Question

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1) Prove that n²-n is divisible by 2 for every positive integer n.



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



3) Prove that the product of two consecutive positive integers is divisible by 2.


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Answer 1

heyaa.........


1.Suppose the positive integer is n.


∴ n = 2q or n = 2q + 1         where q∈Z.

CASE 1:-  

n = 2q

∴ n² - n = (2q)² - 2q

             = 4q² - 2q

             = 2(2q² - q)


CASE 2:-

n = 2q + 1

∴n² - n = (2q + 1)² - (2q + 1)

           = 4q² + 4q + 1 - 2q - 1

           = 4q² + 2q

           = 2(2q² + q)


Thus, in any case, n² - n is divisible by 2.


Thus, n² - n is divisible by 2 for every positive integer n.




2.
let n be any positive integer and b=3

n =3q+r

where q is the quotient and r is the remainder

0_ <r<3

so the remainders may be 0,1 and 2

so n may be in the form of 3q, 3q=1,3q+2



# CASE-1


IF N=3q

n+4=3q+4

n+2=3q+2

here n is only divisible by 3


@ CASE 2

if n = 3q+1

n+4=3q+5

n+2=3q=3

here only n+2 is divisible by 3


$ CASE 3

IF n=3q+2

n+2=3q+4

n+4=3q+2+4

=3q+6

here only n+4 is divisible by 3




HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE


3.
Let one integer be y and other will be y + 1...So


y × (y + 1)


= y × y + y × 1


= y² + y


Now take 2 cases


# Case 1 - y is odd


3 × (3 + 1)


= 3 × 3 + 3


= 9 + 3 = 12, which is divisible by two


# Case 2 - y is even


2 × (2 + 1)


= 2 × 2 + 2


= 4 + 2


= 6, which is divisible by two


∴     hence Proven

hope this explaintion help to u---::)


tysm..#gozmit

Answer 2

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