1.Euclid's division lemma

Euclid's Division Lemma: An Introduction

According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + r where 0 ≤ r < b.

The basis of the Euclidean division algorithm is Euclid’s division lemma. To calculate the Highest Common Factor (HCF) of two positive integers a and b we use Euclid’s division algorithm. HCF is the largest number which exactly divides two or more positive integers. That means, on dividing both the integers a and b the remainder is zero.

Let us now get into the working of this Euclidean algorithm.

Euclid’s Division Lemma Algorithm

Consider two numbers 78 and 980 and we need to find the HCF of these numbers. To do this, we choose the largest integer first, i.e. 980 and then according to Euclid Division Lemma, a = bq + r where 0 ≤ r < b;

980 = 78 × 12 + 44

Now, here a = 980, b = 78, q = 12 and r = 44.

Now consider the divisor 78 and the remainder 44, apply Euclid division lemma again.

78 = 44 × 1 + 34

Similarly, consider the divisor 44 and the remainder 34, apply Euclid division lemma to 44 and 34.

44 = 34 × 1 + 10

Following the same procedure again,

34 = 10 × 3 + 4

10 = 4 × 2 + 2

4 = 2 × 2 + 0

As we see that the remainder has become zero, therefore, proceeding further is not possible. Hence, the HCF is the divisor b left in the last step. We can conclude that the HCF of 980 and 78 is 2.

Let us try another example to find the HCF of two numbers 250 and 75. Here, the larger the integer is 250, therefore, by applying Euclid Division Lemma a = bq + r where 0 ≤ r < b, we have

a = 250 and b = 75

⇒ 250 = 75 × 3 + 25

By applying the Euclid’s Division Algorithm to 75 and 25, we have:

75 = 25 × 3 + 0

As the remainder becomes zero, we cannot proceed further. According to the algorithm, in this case, the divisor is 25. Hence, the HCF of 250 and 75 is 25.

Example

Example: Find the HCF of 81 and 675 using the Euclidean division algorithm.

Solution: The larger integer is 675, therefore, by applying the Division Lemma a = bq + r where 0 ≤ r < b, we have

a = 675 and b = 81

⇒ 675 = 81 × 8 + 27

By applying Euclid’s Division Algorithm again we have,

81 = 27 × 3 + 0

We cannot proceed further as the remainder becomes zero. According to the algorithm, in this case, the divisor is 27. Hence, the HCF of 675 and 81 is 27.

This algorithm has got many practical applications in finding the properties of numbers. There is a lot more left to learn in real numbers. Enrich your knowledge by visiting our website www.byjus.com and download BYJU’S-the learning app and learn anywhere.

2.Arithmetic Fundamental Theorem

Fundamental Theorem Of Arithmetic

Fundamental Theorem of Arithmetic states that every integer greater than 1 is either a prime number or can be expressed in the form of primes. In other words, all the natural numbers can be expressed in the form of the product of its prime factors. To recall, prime factors are the numbers which are divisible by 1 and itself only. For example, the number 35 can be written in the form of its prime factors as:

35 = 7 × 5

Here, 7 and 5 are the prime factors of 35

Similarly, another number 114560 can be represented as the product of its prime factors by using prime factorization method,

114560 = 27 × 5 × 179

So, we have factorized 114560 as the product of the power of its primes.

 

Therefore, every natural number can be expressed in the form of the product of the power of its primes. This statement is known as the Fundamental Theorem of Arithmetic, unique factorization theorem or the unique-prime-factorization theorem.

Fundamental Theorem of Arithmetic

Proof for Fundamental Theorem of Arithmetic

In number theory, a composite number is expressed in the form of the product of primes and this factorization is unique apart from the order in which the prime factor occurs.

From this theorem we can also see that not only a composite number can be factorized as the product of their primes but also for each composite number the factorization is unique, not taking into consideration order of occurrence of the prime factors.

In simple words, there exists only a single way to represent a natural number by the product of prime factors. This fact can also be stated as:

The prime factorization of any natural number is said to be unique for except the order of their factors.

In general, a composite number “a” can be expressed as,

a = p1 p2 p3 ………… pn, where p1, p2, p3 ………… pn are the prime factors of a written in ascending order i.e. p1≤p2≤p3 ………… ≤pn.

Writing the primes in ascending order makes the factorization unique in nature.

Fundamental Theorem of Arithmetic Examples

Example Question: In a formula racing competition the time taken by two racing cars A and B to complete 1 round of the track is 30 minutes and 45 minutes respectively. After how much time will the cars meet again at the starting point?

Solution:

As the time taken by car B is more compared to that of A to complete one round therefore it can be assumed that A will reach early and both the cars will meet again when A has already reached the starting point. This time can be calculated by finding the L.C.M of the time taken by each.

30 = 2 × 3 × 5

45 = 3 × 3 × 5

The L.C.M is 90.

Thus, both cars will meet at the starting point after 90 minutes.

3.Irrational and Rational Numbers

decimal expansions of rational numbers in terms of terminating/non-terminating recurring decimals

Decimal Expansion Of Rational Numbers

Before going into a representation of the decimal expansion of rational numbers, let us understand what rational numbers are. Any number that can be represented in the form of p/q, such that p and q are integers and q ≠ 0 are known as rational numbers. So when these numbers have been simplified further, they result in decimals. Let us learn how to expand such decimals here.

Examples: 6 , -8.1,\frac{4}{5} etc. are all examples of rational numbers.

 

How to Expand Rational Numbers in Decimals?

The real numbers which are recurring or terminating in nature are generally rational numbers.

Decimal - Expansion Of Rational Numbers

For example, consider the number 33.33333……. It is a rational number as it can be represented in the form of 100/3. It can be seen that the decimal part .333…… is the non-terminating repeating part, i .e. it is a recurring decimal number.

Also the terminating decimals such as 0.375, 0.6 etc. which satisfy the condition of being rational (0.375 = \frac{3}{2^3} ,0.6 = \frac{3}{5}).

Consider any decimal number. For e.g. 0.567. It can be written as 567/1000 or \frac{567}{10^3} . Similarly, the numbers 0.6689,0.032 and .45 can be written as \frac{6689}{10^4} ,\frac{32}{10^3} and \frac{45}{10^2} respectively in fractional form.

Thus, it can be seen that any decimal number can be represented as a fraction which has denominator in powers of 10. We know that prime factors of 10 are 2 and 5, it can be concluded that any decimal rational number can be easily represented in the form of \frac{p}{q}, such that p and q are integers and the prime factorization of q is of the form 2^x~ 5^y, where x and y are non-negative integers.

This statement gives rise to a very important theorem.

Theorems

Theorem 1: If m be any rational number whose decimal expansion is terminating in nature, then m can be expressed in form of \frac{p}{q}, where p and q are co-primes and the prime factorization of q is of the form 2^x~ 5^y, where x and y are non-negative integers.

The converse of this theorem is also true and it can be stated as follows:

Theorem 2: If m is a rational number, which can be represented as the ratio of two integers i.e. \frac{p}{q} and the prime factorization of q takes the form 2^x~ 5^y, where x and y are non-negative integers then, then it can be said that m has a decimal expansion which is terminating.

Consider the following examples:

  1. \frac{7}{8} = \frac{7}{2^3} = \frac{7~×~5^3}{2^3~×~5^3} = \frac{875}{10^3}
  2. \frac{3}{80} = \frac{3}{2^4~×~5} = \frac{3~×~5^3}{2^4~×~5^4} = \frac{375}{10^4}

Moving on, to decimal expansion of rational numbers which are recurring, the following theorem can be stated:

Theorem 3: If m is a rational number, which can be represented as the ratio of two integers i.e. \frac{p}{q} and the prime factorization of q does not takes the form 2^x~ 5^y, where x and y are non-negative integers. Then, it can be said that m has a decimal expansion which is non-terminating repeating (recurring).

Consider the following examples:

  1. \frac{1}{6} = 0.1666…. = 0.1\overline{6}
  2. \frac{7}{12} = 0.58333… = 0.58\overline{3}
  3. \frac{9}{11} = 0.8181… = 0.\overline{81}

Rational Number to decimal Examples

Case 1: Remainder equal to zero

Example: Find the decimal expansion of 3/6.

Decimal Expansion of Rational Numbers - Example 1

Here, the quotient is 0.5 and the remainder is 0. Rational number 3/6 results in a terminating decimal.

Case 2: Remainder not equal to zero 

Example: Express 5/13 in decimal form.

Decimal Expansion of Rational Numbers Example -2

Here, the quotient is 0.384615384 and the remainder is not zero. Notice that the number…384 after the decimal is repeating. Hence, 5/13 gives us a non-terminating recurring decimal expansion. And this can be written as 5/13=recurring

A rational number gives either terminating or non-terminating recurring decimal expansion. Thus, we can say that a number whose decimal expansion is terminating or non-terminating recurring is rational. 

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Quadratic Equations

Quadratic Equation

When we equate a quadratic polynomial to a constant, we get a quadratic equation.

Any equation of the form p(x)=c, where p(x) is a polynomial of degree 2 and c is a constant, is a quadratic equation.

The standard form of a Quadratic Equation

The standard form of a quadratic equation is ax2+bx+c=0, where a,b and c are real numbers and a0.
‘a’ is the coefficient of x2. It is called the quadratic coefficient. ‘b’ is the coefficient of x. It is called the linear coefficient. ‘c’ is the constant term.

 

Solving QE by Factorisation

Roots of a Quadratic equation

The values of x for which a quadratic equation is satisfied are called the roots of the quadratic equation.

If α is a root of the quadratic equation ax2+bx+c=0, then aα2+bα+c=0.

A quadratic equation can have two distinct real roots, two equal roots or real roots may not exist.

Graphically, the roots of a quadratic equation are the points where the graph of the quadratic polynomial cuts the x-axis.

Consider the graph of a quadratic equation x24=0:

Quadratic Equations for Class 10 -1

Graph of a Quadratic Equation

In the above figure, -2 and 2 are the roots of the quadratic equation x24=0
Note:

  • If the graph of the quadratic polynomial cuts the x-axis at two distinct points, then it has real and distinct roots.
  • If the graph of the quadratic polynomial touches the x-axis, then it has real and equal roots.
  • If the graph of the quadratic polynomial does not cut or touch the x-axis then it does not have any real roots.

Solving a Quadratic Equation by Factorization method

Consider a quadratic equation 2x25x+3=0
2x22x3x+3=0
This step is splitting the middle term
We split the middle term by finding two numbers (-2 and -3) such that their sum is equal to the coefficient of x and their product is equal to the product of the coefficient of x2 and the constant.
(-2) + (-3) = (-5)
And (-2) × (-3) = 6
2x22x3x+3=0
2x(x1)3(x1)=0
(x1)(2x3)=0
In this step, we have expressed the quadratic polynomial as a product of its factors.
Thus, x = 1 and x =3/2 are the roots of the given quadratic equation.
This method of solving a quadratic equation is called the factorisation method.

 

 

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Solving QE by Completing the Square

Solving a Quadratic Equation by Completion of squares method

In the method of completing the squares, the quadratic equation is expressed in the form (x±k)2=p2.

Consider the quadratic equation 2x28x=10
(i) Express the quadratic equation in standard form.
2x28x10=0

(ii) Divide the equation by the coefficient of x2 to make the coefficient of x2 equal to 1.
x24x5=0

(iii) Add the square of half of the coefficient of x to both sides of the equation to get an expression of the form x2±2kx+k2.
(x24x+4)5=0+4

(iv) Isolate the above expression, (x±k)2 on the LHS to obtain an equation of the form (x±k)2=p2
(x2)2=9

(v) Take the positive and negative square roots.
x2=±3

x=1 or x=5

 

Solving QE Using Quadratic Formula

Quadratic Formula

Quadratic Formula is used to directly obtain the roots of a quadratic equation from the standard form of the equation.

For the quadratic equation ax2+bx+c=0,

x= [-b± √(b2-4ac)]/2a

By substituting the values of a,b and c, we can directly get the roots of the equation.

 

Discriminant

For a quadratic equation of the form ax2+bx+c=0, the expression b24ac is called the discriminant, (denoted by D), of the quadratic equation.
The discriminant determines the nature of roots of the quadratic equation based on the coefficients of the quadratic equation.

 

 

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Solving using Quadratic Formula when D>0

Solve 2x27x+3=0 using the quadratic formula.
(i) Identify the coefficients of the quadratic equation. a = 2,b = -7,c = 3
(ii) Calculate the discriminant, b24ac
D=(7)24×2×3= 49-24 = 25
D> 0, therefore, the roots are distinct.
(iii) Substitute the coefficients in the quadratic formula to find the roots

x= [-(-7)± √((-7)2-4(2)(3))]/2(2)

x=(7 ±5)/4

x=3 and x= 1/2 are the roots.

Nature of Roots

Based on the value of the discriminant, D=b24ac, the roots of a quadratic equation can be of three types.

Case 1: If D>0, the equation has two distinct real roots.

Case 2: If D=0, the equation has two equal real roots.

Case 3: If D<0, the equation has no real roots.

 

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To know more about Nature of Roots, visit here.

Be More Curious

Graphical Representation of a Quadratic Equation

The graph of a quadratic polynomial is a parabola. The roots of a quadratic equation are the points where the parabola cuts the x-axis i.e. the points where the value of the quadratic polynomial is zero.

Now, the graph of x2+5x+6=0 is:

Quadratic Equations for Class 10 -2

In the above figure, -2 and -3 are the roots of the quadratic equation
x2+5x+6=0.

For a quadratic polynomial ax2+bx+c,

If a>0, the parabola opens upwards.
If a<0, the parabola opens downwards.
If a = 0, the polynomial will become a first-degree polynomial and its graph is linear.

The discriminant, D=b24ac

Quadratic Equations for Class 10 -3

Nature of graph for different values of D.

If D>0, the parabola cuts the x-axis at exactly two distinct points. The roots are distinct. This case is shown in the above figure in a, where the quadratic polynomial cuts the x-axis at two distinct points.

If D=0, the parabola just touches the x-axis at one point and the rest of the parabola lies above or below the x-axis. In this case, the roots are equal.
This case is shown in the above figure in b, where the quadratic polynomial touches the x-axis at only one point.

If D<0, the parabola lies entirely above or below the x-axis and there is no point of contact with the x-axis. In this case, there are no real roots.
This case is shown in the above figure in c, where the quadratic polynomial neither cuts nor touches the x-axis.

Formation of a quadratic equation from its roots

To find out the standard form of a quadratic equation when the roots are given:
Let α and β be the roots of the quadratic equation ax2+bx+c=0. Then,
(xα)(xβ)=0
On expanding, we get,
x2(α+β)x+αβ=0, which is the standard form of the quadratic equation. Here, a=1,b=(α+β) and c=αβ.

Sum and Product of Roots of a Quadratic equation

Let α and β be the roots of the quadratic equation ax2+bx+c=0. Then,
Sum of roots =α+β=-b/a
Product of roots =αβ= c/a

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