorem, x and y will have 2 as a common factor. But, it is opposite to fact that x and y are co-prime.

(vii) Hence, we can conclude âˆš2 is irrational.

** For Example:Â **Prove that âˆš3 is irrational.

We shall start by assuming âˆš3 as rational. In other words, we need to find integers x and y such that âˆš3 = x/y.

Let x and y have a common factor other than 1, and so we can divide by that common factor and assume that x and y are coprime. So, yâˆš3 = x.

Squaring both side, we get, 3y

Thus, x

Hence, x = 3z for some integer z.

Substituting a, we get, 3x

Now, from theorem, x and y will have 3 as a common factor. But, it is opposite to fact that x and y are co-prime.

Hence, we can conclude âˆš3 is irrational.

** For Example:Â **Prove that 6 + âˆš2 is irrational.

Let us assume 6 + âˆš2 to be rational.

Therefore, we must find two integers a, b (b â‰ 0) such that

6 + âˆš2 = a/b i.e. âˆš2 = a/b â€“ 6.

Since, a and b are integers, a/b â€“ 6 is also rational and hence âˆš2 must be rational.

Now, this contradicts the fact that âˆš2 is irrational.

Hence, 6 + âˆš2 is irrational.

**(5) Revisiting Rational Numbers and Their Decimal Expansions:
**

** Theorem 2:** Let x = p/q be a rational number, such that the prime factorisation of q is of the form 2

** Theorem 3:** Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2

The prime factorisation of 6/15 can be written as

6/15 = (2 x 3)/(3 x 5) = 2/5

Here, the denominator is of the form 5

Hence, decimal expansion of 6/15 is terminating.

** For Example:Â **Write down the decimal expansions of 17/8.

The decimal expansion of 17/8 is

** For Example:Â **The following real number has decimal expansions as given below. Decide whether it is rational or not. If it is rational, and of the form, p/q what can you say about the prime factors of q?

Here, as the decimal expansion is non-terminating recurring, the given number is a rational number of the form p/q.

Moreover, q is not of the form 2

Not so helpful

Nice

Thanks for sharing this information.

Seems to be very helpful thanks for making our study easy

These are so helpful for me........

Thank you so much

Very nice

Thank you

Thank you so much

Very helpful sir

Thanks

thank you so much

Very nice really very helpful in making notes

Thanks alot

Thankyou so much for helping me This notes is so useful,

nice