![]() We can write three consecutive integers as \(n\), \(n 1\) and \(n 2\), so the sum of three consecutive integers can be written as: \(n (n 1) (n 2)\) This shows the sum of three consecutive integers is a multiple of 3 in these cases, but to prove it is true in all cases, we can use algebra. Prove that the sum of three consecutive integers is a multiple of 3. If \(n\) is an integer, then the consecutive integers starting at \(n\) are \(n\), \(n 1\), \(n 2\), \(n 3\) and so on. For example, 15, 16, 17 are consecutive integers. ![]() Consecutive integersĬonsecutive integers are whole numbers that follow each other without gaps. Since \(n\) is an integer, the term in the bracket, \(3n 2\) will also be an integer, so we have shown that \(3a 2\) is always an odd number, as it can be written in the form \(2m 1\), where \(m\) is an integer. The first two terms of this expression, \(6n 4\) have a common factor of 2, so factorising gives: To obtain the “ 1” term, we can re-write the expression as: We need to show that this is an odd number we need to rearrange the expression into the form \(2m 1\), where m is an integer. Substituting this expression and simplifying gives Since a is an odd number, we can write \(a = 2n 1\), where \(n\) is an integer. Prove that \(3a 2\) is always an odd number. So the product of two odd numbers is always odd because \((2n 1)(2m 1) = 2(2nm n m) 1\). ![]() This means that the expression \(2(2nm n m) 1 \) represents an odd number, as it is 2 multiplied by an integer plus 1. Since \(n\) and \(m\) are integers, the expression inside the bracket, \(2nm n m\), will also be an integer. The first three terms have a common factor of 2, so the expression can be re-written as: Multiplying the two odd numbers together gives: To prove that it is true for all odd numbers, we can write two odd numbers as \(2n 1\) and \(2m 1\), where \(n\) and \(m\) are integers. Product is the value obtained by multiplying. Prove that the product of two odd numbers is always odd. This shows that whenever two even numbers are added, the total is also an even number because \(2n 2m = 2(n m)\). Since \(n\) and \(m\) are both integers, then \(n m\) will also be an integer, so the expression \(2(n m)\) represents an even number. This can be factorised to give \(2n 2m = 2(n m)\) We cannot use the same letter for both expressions, otherwise they would represent the same even number.Īdding the two even numbers gives \(2n 2m\). The second even number can be written as \(2m\), where \(m\) is also an integer. Write the first of the two even numbers as \(2n\), where \(n\) is an integer. ![]() This shows that the statement is true for these examples, but to prove that it is true all the time we must use algebra. Prove that whenever two even numbers are added, the total is also an even number. The expressions \(2n - 1\) and \(2n 1\) can represent odd numbers, as an odd number is one less, or one more than an even number. If \(n\) is an integer (a whole number), then the expression \(2n\) represents an even number, because even numbers are the multiples of 2. Using letters to stand for numbers means that we can make statements about all numbers in general, rather than specific numbers in particular. A mathematical proof is a sequence of statements that follow on logically from each other that shows that something is always true. ![]()
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