**Notes**
**NCERT Solutions**
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**Revision Notes**
**Question 1)** Find the nature of the product (√2 -√3) ( √3 + √2) ?

Solution
(√2 -√3) ( √3 + √2)

=(√2 -√3) ( √2 + √3)

=2-3=-1

SO it is rational number

**Question 2)** Prove that the sum of a rational number and an irrational number is always irrational.

Solution
We will try to prove it contradiction method

Let z be the irration number, p/q is the rational number. We assume the sum is a rational number a/b

Then

z+ p/q = a/b

z= a/b - p/q = (aq-bp)/bq

Now aq,bp and bq are all integers as a,b,p,q are integers.So (aq-bp)/bq is a rational number but x is irrational number.So our assumption is wrong

The sum is a irrational number

**Question 3)** Prove that √5 is an irrational number.

Solution
We will try to prove it contradiction method

Let √5 be rational

then it must in the form of p/q [q is not equal to 0][p and q are co-prime]

√5=p/q

√5 X q = p

squaring on both sides

5q^{2} = p^{2}

p^{2} is divisible by 5

So p is divisible by 5

p = 5c [c is a positive integer] [squaring on both sides ]

p^{2} = 25c^{2}

So we can obtain below from both the above equation

5q^{2} = 25c^{2}

q^{2} = 5c^{2}

So q is divisble by 5

thus q and p have a common factor 5

there is a contradiction

so √5 is an irrational

**Question 4)** Show that 3 + 5√2 is an irrational number. Is sum of two irrational numbers always an irrational number?

**Question 5)** Prove that √3 is an irrational number and hence show that 2√3 is also an irrational number.

**Question 6)** Prove that 5 - √3 is an irrational number.

**Question 7)** Prove that 2√5 is an irrational number.

**Question 8)** Show that (√3+ √5)

^{2} is an irrational number.

**Question 9)** Prove that 4 - √5 is an irrational number.

**Question 10)** Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution
For a and b any two positive integer, we can always find unique integer q and r such that

a=bq + r , 0 ≤ r < b

Now on putting b=3 ,we get

a=3q+ r , 0 ≤ r < 3

a=3q , a^{3}=27q^{3} a^{3} =9m where m=3q^{3}

a=3q+1 , a^{3}=27q^{3} +27q^{2} +9q +1=9(3q^{3}+3q+q) +1 =9m+1 where m=3q^{3}+3q+q

a=3q+2 , a^{2}=27q^{3} +54q^{2} +36q +8=9(3q^{3}+6q+4q) +1 =9m+1 where m=3q^{3}+6q+4q

**Question 11)** Prove that √2 + 1/√2 is an irrational number

**Question 12)** Prove that for any positive integer n, n

^{3} – n is divisible by 6.

Solution
n^{3} - n =n(n^{2} -1) = n(n-1)(n+1)

Any Number can be represented in the form 3q,3q+1 and 3q+2

for n=3q, n(n-1)(n+1)= 3q(3q-1)(3q+1) Divisible by 3

for n=3q+1,n(n-1)(n+1)= (3q+1)(3q)(3q+2) Divisible by 3

for n=3q+2,n(n-1)(n+1)= (3q+2)(3q+1)(3q+3)=3(3q+2)(3q+1)(q+1) Divisible by 3

So product n(n-1)(n+1) is divisble by 3

Similarly

Any Number can be represented in the form 2m,2m+1

for n=2m, n(n-1)(n+1)= 2m(2m-1)(2m+1) Divisible by 2

for n=2m+1, n(n-1)(n+1)= (2m+1)(2m)(2m+1) Divisible by 2

So this is divisble by both 2 and 3 =6

**Question 13)** If n is rational and √m is irrational, then prove that (n + √m) is irrational.

**Question 14)** Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer
Solution
n,n+4,n+8,n+12,n+16 be integers.

where n can take the form 5q, 5q+1 ,5q+2 , 5q + 3 , 5q + 4.

**Case I** when n=5q

Then n is divisible by 5.

but neither of 5q+4 ,5q+8 , 5q + 12 , 5q + 16 is divisible by 5.

**Case II** when n=5q+1

Then n is not divisible by 5.

n+4 = 5q+1+4 = 5q+5=5(q +1),

which is divisible by 5.

but neither of 5q+1 ,5q+9 , 5q + 13 , 5q + 17 is divisible by 5.

**Case III** when n=5q+2

Then n is not divisible by 5.

n+8 = 5q+2+8 =5q+10=5(q+2),

which is divisible by 5.

but neither of 5q+2 ,5q+6 , 5q + 14 , 5q + 18 is divisible by 5.

**Case IV** when n=5q+3

Then n is not divisible by 5.

n+12 = 5q+3+12 =5q+15=5(q+3),

which is divisible by 5.

but neither of 5q+3 ,5q+7 , 5q + 11 , 5q + 19 is divisible by 5.

**Case V** when n=5q+4

Then n is not divisible by 5.

n+16 = 5q+4+16 =5q+20=5(q+4),
which is divisible by 5.

but neither of 5q+4 ,5q+8 , 5q + 12 , 5q + 16 is divisible by 5.

Hence, one of n, n+4,n+8,n +12 and n+16 is divisible by 5.

**Question 15)** Prove that √11 is irrational.

**Question 16)** Show that 3√2 is irrational.

**Question 17)** Show that 4

^{n} can never end with the digit zero for any natural number n.

Solution
A number ending with zero must have 2 and 5 as factors like 10=2X5 550=2X5X5X11

Here 4^{n} =(2X 2)^{n}

So it does not have any 5

Therefore we can say that it will not with zero

**Question 18)**The product of a non-zero rational and an irrational number is

(A) always irrational

(B) always rational

(C) rational or irrational

(D) one

Solution
The product is always irrational

We can prove it like below with contradiction method

Let y be irrational number and a/b is rational and Product is rationap p/q

then

ya/b = p/q

y= bp/aq

bp,aq are integers,so ratio is rational but y is irrational so our assumption is wrong. Product is always irrational

**Question 19)** Prove that √p + √q is irrational, where p, q are primes.

**Question 20)** Prove that one of any three consecutive positive integers must be divisible by 3.

Solution
Let three consecutive positive integers are n,n+1,n+2 where n is any positive inetger

Any Number can be represented in the form 3q,3q+1 and 3q+2

for n=3q, n=3q So it Divisible by 3

for n=3q+1,n+2=3q+3 Divisible by 3

for n=3q+2,n+1=3q+3 Divisible by 3

So one of any three consecutive positive integers must be divisible by 3.

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