Class 10 real numbers extra questions

Those rational number which can be expressed in form x/2^m X5ⁿ    are terminating expression and those can not be are non terminating decimal expression
Terminating decimal:  (a), (d)
Non terminating repeating decimal: (b), (c), (e), (f), (g).(h) ,(i)

a.
Using Euclid theorem
$867=225 \times 3 +192$
$225=192 \times 1 +33$
$192=33 \times 5+ 27$
$33=27 \times 1+6$
$27=6 \times 4+3$
$6=3 \times 2+0$
So solution is 3
b. 8

a. Let’s take this as rational number
$\frac{a}{b}=2+\sqrt{3}$
Or
$\frac{a-2b}{b}=\sqrt{3}$
Since a rational number can’t be equal to irrational number, our assumption is wrong
b. Let’s take this as rational number
q=3√3
q/3=√3
Since a rational number can’t be equal to irrational number, our assumption is wrong

1. True
2. True
3. False, as it is written as 6
4. True ,as any integer can be expressed in the form p/q
5. True
6. False,There are finite integer between two integers
7. False
8. False
9. False
10. True

Answer is (d)
LCM X HCF=aXb

Answer is (c)

a

(a)

(d)

(a)
$ 27= 3 \times 3 \times 3$
$81=3 \times 3 \times 3 \times 3$
HCF=27
LCM=81
b.
$120=2 \times 2 \times 3 \times 2 \times 5$
$144=2 \times 2 \times 3 \times 2 \times 2 \times 3$
$HCF=2^3 \times 3=24$
LCM=720
c.
$29029= 29 \times 13 \times 11 \times 7$
$580=29 \times 5 \times 4$
HCF=29
LCM=$29 \times 13 \times 11 \times 7 \times 4 \times 5=580580$

$p^2+16=q^2$
$(q-p)(q+p)=16$
So we have
Case 1
$q-p=8$ and $q+p=2$ which gives p=3
Case 2
$q-p=4$ and $q+p=4$ which gives p=0
Case 3
$q-p=2$ and $q+p=8$ which gives p=3
So the answer is 3 only

$(\sqrt {2} - \sqrt {3}) ( \sqrt {3} + \sqrt {2})$
$=(\sqrt {2} - \sqrt {3}) ( \sqrt {2} + \sqrt {3})$
=2-3=-1
So, it is rational number

A number ending with zero must have 2 and 5 as factors like 10=2X5 550=2X5X5X11
Here 4ⁿ =(2X 2)ⁿ
So it does not have any 5
Therefore we can say that it will not with zero

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   , a³=27q³  a³ =9m  where m=3q³
a=3q+1  , a³=27q³ +27q² +9q +1=9(3q³+3q+q) +1 =9m+1 where m=3q³+3q+q
a=3q+2  , a²=27q³ +54q² +36q +8=9(3q³+6q+4q) +1 =9m+1 where m=3q³+6q+4q

We can solve it using contradiction method.
Let $3 + 5 \sqrt {2} $ is a rational number, then
$3 + 5 \sqrt {2} = \frac {p}{q}$
$3q+ 5q\sqrt {2} = p$
$\sqrt {2} = \frac {p-3q}{5q}$
Clearly,RHS is a rational number ,but LHS is not.
So,our contradiction is wrong.
$3 + 5 \sqrt {2} $ is an irrational number

sum of two irrational numbers is not always an irrational number
Example
$3 + 5 \sqrt {2} $ and $3 - 5 \sqrt {2} $ are irrational number, but the sum is rational number
$3 + 5 \sqrt {2} + 3 - 5 \sqrt {2}= 6$

Since we cannot clearly express in p/q form,it is difficult to say,So lets us assume this is rational number
then
$\sqrt {3}=\frac {p}{q}$

where p and q are co primes.
or
$q\sqrt {3}=p$
squaring both sides
$3q^2=p^2$

So 3 divides p²,from theorem we know that,
2 will divide p also. p=3c

$3q^2=9c^2$
or
$q^2=3c^2$
So q divided by 3 also

So both p and q are divided by 3 which is contradiction from we assumed
So $\sqrt {3}$ is irrational number

let $5 - \sqrt {3}$ be rational number
then,
$5 - \sqrt {3}=\frac {p}{q}$
$\sqrt {3}= \frac {5q -p}{q}$
Clearly RHS is rational number but LHS is not,
So our assumption is wrong.
$5 - \sqrt {3}$ is an irrational number

Let $(\sqrt {3} + \sqrt {5})^2= 8+ 2 \sqrt {15}$ is a rational number
Then
$8+ 2 \sqrt {15}=\frac {p}{q}$
$\sqrt {15}= \frac {p -8q}{2q}$
Clearly RHS is rational number but LHS is not,
So our assumption is wrong.
$(\sqrt {3} + \sqrt {5})^2$ is an irrational number.

We will try to prove it contradiction method
Let $\sqrt {5}$ be rational
then it must in the form of p/q [q is not equal to 0][p and q are co-prime]
$\sqrt {5}=\frac {p}{q}$
$\sqrt {5}$ \times q = p$
squaring on both sides
$5q^2 = p^2$
So, p² is divisible by 5
Now from the theorem, p is divisible by 5
let $p = 5c$ [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 divisible by 5
thus q and p have a common factor 5
there is a contradiction
so $\sqrt {5}$ is an irrational number

n³ - n =n(n² -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 divisible 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 divisible by both 2 and 3 =6

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.

We will try to prove it contradiction method
Let z be the irrational 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

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 rational 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

Let three consecutive positive integers are n,n+1,n+2 where n is any positive integer
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.

(1) Composite
(2) prime
(3) remainder
(4) Euclid
(5) quotient
(6) Terminating
(7) Rational