We provide you here with notes for Class 10 mathematics Chapter 1 Real Numbers, these notes for Class 10 Mathematics Real number provide you explanations of Euclid’s Division Lemma, The Fundamental Theorem of Arithmetic among other topics. They also contain solved examples for each type of mathematical problem present in Chapter 1 Real Numbers. These notes will help the student to understand and remember the concepts easily.
REAL NUMBERS: all the rational and irrational numbers together form real numbers.
RATIONAL NUMBERS: numbers which can be expressed in the form p/q , where q ≠ 0 and p and q are co – primes.
IRRATIONAL NUMBERS: numbers which are not rational. Those numbers which cannot be expressed in the form p/q.
EUCLID’S DIVISION LEMMA: For any two given integers a and b, there exist unique integers q and r such that a = bq + r, such that
[0 ≤ r < b] where,
a = dividend
b = divisor
q = quotient
r = remainder
→ dividend = (divisor × quotient) + remainder
Example: Use euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m.
Solution: Let a be any positive integer. On dividing a by 3, let quotient be (q) and remainder be (r).
Since, b=divisor=3 → r=0, 1 or 2 → a= 3q+r
Case 1: [when r=0]
a=3
squaring both sides…
a2=9q2 → a2=3(3q2)
a2=3m where m=3q2 is an integer.
Case 2: [when r=1]
a=3q+1
squaring both sides…
a2=9q2+1+6q → a2=3(3q2+2q)+1
a2=3m+1 where m=3q2+2q is an integer.
Case 3:[when r=2]
a=3q+2
squaring both sides….
a2=9q2+4+12q → a2=9q2+12q+3+1
[splitting 4= 3+1]
a2=3(3q2+4q+1)+1
a2=3m+1 where m= 3q2+4q+1 is an integer.
Hence, the square of any positive integer is of the form 3m or 3m+1 for some integer m.
EUCLID’S DIVISION ALGORITHM
Euclid’s division algorithm is a procedure which is used to find HCF of two positive integers. The steps are as follows:
- Suppose the two numbers are a and b and apply euclid’s division lemma on them such that a=bq+r where 0 ≤r<b.
- If the remainder r=0, then the HCF of a and b is b. If the remainder r ≠ 0, apply euclid’s division lemma to b and r.
- Repeat the above step until the remainder is 0, at this stage, the divisor will be the required HCF.
Example: Use euclid’s division algorithm to find HCF of 462 and 246
Solution: Applying euclid’s division lemma on 462 and 246, we get c 462 = 246 × 1+216
Here, r≠0
Applying euclid’s division lemma on 246 and 216, we get
246= 216 × 1 +30
Here, r≠0
Applying euclid’s division lemma on 216 and 30, we get
216=30×7+6
Here, r≠0
Applying euclid’s division lemma on 30 and 6, we get
30=6×5+0
Here, r=0
Hence, the HCF of 462 and 216 is 6.
THE FUNDAMENTAL LAW OF ARITHEMATIC
Theorem → Every composite number can be expressed as the product of its primes, and this factorization is unique, apart from the order in which the prime factors occur.
Example → Express 10829 as the product of its prime factors.
Hence → 7×7×13×17=10829
Note → The product of two positive integers is equal to the product of their HCF and LCM.
a × b = HCF(a,b) × LCM(a,b)
Theorem → Let p be a prime number. If p divides a2, then p divides a.
Example: Prove that √a is irrational, where a is prime number.
Solution: Let us assume that √a is rational number.
√a = p/q where q ≠ 0, p and q are co-primes.
Squaring both sides…
a= p2/q2 → p2=aq2 ………..(1)
→ p2 is divisible by a.
→ p is divisible by a.
Let p=ac for some integer c.
Substituting p=ac in (1)
a2c2=aq2 → q2 =ac
→ q2 is divisible by a.
→ q is divisible by a.
But, this contradicts to the fact that p and q are co-primes, i.e..,
they cannot have common factor other than 1.
This contradiction is due to the wrong assumption that √a is rational. Hence, √a is irrational.
Example: Prove that √2 is irrational.
Solution: Let us assume that √2 is rational number.
√2 = p/q where q≠0, p and q are co-primes.
Squaring both sides…
2=p2/q2 → p2=2q2 ……(1)
→ p2 is divisible by 2.
→ p is divisible by 2.
Let p=2c for some integer c.
Substituting p=2c in (1)
2c2=q2
→ q2 is divisible by 2.
→ q is divisible by 2.
But, this contradicts to the fact that p and q are co-primes, i.e.., they cannot have common factor other than 1.
This contradiction is due to the wrong assumption that √2 is rational.
Hence, √2 is irrational.
Note:
- If there is a rational number x=p/q where prime factorization of q=2m5n where m and n are positive integers, then decimal expansion of x will terminate.
- If there is a rational number x=p/q where prime factorization of q≠2m5n, then decimal expansion of x will not terminate.
Example: Express 0.535353….. in the form of p/q.
Solution: Let x=0.535353……. → (1)
→100x=53.535353….. → (2)
Subtracting (1) from (2)…
99x=53
x= 53/99
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