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8η τάξη
Course: 8η τάξη > Ενότητα 1
Μάθημα 3: Irrational numbersWorked example: classifying numbers
Deciding which number set 3.4028 belongs to: integers, rationals, or irrationals? Δημιουργήθηκε από Σαλ Καν και Monterey Ινστιτούτο Τεχνολογίας και Εκπαίδευσης.
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What number sets does
the number 3.4028 repeating belong to? And before even answering the
question, let's just think about what this represents. And especially what this
line on top means. So this line on top means
that the 28 just keep repeating forever. So I could express this number
as 3.4028, but the 28 just keep repeating. Just keep repeating on and
on and on forever. I could just keep writing
them forever and ever. And obviously, it's just easier
to write this line over the 28 to say that it
repeats forever. Now let's think about what
number sets it belongs to. Well, the broadest number set
we've dealt with so far is the real numbers. And this definitely belongs
to the real numbers. The real numbers is essentially
the entire number line that we're used to using. And 3.4028 repeating sits
someplace over here. If this is negative 1, this
is 0, 1, 2, 3, 4. 3.4028 is a little bit more
than 3.4, a little bit less than 3.41. It would sit right over there. So it definitely sits
on the number line. It's a real number. So it definitely is real. It definitely is
a real number. But the not so obvious question
is whether it is a rational number. Remember, a rational number is
one that can be expressed as a rational expression
or as a fraction. If I were to tell you that p is
rational, that means that p can be expressed as the
ratio of two integers. That means that p can be
expressed as the ratio of two integers, m/n. So the question is, can I
express this as the ratio of two integers? Or another way to think
of it, can I express this as a fraction? And to do that, let's actually
express it as a fraction. Let's define x as being
equal to this number. So x is equal to 3.4028
repeating. Let's think about
what 10,000x is. And the only reason why I want
10,000x is because I want to move the decimal point all the
way to the right over here. So 10,000x. What is that going
to be equal to? Well every time you multiply
by a power of 10, you shift the decimal one to the right. 10,000 is 10 to the
fourth power. So it's like shifting
the decimal over to the right four spaces. 1, 2, 3, 4. So it'll be 34,028. But these 28's just
keep repeating. So you'll still have the 28's
go on and on, and on and on, and on after that. They just all got shifted to the
left of the decimal point by five spaces. You can view it that way. That makes sense. It's nearly 3 and 1/2. If you multiply by 10,000,
you get almost 35,000. So that's 10,000x. Now, let's also think
about 100x. And my whole exercise here is
I want to get two numbers that, when I subtract them and
they're in terms of x, the repeating part disappears. And then we can just treat them
as traditional numbers. So let's think about
what 100x is. 100x. That moves this decimal point. Remember, the decimal point
was here originally. It moves it over to the
right two spaces. So 100x would be 300-- Let
me write it like this. It would be 340.28 repeating. We could have put the 28
repeating here, but it wouldn't have made
as much sense. You always want to write it
after the decimal point. So we have to write 28 again to
show that it is repeating. Now something interesting
is going on. These two numbers, they're
just multiples of x. And if I subtract the bottom one
from the top one, what's going to happen? Well the repeating part
is going to disappear. So let's do that. Let's do that on both sides
of this equation. Let's do it. So on the left-hand side of this
equation, 10,000x minus 100x is going to be 9,900x. And on the right-hand side,
let's see-- The decimal part will cancel out. And we just have to figure out
what 34,028 minus 340 is. So let's just figure this out. 8 is larger than 0, so we
won't have to do any regrouping there. 2 is less than 4. So we will have to do some
regrouping, but we can't borrow yet because we
have a 0 over there. And 0 is less than 3, so we
have to do some regrouping there or some borrowing. So let's borrow from
the 4 first. So if we borrow from the 4, this
becomes a 3 and then this becomes a 10. And then the 2 can now
borrow from the 10. This becomes a 9 and
this becomes a 12. And now we can do
the subtraction. 8 minus 0 is 8. 12 minus 4 is 8. 9 minus 3 is 6. 3 minus nothing is 3. 3 minus nothing is 3. So 9,900x is equal to 33,688. We just subtracted 340
from this up here. So we get 33,688. Now, if we want to solve
for x, we just divide both sides by 9,900. Divide the left by 9,900. Divide the right by 9,900. And then, what are
we left with? We're left with x is equal
to 33,688 over 9,900. Now what's the big
deal about this? Well, x was this number. x was
this number that we started off with, this number that
just kept on repeating. And by doing a little bit of
algebraic manipulation and subtracting one multiple of it
from another, we're able to express that same exact
x as a fraction. Now this isn't in simplest
terms. I mean they're both definitely divisible by 2
and it looks like by 4. So you could put this in lowest
common form, but we don't care about that. All we care about is the fact
that we were able to represent x, we were able to represent
this number, as a fraction. As the ratio of two integers. So the number is
also rational. It is also rational. And this technique we
did, it doesn't only apply to this number. Any time you have a number that
has repeating digits, you could do this. So in general, repeating
digits are rational. The ones that are irrational are
the ones that never, ever, ever repeat, like pi. And so the other things, I think
it's pretty obvious, this isn't an integer. The integers are the
whole numbers that we're dealing with. So this is someplace in
between the integers. It's not a natural number or a
whole number, which depending on the context are viewed
as subsets of integers. So it's definitely
none of those. So it is real and
it is rational. That's all we can
say about it.