Question

Show that any positive odd integer is of the form 3m or 3m+1 or 3m+2 ,where m is some integer
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Answer 1

Their is some mistake in question . I m Providing both questions and their solution for odd integer and , which numbers has form , 3m, 3m + 1 or 3m +1 ?

Question 1) Show that any positive odd integer is of the form 6q+1, 6q+3 or 6q+5, where q is some integer ?

Solution :-

By Euclid division , we know that ,

→ a = bq + r , {0 ≤ r < b }

Let us assume that, a = any positive integer , b = 6 .

So,

→ a = 6q + r .

here, r is greater or equal to 0 and less than 6. So, r can be 0,1,2,3,4 and 5 .

Putting values of r one by one now , we get :-

when r = 0

→ a = 6q + 0

→ a = 6q

6 is divisible by 2, so it is an even number.

when r = 1

→ a = 6q + 1.

6 is divisible by 2, but 1 is not divisible by 2, so it is an odd number.

when r = 2

→ a = 6q + 2 .

6 is divisible by 2, and 2 also divisible by 2, so it is an odd number.

when r = 3

→ a = 6q + 3.

6 is divisible by 2, but 3 is not divisible by 2, so it is an odd number.

when r = 4

→ a = 6q + 4.

6 is divisible by 2, and 4 is also divisible by 2, so it is an even number.

when r = 5

→ a = 6q + 5.

6 is divisible by 2, but 5 is not divisible by 2, so it is an odd number.

Hence, we can conclude that, Any positive odd integer is of the form 6q+1, 6q+3 or 6q+5.

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Question 2) :- Show that the square of any positive integer is of the form 3m or 3m+1 where m is some integer. ?

Solution :-

By Euclid division , we know that ,

→ a = bq + r , {0 ≤ r < b }

Let us assume that, a = any positive integer , b = 3 .

So,

→ a = 3q + r .

here, r is greater or equal to 0 and less than 3. So, r can be 0,1, 2 .

Now, putting values of r one by one :-

Case 1) :- r = 0

→ a = 3q + 0

→ a = 3q

Squaring both sides,

→ a² = (3q)²

→ a² = 9q²

→ a² = 3(3q²)

→ a² = 3m { where m = 3q².}

Case 2) :- r = 1.

→ a = 3q + 1

Squaring both sides,

→ a² = (3q + 1)²

→ a² = (3q)² + 1 + 2 * (3q) * 1

→ a² = 9q² + 6b + 1

→ a² = 3(3q² + 2q) + 1

→ a² = 3m + 1 { where m = 3q² + 2q .}

Case 2) :- r = 2.

→ a = 3q + 1

Squaring both sides,

→ a² = (3q + 2)²

→ a² = 9q² + 4 + 2 * 2 * 3q

→ a2 = 9q² + 12q + 3 + 1

→ a² = 3(3q² + 4q + 1) + 1

→ a² = 3m + 1 { where m = 3q² + 4q + 1.}

Hence, we can conclude that, square of any positive integer is of the form 3m or 3m+1.