Prove that the sum of three successive even integers is always a multiple of 6.
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Answer:
Step-by-step explanation:
<u>Errors in Algebraic Operations </u>
It's usual that students make mistakes when misunderstanding the application of algebra's basic rules. Here we have two of them
- When we change the signs of all the terms of a polynomial, the expression must be preceded by a negative sign
- When multiplying negative and positive quantities, if the number of negatives is odd, the result is negative. If the number of negatives is even, the result is positive.
- Not to confuse product of fractions with the sum of fractions. Rules are quite different
The first expression is
Let's arrange into format:
We can clearly see in all of the factors in the expression the signs were changed correctly, but the result should have been preceeded with a negative sign, because it makes 3 (odd number) negatives, resulting in a negative expression. The correct form is
Now for the second expression
Let's arrange into format
It's a clear mistake because it was asssumed a product of fractions instead of a SUM of fractions. If the result was correct, then the expression should have been