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Question 1. Asked on :29 June 2021:11:08:46 AM

Prove that √5 is an irrational number.

-Added by Master Purushottam Mathematics » Real Numbers

Master Purushottam

Let us assume that √5 is a rational number.

we can write it as √5 = pq

Now we can reduce p and q and we get reduced value of p and q as a and b which are co-prime have no common factor other than 1.

So we have √5 = ab

⇒ √5b = a

⇒ 5b2 = a2   [squaring both sides]  ............ (i)

⇒ b2 = a25

5 divides a2 so a is also divisible by 5   (theorem 1.3) ..... (ii)

so 5 is a factor of a.

Now taking a = 5c for some integer c

⇒ 5b2 = (5c)2

⇒ 5b2 = 25 c2

⇒ b2 = 5 c2

⇒ c2 = b25

5 divides b2 so b is also divisible by 5 (theorem 1.3)  ...... (iii)

from (ii) and (iii) we get that

a and b are divisible by 5 this implies that 5 is common factor of a and b, while we assume that a and b are co-prime having no common factor other than 1.

This contradiction has risen due to our wrong assumption so √5 cann't be

expressed as pq, Hence √5 is an irrational number.

-Answered by Master Purushottam On 29 June 2021:11:10:35 AM

Question 2. Asked on :24 June 2021:06:26:16 PM

Show that any odd positive integer is of the form 4q + 1 or 4q + 3, where q is some integer.

-Added by Master Mind Mathematics » Real Numbers

Master Mind

Let the a is the any odd positive interger

To Show that: a = 4q + 1 or 4q + 3

Using Euclid's division algorithem

a = bq + r

∴ b = 4

so, possible remainder will be 0, 1, 2, 3 ( 0 ≤ r < b)

Hence, a is an odd positive integer therefore remainder will be also odd.

∴ r = 1 or 3

Now expressing a in the form of bq + r

a = 4q + 1 or 4q + 3

Hence proved

-Answered by Master Mind On 24 June 2021:06:28:42 PM

Question 3. Asked on :24 June 2021:06:10:31 PM

If the HCF of 210 and 55 is expressible in the form 210 × 5 + 55y then find y.

-Added by Master Mind Mathematics » Real Numbers

Master Mind
HCF of 210 and 55 using Euclid's division algorithem

210 = 55 × 3 + 45

55 = 45 × 1 + 10

45 = 10 × 4 + 5

10 = 5 × 2 + 0

∴ HCF = 5

Now 210 × 5 + 55y = 5

1050 + 55y = 5

55y = 5 - 1050

55y = - 1045

y = - 104555 = - 19

y = - 19

-Answered by Master Mind On 24 June 2021:06:15:15 PM

Question 4. Asked on :17 August 2019:09:22:47 AM

Prove that √2 is irrational?

-Added by Akki chauhan Mathematics » Real Numbers

Himanshi Verma
Let √2 be a rational number

Therefore, √2= p/q  [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0

On squaring both sides, we get
p²= 2q²           ...(1)
Clearly, 2 is a factor of 2q²
⇒ 2 is a factor of p²          [since, 2q²=p²]
⇒ 2 is a factor of p

Let p =2 m for all m ( where  m is a positive integer)

Squaring both sides, we get
p²= 4 m²          ...(2)
From (1) and (2), we get
2q² = 4m²      ⇒      q²= 2m²
Clearly, 2 is a factor of 2m²
⇒       2 is a factor of q²          [since, q² = 2m²]
⇒       2 is a factor of q

Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1

Therefore, Our supposition is wrong

Hence √2 is not a rational number i.e., irrational number.

-Answered by Himanshi Verma On 18 August 2019:11:09:55 AM

Question 5. Asked on :17 June 2019:11:25:45 PM

Show that one and only one out of n, n+2 or n+4 is divisible by 3, where n is any positive integer.

Solution: Using Euclid's division lemma any positive integer can be written in the form of a = bq + r where r = 0, 1, 2 ...... and q is quotients.

Let the number which is divisible by 3 be 3q + 0 or 3q + 1 or 3q + 2 where [0 <= r < b]

Now n = 3q or n = 3q + 1 or n = 3q + 2

Case I,

When n = 3q  .......... (i)

⇒ n = 3(q) where n is divisible by 3

Adding 2 both sides in equ. (i)

We have,

n + 2 = 3q + 2 Where n + 2 is not divisible by 3

Now adding 4 both side in equ. (i)

We have,

n + 4 = 3q + 4 Where n + 4 is not divisible by 3

Case II

Taking n = 3q + 1 ........ (ii) where n is not divisible by 3

Adding 2 both sides in equ. (ii)

We have,

n + 2 = 3q + 1 + 2 = 3q + 3

n + 2 = 3(q + 1) where n + 2 is divisible by 3

Now adding 4 both sides in equ. (ii)

n + 4 = 3q + 1 + 4 = 3q + 5 where n + 4 is not divisible by 3

Case III

taking n = 3q + 2 .... (iii) where n is not divisible by 3

Adding 2 both sides in equ. (iii)

We have,

n + 2 = 3q + 2 + 2 = 3q + 4 where n + 2 is not divisible by 3

Now adding 4 both sides in equ. (iii)

We have,

n + 4 = 3q + 2 + 4 = 3q +6

n + 4 = 3(q + 2) where n + 4 is divisible by 3

Hence in all these three cases we have seen that either one and only one n or n + 2 or n + 4 is divisible by 3.

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