euclid's division lemma for real numbers
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euclid's division lemma for real numbers

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.

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euclid's division lemma for real numbers

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Mar 27, 2011
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by: Staff


The question:

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.



The answer:

When any positive integer is divided by 3, the results are always another integer “h” plus a remainder.

The REMAINDER of dividing a positive integer divided by 3 is ALWAYS: 0, 1, or 2

In other words, every integer is of the form: h + 0, or h + 1, or h + 2

Try it:

1 ÷ 3 = 0 (the “h” = 0) plus a remainder of 1

2 ÷ 3 = 0 (the “h” = 0) plus a remainder of 2

3 ÷ 3 = 1 (the “h” = 1) plus a remainder of 0

4 ÷ 3 = 1 (the “h” = 1) plus a remainder of 1

5 ÷ 3 = 1 (the “h” = 1) plus a remainder of 2

6 ÷ 3 = 2 (the “h” = 2) plus a remainder of 0
.
.
.


In other words, every POSITIVE INTEGER only has one of THREE forms:

h + 0

h + 1

h + 2


Since “h” is the result of dividing by 3, then “h” can be rewritten as 3 times another integer “k”, as follows:

h + 0 = 3k+0

h + 1 = 3k+1

h + 2 = 3k+2


“m” = another integer.

Squaring each of the three forms an integer can take (3k+0, 3k+1, and 3k+2), the results are:

(3k + 0)²
= 9k²
= 3*3k²
= 3*(another integer) = 3m


(3k + 1)²
= 9k² + 6k + 1
= 3*k(3k + 2) + 1
= 3*(another integer) + 1
= 3m + 1


(3k + 2)²
= 9k² + 12k + 4
= 3*(3k² + 4k) + 1
= 3*(another integer) + 1
= 3m + 1


Notice that while a positive integer can have three forms (h + 0, h + 1, or h + 2), the SQUARE of a positive integer has ONLY TWO POSSIBLE FORMS (3m and 3m + 1).


In conclusion, the square of a positive integer can only have two forms: 3m, or 3m + 1




Thanks for writing.


Staff
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