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Question 1. Asked on :17 August 2019:09:22:47 AM

 Prove that √2 is irrational?

-Added by Akki chauhan Mathematics » Real Numbers

Answer:

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 2. 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. 

-Added by ATP Admin Mathematics » Real Numbers

Answer:

ATP Admin

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. 

 

 



-Answered by ATP Admin On 17 June 2019:11:28:43 PM


 

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