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I know you've been working with square roots for a while now, but let's back up
for a little bit and pretend that we're just starting out with square roots for
the first time, that we don't know much about them. All we know is that the
Pythagorean Theorem, that the square of the longest side in a right triangle is
equal to the sum of the squares of the other two sides.
So I look at this little right triangle here, each side has a length of 1, and so
I want to find the length of this hypotenuse right here
using the Pythagorean Theorem, so I think I'll call this length right here, "x"
and so what we have is this:
x² must be equal to 1² + 1²
x² must be equal to 1² + 1²
by the Pythagorean Theorem,
that means that x² = 1 + 1
that means that x² = 1 + 1, because 1² is just 1,
and 1 + 1 = 2, so x²
x² = 2.
Now here's the problem:
there's no
fraction and there's no decimal whose square is equal to 2. No matter how
hard I try, I can't find a fraction or a decimal that I can square to get 2.
If you think you know one, write it down, square it, and see what happens. You might
get closer to 2, but you'll never get exactly 2.
But I can see that here I have a line segment, right? That's going to have that length,
whatever length that is, whatever number is associated with it,
when I square it, I have to get 2.
So I'm going to make up new notation for that number whose square is 2, and
that notation looks like this:
x =
√(2).
That's the notation I use to represent the number whose square is 2
and I only use new notation here
when there is no old notation I could use.
And there's no fraction and no decimal whose square is 2.
√(2) is an irrational number. So, that tells me that the length
of this hypotenuse right here
is going to be
√(2).
Now what I want to do is I want to add another line segment onto this end of
this hypotenuse,
add a right angle, and I want to make its length equal to 1 also. So I'm going to do that.
add a right angle, and I want to make its length equal to 1 also. So I'm going to do that.
So I have a little graph paper here that I've made up to use as the ruler. There's four
squares per inch,
squares per inch,
or, I'm going to say there's four squares per inch, so one unit on here is
going to be four squares,
and all the lines meet each other at right angles.
So, if I simply line this up
so that one of these straight lines right here is right along that hypotenuse,
then all I have to do is go out to that point
then all I have to do is go out to that point
and move
1, 2, 3, 4 units
and what I have is a line segment
of length = 1
that meets
that hypotenuse at a right angle.
Now the question is, what's the length of this diagonal?
So when I draw this diagonal in right here,
what's going to be the length of that hypotenuse this right triangle?
So, I'm going to do the same thing I just did. I'm going to use the Pythagorean Theorem,
So, I'm going to do the same thing I just did. I'm going to use the Pythagorean Theorem,
let's call it "x" and so whenever x is
x²
is going to be equal to this side squared,
so √(2)²
so √(2)²
+ 1²
Okay, well √(2)², here's the number I don't know much about because I just
made up this notation down here.
As a matter of fact, I really only know one thing about this number right here,
and that is
if I square it, I end up with 2.
Well, look: I'm squaring it right here so I must get 2.
So x²
= 2
+ 1² which equals 1
x² =
3.
Now, I have the same problem I just had, there's no
fraction and there's no decimal whose square is 3,
so I have to use my new notation.
x² = √(3)
That's how I write it, and that stands for the number, the
positive number, whose square is three. So I'm going to fill this in now with √(3).
That's the length of that hypotenuse,
okay? I'm going to do the same thing again on the end of this
hypotenuse right here I want to add a line segment at a right angle
whose length is 1. So I'll line this up again
right along one of these lines so I know I'm going to get a right angle there, and
then I'll just move out 1,
2, 3,
4 units,
and that's what I'm calling 1 unit right here.
So that's 1, and now the question is,
what is the length of
this hypotenuse in this new
right triangle right here?
Well, I'll do the same thing I just did and I'll call it "x"
and so I have this x²
is equal to this length right here, which is √(3)
with my new notation, √(3)², plus this length, which is 1²
with my new notation, √(3)², plus this length, which is 1²
so x² =
again I don't know much about this number, in fact, the only thing I do know
about it is that if I square it, I get it 3. So that's 3
+ 1
x² = 4,
and now I have a number that I can square and get 4, and that number is 2.
So x = 2,
but I could also use my new notation
and write this down; √(4).
So, for this number right here, I could write √(4)
or just 2.
Okay, let's do it one more time
onto the end of this
diagonal right here,
I want to add
a line segment of length 1 that meets that
hypotenuse at a right angle, so I'll line up that line, and then I'll just move out
1,
2, 3,
4 units,
I know that this is 1 right here,
so what is the length of
this diagonal
right here?
So what do you think that length is going to be?
Using a little inductive reasoning we can guess that that's going to be
√(5). Let's try it and see.
This I'll call x, so x² =
This I'll call x, so x² =
this length right here, which is 2²
this length right here, which is 2²
plus this length right here which is 1²,
x² =
4 + 1
x² = 5,
and 5 is another one of those numbers
that doesn't have a number, a fraction or a decimal, to square to get it,
so use my new notation, x = √(5).
so use my new notation, x = √(5).
Here you can see I've got this spiral moving out, and each one of these
diagonals turns out to be the square root of one of the counting numbers like
that.
So I'm going to go over and get a few more pens for some
different colors and kind of finish this out right here, and then come back and
show you what it looks like. I'll be right back.
Okay, so if I continue what I'm doing and just keep going like this, what I end
up with is this
nice spiral which actually just keeps going forever and ever, and has
diagonals, or little spokes here you might say, that are the square root of all
the counting numbers.
What I like about this is it gives us a way to visualize all these irrational
numbers.
I mean at first, when we come across irrational numbers it seems kind of odd that
there's no fraction or decimal to square to get them,
but that turns out to be true, and that's why we have to use this new notation, the
square root notation,
to represent the number whose square is
2, 3, 5, 6, 7, 8, so on, and so forth,
because you can see that there are line segments that have those lengths so we
need a number to associate with those lengths
but there's no fraction or decimal that will square to get these things, so
we make up this new notation.
But this is a very attractive spiral right here, I think you can see that
looks maybe like one of those chambered nautilus things that you see in nature
something like that,
it's kind of fun to take this and expanded it out and see how it wraps around and just keeps
going.
But in any case, there's a look at the Pythagorean Theorem and a way we can use
it to visualize
a lot of the irrational numbers that we work with.