Prove that the square of any positive integer is from 3m or 3m +1 but not in the form of 3m+2
Answers
Answered by
165
Hi ,
Let a be any odd positive integer.
We apply the division lemma with
a and b = 3 .
Since 0 ≤ r < 3 ,
the possible remainders are 0 , 1
and 2 .
That is , a can be 3q , or 3q + 1 , or
3q + 2 , where q is the quotient .
Now ,
a² = ( 3q )² = 9q²
Which can be written in the form
= 3 ( 3q² )
= 3m , [ since m = 3q² ]
It is divisible by 3 .
Again ,
a² = ( 3q + 1 )²
= 9q² + 6q + 1
= 3( 3q² + 2q ) + 1
= 3m + 1 [ since m = 3q² + 2q ,
3( 3q² + 2q ) is divisible by 3 ]
Lastly ,
a² = ( 3q + 2 )²
= 9q² + 12q + 4
= ( 9q² + 12q + 3 ) + 1
= 3( 3q² + 4q + 1 ) + 1
Which can be written in the form
3m + 1 , since
3( 3q² + 4q + 1 ) is divisible by 3.
Therefore ,
The square of any positive integer
is either of the form 3m or 3m + 1
for some integer m .
I hope this helps you.
: )
Let a be any odd positive integer.
We apply the division lemma with
a and b = 3 .
Since 0 ≤ r < 3 ,
the possible remainders are 0 , 1
and 2 .
That is , a can be 3q , or 3q + 1 , or
3q + 2 , where q is the quotient .
Now ,
a² = ( 3q )² = 9q²
Which can be written in the form
= 3 ( 3q² )
= 3m , [ since m = 3q² ]
It is divisible by 3 .
Again ,
a² = ( 3q + 1 )²
= 9q² + 6q + 1
= 3( 3q² + 2q ) + 1
= 3m + 1 [ since m = 3q² + 2q ,
3( 3q² + 2q ) is divisible by 3 ]
Lastly ,
a² = ( 3q + 2 )²
= 9q² + 12q + 4
= ( 9q² + 12q + 3 ) + 1
= 3( 3q² + 4q + 1 ) + 1
Which can be written in the form
3m + 1 , since
3( 3q² + 4q + 1 ) is divisible by 3.
Therefore ,
The square of any positive integer
is either of the form 3m or 3m + 1
for some integer m .
I hope this helps you.
: )
Answered by
50
Step-by-step explanation:
let ' a' be any positive integer and b = 3.
we know, a = bq + r , 0 < r< b.
now, a = 3q + r , 0<r < 3.
the possibilities of remainder = 0,1 or 2
Case I - a = 3q
a² = 9q² .
= 3 x ( 3q²)
= 3m (where m = 3q²)
Case II - a = 3q +1
a² = ( 3q +1 )²
= 9q² + 6q +1
= 3 (3q² +2q ) + 1
= 3m +1 (where m = 3q² + 2q )
Case III - a = 3q + 2
a² = (3q +2 )²
= 9q² + 12q + 4
= 9q² +12q + 3 + 1
= 3 (3q² + 4q + 1 ) + 1
= 3m + 1 where m = 3q2 + 4q + 1)
From all the above cases it is clear that square of any positive integer ( as in this case a² ) is either of the form 3m or 3m +1.
Hence, it is solved .
THANKS
#BeBrainly .
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