Video Transcript
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 example.
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 together.
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 proof.
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 here.
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 one.
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 four
𝑛, 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.
