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Real Numbers Solutions
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1]
Proof: Suppose a is any positive integer and b = 3. By using Euclid’s algorithm we get a = 3q + r, for some integer q 0 , and r = 0,1,2, because of 0 r 3.
Now substituting the value of r, we get,
If r = 0, then a = 3q
Similarly, for r = 1 and 2, the value of a is 3q + 1 and 3q + 2respectively.
Now, we can find the square of values 3q , 3q + 1 and 3q + 2 . So, we have the following cases.
Case I: When a = 3q
(3q)² = 9q², if we divide by 3, then the remainder is zero.
we can write it as 3 x (3q²). So it is in the form of 3m, where m = 3q².
Case II: When a = 3q +1
(3q + 1)² = (3q)² + 2 3q 1 + 1² [ (a + b)² = a² + 2ab + b²]
= 9q² + 6q + 1, if we divide by 3, then the remainder is 1.
We can write it as 3(3q² + 2q) + 1. So it is in the form of 3m + 1 and the value of m is 3q² + 2q.
Case III: When a = 3q + 2
(3q+2)² = (3q)² + 2 x 3q x 2 + 2². [ (a + b)² = a² + 2ab + b²]
= 9q² + 12q + 4, if we divide by 3, then the remainder is 1.
We can write it as 3(3q² + 4q +1) + 1. So it is in the form of 3m + 1 and the value of m will be 3q² + 4q +1.
Square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
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