Chapter 1: Real Numbers

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Real Numbers Solutions

Question:

Use Euclidโ€™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.

Answer:

[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  \geq 0 , and  r = 0,1,2, because of \leq  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 [ \because (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 3q 2  + 2². [ \because (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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