Honors Calculus I - Irrational Numbers - Math Worksheet With Answers Page 2
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1
2
3side must also be even. That is
is even. But
is a whole number, and so Fact 1 again
tells us that
must be even. But if
and
are both even, they must have a factor of 2 in
common. This gives our final contradiction, and so the initial assumption of the rationality
of
2 is false.
2.
Show that
is not rational.
Suppose that there are whole numbers
and
such that
=
. This means that
24 5 =
2, and so we have that
2 is also rational. But this contradicts part 1 above, so
cannot be rational.
Prove that there are infinitely many irrational numbers.
There are infinitely many rational numbers
. The argument used for
above also shows
that
is irrational for each
. Thus we have infinitely many irrationals of the form
.
3.
How might you prove that
2 is not a rational number?
You would give an (almost verbatim; some equations will have 8’s instead of 4’s, and squares
will be replaced by cubes) version of the proof of the irrationality of
2 in 1 above, except
that you would appeal to Fact 2 instead of Fact 1.
Fact 2. If the cube of a whole number is even, then the whole number must also be even.
Proof. As in the proof of Fact 1, if
= 2
+ 1 is an odd whole number (where
is another
whole number), then
= (2
+ 1) = 8
+ 12
+ 6
+ 1 = 2(4
+ 6
+ 3 ) + 1
is also an odd number. Thus the only way that
can be even is for
to be even.
4.
How might you prove that
3 is not a rational number?
The proof is very similar to the proof of irrationality of
2 in 1 above, but this time we
cannot talk about even numbers. Instead we have to look at multiples of three. Fact 3 below
is the analog of Fact 1 which will be used in the proof.
Fact 3. Suppose that the square of a whole number is a multiple of 3, then the whole number
must also be a multiple of 3.
Proof. Let
be a whole number whose square is a multiple of 3. We show that
also must
be a multiple of 3, by ruling out the cases where
is not a multiple of three. There are two
cases to consider: either
has a remainder of 1 when divided by 3, or
has a remainder of
2 when divided by 3.
In the first case, we can write
= 3
+ 1 for some other whole number
. Then
= (3
+ 1) = 9
+ 6
+ 1 = 3(3
+ 2 ) + 1
has a remainder of 1 and so is not a multiple of 3.
In the second case, we can write
= 3
+ 2 for some other whole number
.Then
= (3
+ 2) = 9
+ 12
+ 4 = 3(3
+ 4
+ 1) + 1
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