In this video, we’re going to look at how to write out a series of algebraic
expressions and statements to prove whether or not a statement is true. When setting out algebraic proofs, you must always explain any assumptions
you’re making, define whatever variables you use, and clearly write out a series of statements
which tell a logical story which proves or disproves the starting statement. Let’s look at an
Prove that the sum of — so the sum is a keyword there — the sum of any odd number
and any even number is odd. So there’re a number of things we’ve got to do here. We’ve got to think about an
algebraic expression which encapsulates the idea of an odd number or an even number. And we
then got to sum- we’ve got to add a couple of expressions together and rearrange that expression
in such a way that it proves that it must be an odd number.
Well thinking about it, an even number is an integer which is divisible by two.
And an odd number is an integer which isn’t divisible by two. If we take an integer and double it, the result must be an even number. We’ve
taken an integer and multiplied it by two, so the result must be divisible by two. If we take an
integer and double it and then add one on top of that, then the result must be an odd number
because it won’t be divisible by two.
Our question asks for any odd number and any even number, so these two things
are completely unrelated. So we’re gonna start off with two completely unrelated integers and
we’re gonna call them 𝑚 and 𝑛. So we’re just gonna write this down then.
Let 𝑚 and 𝑛 be integers. Then we’re gonna come up with some algebraic expressions which explain the
idea of relative even, so then two 𝑚 must be an even number. Remember, we’ve doubled an
integer, so that must be an even number. And two 𝑛 plus one must be odd. So that’s defined our expressions for an odd number and an even number. And
now we’ve got to get across this idea of summing them. Well summing them is just adding them
So we need to state that in our proof. Summing these even and odd numbers
gives two 𝑚 plus two 𝑛 plus one.
And we can rearrange that. We’ve got two 𝑚 and we’ve got two 𝑛. So if I put those two terms
together and factor out the two, I’ve got two lots of 𝑚 plus 𝑛. And then we’ve added
one to that. We’ve still got a plus one on the end. And logically if 𝑚 is an integer and 𝑛 is an integer and you add them together,
you must get the result of an integer.
So this first part of our expression here, two brackets 𝑚 plus 𝑛
close brackets or two parentheses 𝑚 plus 𝑛 close parentheses is two times an integer and
that is an even number. And our expression was two 𝑚 plus 𝑛 plus one, so this means we’ve got an even
number plus one which must be an odd number.
So we can see that algebraic proofs require you to do lots of definitions and
lots of explaining. They’re not like a lot of other maths cause they can be quite wordy. They require you to define your variables, in our case 𝑚 and 𝑛 being integers.
They require you to set out your logic. We’ve explained how to come up with even and odd
numbers. And then they require you to come up with some expressions, in this case the sum of an
odd and an even number. And then do a bit of rearranging of those expressions to deliver the
Now there is quite a lot of writing and you could probably get away without
putting all of those words in there. And for things like this means we could — there’s another
word for that, therefore — so therefore, this is true. And we could write that as three dots like
this. So there are sort of a few symbols that we can use in order to save a few words on the page,
but essentially you’ve got to be prepared to tell the story. That’s what proofs are all about.
Okay next example. Prove that the sum of any three consecutive integers is a
multiple of three. Now in this example, we’re not looking for odd and even numbers, but we are
looking for consecutive integers, consecutive numbers. And we’re not looking for odd or even. As
we say, we’re looking for multiples of three.
We need to think how we can express this concept, multiple of three, as an
algebraic expression. And we need to explain each logical step in our proof. So we could start off by defining 𝑛 to be an integer. So if we add one to that and then add one to it again, we’ve got three
consecutive integers 𝑛, 𝑛 plus one, and 𝑛 plus two.
So we can then state our general algebraic expression representing three
consecutive integers: 𝑛, 𝑛 plus one, 𝑛 plus two. And we’re summing them, so we can add them
together: 𝑛 plus 𝑛 plus one plus 𝑛 plus two.
Now we can take out that expression and we can simplify it or rearrange it.
And 𝑛 plus 𝑛 plus 𝑛 is three 𝑛 and one plus two is three.
Now if we look back to the question, we’re looking for an expression to
represent a multiple of three. Well that’s gonna be three times some integer, so looking at a little
expression down here: three 𝑛 plus three. If we factor out the three, if what’s inside the
parentheses is an integer, then we have a multiple of three. So taking out that common factor of three outside the parentheses, we’re left with
𝑛 plus one inside.
Now let’s just set up this next bit and define what a multiple of three is. Well
it’s just an integer multiplied by three after all. Remember, we defined 𝑛 to being an integer at the beginning of our proof, so
𝑛 plus one is an integer plus one is another integer. So our expression is three times an integer, therefore, we have a multiple of three. Now
this isn’t just our final answer. In these proofs, the whole argument, everything you write down,
makes up the final answer. You can’t just write the final answer down. You have to have all
those steps along the way. In some format or other, the exact wording might be slightly
different, but you do need to include all those steps.
In our last example, we are back to odd numbers, but were also looking at
multiples of four. So we’ve gotta unpick the wording to see how we can represent the process with
algebra, but it’s still important to explain each logical step in the proof. Prove that the
difference between the squares of any two consecutive odd numbers — so we’ve got consecutive odd
numbers. We were looking for the difference between those squares — is always a multiple of four.
So we’re gonna be looking to define an odd number, which is gonna mean
defining a variable which is an integer and then doubling it and adding one or subtracting one.
We’re gonna be looking at an integer times four, a multiple of four, and the difference between the
squares, so we’re gonna be squaring these odd numbers and then subtracting one from the other. So we’re gonna start off again, let 𝑛 be an integer. By doubling an integer, we’ve definitely got an even number. And if we had one to
that, we’ve definitely got an odd number. So we’ve got our first odd number that we’re looking at
And then the next consecutive odd number are for the two 𝑛 plus one would be
just two bigger than that, so two 𝑛 plus one plus two, which is two 𝑛 plus three, obviously. Now we could equally have to find two consecutive odd numbers as being
two 𝑛 minus one and two 𝑛 plus one. And that would make the algebra slightly
different, although the steps will be basically the same. I am gonna stick with two 𝑛 plus one and
two 𝑛 plus three for now, but remember you could do this in a slightly alternative way.
And then the difference between the squares of those consecutive odd numbers
is simply the biggest of the two squared minus the smallest of the two squared. I mean I could
put those the other way round, but this is gonna make them that slightly easier. It’s just gonna leave us
with a positive result. And now I’m just going to expand and simplify those- that expression. So
two 𝑛 plus three all squared is two 𝑛 plus three times two 𝑛 plus three. And two 𝑛 plus one all squared is two 𝑛 plus one times two 𝑛 plus
So multiplying each term in the first parentheses by each term in the second
parentheses gives us four 𝑛 squared plus six 𝑛 plus six 𝑛 plus nine. And the second part, the second
term in that expression, gives us four 𝑛 squared plus two 𝑛 plus two 𝑛 plus one. Now I’ve included
brackets here and here because we’re taking away the whole of that expression. So we need to be
really really careful with those signs, so much better to include brackets in these places to
try and stop us from making mistakes later on in the calculation.
So just tidying up what’s inside the parentheses slightly, I’ve got four 𝑛
squared plus twelve 𝑛. That’s six 𝑛 plus six 𝑛 plus nine minus four 𝑛 squared plus four 𝑛 which was two 𝑛 plus two 𝑛 plus
one. Now remember, we were subtracting the entire second term. So that’s- we’re
taking away four 𝑛 squared, we’re taking away four 𝑛, and we’re taking away one. So simplifying that big term then, I’ve got four 𝑛 squared take away four 𝑛
squared, well that’s nothing. So those two cancel out. I’ve got twelve 𝑛 take away
which is eight 𝑛. And I’ve got nine take away one, which is eight. So that whole expression there can be simplified down to eight 𝑛 plus eight.
Right, now I’ve wrapped all that lot out. I’m not suggesting that you don’t show
that in your working out. It’s just that I’ve run out of space on the page here, so that is important to
include all your steps of working out in your solution. So for the last step then, we’re looking to show algebraically that that
expression can be a multiple of four. So we want four times an integer, so I’m gonna factor out four. And we’ve gotta try and prove that what’s inside this bracket here is an integer.
And if that is an integer, then we’ve got four times an integer. That’s the definition of a
multiple of four. And since 𝑛 is an integer, if we double an integer, we’re still gonna get an
integer. And if we add another integer to it, we’ve also still got an integer. So we can say that four brackets two 𝑛 plus two close bracket is four times
an integer, i.e. a multiple of four and that’s that proved.
So in our algebraic proofs, we need to make sure that we define any variables
that we’re going to use, for example let 𝑛 be an integer. You also have to state algebraically how you can express what you’re trying to
prove. So for example, if 𝑛 is an integer, then two 𝑛 must be even. Or if 𝑛 is an integer, two 𝑛 plus
one or two 𝑛 minus one must be an odd number or four times an integer is a multiple of four and so on.
And finally, you need to express the statement you’re trying to prove
algebraically and then rearrange it to support the statement. So for example, we had three 𝑛
plus three. Well if we factored out the three, we knew that 𝑛 was an integer. So
𝑛 plus one was an integer, so we had three times an integer which proves that that must be a
multiple of three.
So overall, algebraic proofs can be a little bit wordy, but they’re such
powerful tools in mathematics that you really do need to master them and use them and take
pride in telling a fantastic logical mathematical story.