Two consecutive integers are
squared. Prove that the difference of these
two squared numbers is always an odd integer.
Remember a proof is a sequence of
statements that show that something is always true. It’s simply not enough to pick some
arbitrary values and show that it works for these numbers. It must work for every conceivable
value. Here that’s two consecutive
integers. In order to prove the statement
that the difference of these two squared numbers is always an odd integer, we’ll
need to use algebra.
We can define any integer which is
just a whole number as 𝑛. Consecutive integers are whole
numbers that follow on from one another with no gaps. So two, three, four, five, and so
on are all consecutive integers. That means our next integer would
be found by adding one. The next integer we’re interested
in then is 𝑛 plus one.
Next, we’re told that each of these
integers are squared. Squaring the first integer simply
gives us 𝑛 squared. However, when we square 𝑛 plus
one, we do need to be little bit careful. We write it as 𝑛 plus one in
brackets all squared. When we square something, we
multiply it by itself. So we’re going to write this
expression as 𝑛 plus one multiplied by 𝑛 plus one.
We can then use the FOIL method to
expand this pair of brackets. Multiplying the first term in the
first bracket by the first term in the second bracket is 𝑛 times 𝑛. That’s 𝑛 squared. The outer terms are 𝑛 and one. 𝑛 multiplied by one is simply
𝑛. The inner terms are one and 𝑛. One multiplied by 𝑛 is 𝑛. And the last two terms in the
brackets are one and one. One multiplied by one is one. Don’t forget to simplify by
collecting like terms. 𝑛 plus one all squared is
therefore 𝑛 squared plus two 𝑛 plus one.
Finally, we’re told to find the
difference of these integers. Difference means subtract. So we’ll subtract the smaller
number from the larger number. That’s 𝑛 squared plus two 𝑛 plus
one minus 𝑛 squared. 𝑛 squared minus 𝑛 squared is
zero. So our expression simplifies to two
𝑛 plus one.
But how do we prove that this is an
odd integer? Well, first, we define our original
integer to be 𝑛. That means that any multiple of 𝑛
must also be an integer. Further, any multiple of two is an
integer that appears in the two times tables — it’s an even number. That means then that two 𝑛 is an
even integer. If we take any even number and add
one, we get an odd number. That means then that two 𝑛 plus
one must be an odd integer. And we have proved our
statement. The difference of the two squared
numbers is always an odd integer.