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    Exercise 10.1.2

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6th class > > Exercise 10.1.2

Exercise 10.1.2

1. Prove that 5 is irrational

This means that √5 can be expressed as a fraction ab ,√5=ab.
Square both sides of the equation to get rid of the square root 5 = a2/b2.
Multiply both sides by b2 to clear the fraction : 5b2 = a2.
Substitute a=5k into the equation : 5b2 = a2 : 5b2 = 5k2 : b2 = 5k2.
At this point, we have shown that both a and b are multiples of .
Since our assumption that √5 is rational leads to a contradiction, we conclude that √5 is irrational.

2. Prove that 3+2√5 is irrational.

This means that 3+2√5 can be expressed as a fraction ​ab,3+2√5=ab.
Isolate 2√5 on one side of the equation : 2√5=ab-3.
Express the right-hand side with a common denominator : 2√5=a−3bb.
Isolate √5 : √5=a−3b2b.
Since a and b are integers, a−3b2b is a rational number.
Since our assumption that 3+2√5 is rational leads to a contradiction, we conclude that 3+2√5 is irrational.

3. Prove that the following are irrationals.

A rational number is defined as any number that can be expressed as a fraction where both the numerator and the denominator are integers, and the denominator is not zero.

(i) 1/2

Here, the numerator 1 and the denominator 2 are both integers.

Since 12can be expressed as a fraction of integers, it is a rational number, not irrational.

Therefore, 12is .

(ii)75

75 can be expressed as 751.

Here, 75 is an integer, and it can be written as a fraction where the denominator is 1, an integer.

Therefore,75 is a .

(iii)6+2

Calculate = 6 + 2 =.

8 can be expressed as 81.

Here,8 is an integer, and it can be written as a fraction where the denominator is 1, an integer.

Therefore,8 is a .

The graphs of y = p(x) are given in Figs.Below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

The Number of Zeroes is .

The Number of Zeroes is .

The Number of Zeroes is .

The Number of Zeroes is .

The Number of Zeroes is .

The Number of Zeroes is .