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| Theorem 3 | ||||
x
+
y
≤
x + y
≤
x
+
y
+ 1
|
||||
| proof of Theorem 3 | |
|
x =
x
+ α
where the real number α satisfies
0 ≤ α < 1
|
statement 1 | |
|
y =
y
+ β
where the real number β satisfies
0 ≤ β < 1
|
statement 2 | |
|
Therefore
x + y =
x
+
y
+ ( α + β )
|
statement 3 | |
|
By theorem 1,
x + y
=
x
+
y
+
α + β
|
statement 4 | |
| Adding β to the inequality in statement 1 we find 0 ≤ β ≤ α + β < 1 + β | statement 5 | |
| And adding 1 throughout statement 2 we get 1 ≤ 1 + β < 2 | statement 6 | |
| Thus 0 ≤ α + β < 2 | statement 7 | |
Let's split into two cases.
| Case 1 : | Suppose that 0 ≤ α + β < 1 | |
|
Then
α + β
= 0
|
||
|
Statement 3 implies that
x + y
=
x
+
y
|
||
|
This proves the left hand side of the
theorem for this case.
Now let's prove the right hand half: |
||
|
Since
x
+
y
=
x + y
,
|
||
|
1 +
x
+
y
=
1 +
x + y
>
x + y
|
||
| This proves the right half of the theorem for case 1. | ||
| Case 2 : | Suppose that 1 ≤ α + β < 2 | |
|
Then
α + β
= 1
|
||
| So, from statement 3 we have | ||
|
x + y
=
x
+
y
+ 1
>
x
+
y
|
||
|
Therefore,
x + y
≥
x
+
y
|
||
| This proves the left side of the theorem for case 2. | ||
| The right hand side of the theorem also follows from statement 3. | ||
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