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Show that any positive odd integer is of the 3m or 3m +2 where 'm' is some integer.

User Smdsgn
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Final answer:

Any positive odd integer can be of the forms 3m or 3m + 2. Since the remainder when an odd integer is divided by 3 can only be 1 or 2 and 3m + 1 represents an even number, we exclude this form.

Step-by-step explanation:

To show that any positive odd integer is of the form 3m or 3m + 2, where 'm' is some integer, let's consider the division of an integer by 3. When a positive integer is divided by 3, there are three possible remainders: 0, 1, or 2 (as the remainder must be less than the divisor, which is 3). For even integers, the remainders could be either 0 or 2, coinciding with the forms 3m and 3m + 2.

However, since we're dealing with positive odd integers, the remainder cannot be 0 because that would imply the integer is divisible by 3, making it even. Therefore, an odd integer can only have remainders of 1 or 2 when divided by 3, which correlates to the forms mentioned. Specifically:

  • If the remainder is 1, then the integer is of the form 3m + 1. Since the integer is odd, we can exclude this form and focus on:
  • If the remainder is 2, then the integer is of the form 3m + 2.

In conclusion, any positive odd integer can be expressed in the form of 3m + 2 when m is an integer that represents the quotient of dividing the integer by 3 (excluding the remainder).

User Hamed Mahdizadeh
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