1. Mildred Dunbar works as a waitress at Famous Foods.
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Answer:
1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that where
and
are non negative integers, and the fraction
is irreducible, i.e., the numbers
and
have no common factors.
Now, squaring the equality at the beginning we get that
(1)
which is equivalent to . From this we can deduce that 3 divides the number
, and necessarily 3 must divide
. Thus,
, where
is a non negative integer.
Substituting into (1), we get
which is equivalent to
.
Thus, 3 divides and necessarily 3 must divide
. Hence,
where
is a non negative integer.
Notice that
.
The above equality means that the fraction is reducible, what contradicts our initial assumption. So,
is irrational.
2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, , which is equivalent to say that
where
and
are non negative integers. Also, assume that
. So, we want to prove that
. Recall that an integer
can be written as
.
Then,
.
Notice that the product is an integer. Thus, the fraction
is a rational number. Therefore,
.
3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have is irrational and
.
Write and let us suppose that
is a rational number. So, we get that
.
But the subtraction or addition of two rational numbers is rational too. Then, the number must be rational too, which is a clear contradiction with our hypothesis. Therefore,
is irrational.
Since we are translating this graph one unit to the right, replace x with (x-1)
Note: If you are translating a graph to the left by k units, replace x with (x+k), and if by the right, replace x with (x-k)
After simplifying, we now have the following equation:
This is the new equation. Let me know if you need any clarifications, thanks!