The number system involves the collection of varied types, namely – natural numbers denoted by N which doesn’t include the number 0 (1, 2, ….. ), whole numbers denoted by W which includes 0 (0, 1, 2, …..), integers that consist of positive and negative numbers denoted by the letter Z, and rational numbers denoted by Q and that can be represented in the form of p/q. The number system in Class 9 assists the students in understanding the differences between the number system types such as rational and irrational numbers, in which irrational numbers cannot be written in the form of p/q, and also regarding real numbers. A real number is one that can be expressed as a point that is unique in nature, on the number line. Also, each point on the number line denotes a unique real number. Hence, the name, real number line.
The main topics focused on are irrational numbers, real numbers expansion in the decimal form, real numbers. The ncert solutions for Class 9 of Maths chapter 1 consist of the representation of numbers that are real on a number line, arithmetic operations on real numbers and exponent rules related to real numbers. Now let us see Problems Based on Number System Class 9 below.
Problem 1: State whether the following statements are true or false.
a] Each whole number is a natural number.
b] Each integer is a rational number.
c] Each rational number is an integer.
Answer:
a] Each whole number is a natural number.
The above statement is false.
The number zero is a whole number but not a natural number.
b] Each integer is a rational number.
The above statement is true.
Each integer “a” can be written in the form of a/1, and hence it is a rational number.
c] Each rational number is an integer.
The above statement is false.
For example, 3/5 is not an integer.
Problem 2: Check whether the numbers 7√5, 7/√5, √2 + 21, π – 2 are irrational numbers in nature or not.
Answer:
On evaluating the given numbers mathematically,
√5 = 2.36…
√2 = 1.4142….
π = 3.1415…
7√5 = 15.652…
7/√5 = 7√5/(√5 √5) = 7√5/5 = 3.1304…
√2 + 21 = 22.4142…
π – 2 = 1.1415…
The above numbers are non-terminating & non-recurring decimals and hence termed irrational numbers.
Problem 3: Add the numbers 2√2 + 5√3 and √2 – 3√3.
Answer:
2√2 + 5√3 and √2 – 3√3
= (2√2 + √2) + (5√3 – 3√3)
= (2 + 1)√2 + (5 – 3)√3
= 3√2 + 2√3
Problem 4: Multiply 6√5 by 2√5.
Answer:
6√5 by 2√5
6√5 * 2√5
= (6 * 2) * (√5 * √5)
= 12 * 5
= 60
Problem 5: Divide 8√15 by 2√3.
Answer:
8√15 by 2√3
8√15 ÷ 2√3
= [8√3 * √5]/2√3
= 4√5
Problem 6: Simplify the following given expressions.
a] (5 + √7) (2 + √5)
b] (5 + √5) (5 – √5)
c] (√3 + √7)2
d] (√11 – √7) (√11 + √7)
Answer:
a] (5 + √7) (2 + √5)
= 10 + 5√5 + 2√7 + √35
b] (5 + √5) (5 – √5)
= 52 – (√5)2
= 25 – 5
= 20
c] (√3 + √7)2
= (√3 + √7) (√3 + √7)
= (√3)2 + 2√3 √7 + (√7)2
= 3 + 2√21 + 7
= 10 + 2√21
d] (√11 – √7) (√11 + √7)
= (√11)2 – (√7)2
= 11 – 7
= 4
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