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A
rational expression
is a ratio of polynomials:
Examples:
Whenever an expression containing
variables is present in the
denominator of a fraction,
you should be alert to the
possibility that certain values
of the variables might make
the denominator equal to zero,
which is forbidden. This means
that when we are talking about
rational expressions we can
no longer say that the variable
represents “any real number.”
Certain values may have to
be excluded. For example,
in the expression
 ,
we cannot allow the value
x = 0 so
we would parenthetically add
the comment (x ¹ 0),
and for
we would say (x ¹ 3)
. In the case of
we would exclude both x = 1
and x = –1,
since either choice would
make the denominator zero.
We don’t care if the numerator
is zero. If the numerator
is zero, that just makes the
whole rational expression
zero (assuming, of course,
that the denominator is not
zero), just as with common
fractions. Recall that 0/4 = 0,
but 4/0 is undefined.
It is important to keep this
in mind as you work with rational
expressions, because it can
happen that you are trying
to solve an equation and you
get one of the “forbidden”
values as a solution. You
would have to discard that
solution as being unacceptable.
You can also get some crazy
results if you don’t pay attention
to the possibility that the
denominator might be zero
for certain values of the
variable. For example, the
celebrated proof that 0 = 1
and other nonsense.
Proof that 1 = 0
And
other nonsense
Can
you identify the flaw
in this argument?
Let
x = 1.
Then
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Given: |
x = 1 |
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Multiply both
sides by x: |
x2
= x |
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Subtract x
from both sides: |
x2 – x = 0 |
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Factor out an
x: |
x(x – 1) = 0 |
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Divide both sides
by (x ‑ 1): |
x = 0 |
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But x = 1,
so substitute
1 for x
to get: |
1 = 0 |
This
is a very simple variant
of this classic ‘proof’.
Once you see the trick*,
you can construct more
elaborate versions that
do a better job of concealing
the error, and you can
vary it to ‘prove’ other
nonsense such as 1 = 2.
For
example:
Let
x = 1
. Then
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Given: |
x = 1
|
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Multiply both
sides by –1: |
-x = –1 |
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Add x2
to both sides: |
x2 - x = x2 – 1 |
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Factor both sides: |
x(x – 1) = (x – 1)(x + 1) |
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Divide both sides
by (x – 1): |
x = (x + 1) |
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Substituting
1 back in for
x gives
the result: |
1 = 2 |
*The
trick: We are dividing
by zero because if x = 1
then (x – 1) = 0.
Thus, all of these “proofs”
are invalid because
they use an illegal
step. |
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