or alternatively
(
∀
x)(
∀
y)( (x
ε
S
∧
y
ε
S)
→
x+y
ε
4. (10 pts.) Let {a
n
} be defined by the formula
a
n
=5
n+2f
o
r
n = 1,2,3,.
... Define the sequence {b
n
} recursively by b
1
= 7 and
b
n+1
=b
n
+ 5 for n = 1,2,3,.
...
Give a proof by induction that
a
n
n
for n = 1,2,3,.
...
Proof:
Basis Step: Evidently, a
1
(
1
)+2=7=b
1
.
Induction Step: To show (
∀
n
ε
+
)( a
n
n
→
a
n+1
n+1
), assume
a
n
n
for an arbitrary n
ε
+
.
Then using the induction
hypothesis,
a
n
n
→
a
n
+5=b
n
+ 5
[Algebra; Motivation: Use the
recursive definition of {b
n
}.]
→
(
5
n+2
)+5=b
n+1
[Recursive definition of {b
n
},
definition of {a
n
}]
→
5(n+1)+2=b
n+1
[Algebra; Motivation: Use the
definition of {a
n
}]
→
a
n+1
n+1
[The definition of {a
n
}]
Since we have shown that a
n
n
implies a
n+1
n+1
for an
arbitrary n
ε
+
, universal generalization implies the truth of
(
∀
n
ε
+
)( a
n
n
→
a
n+1
n+1
).
Finally, since we have verified the hypotheses of the
Principle of Mathematical induction, using modus ponens, we may
conclude that the proposition (
∀
n
ε
+
)( a
n
n
) is true.
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5. (15 pts.)
Label each of the following assertions with "true"
or "false".
Be sure to write out the entire word.
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 Spring '08
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 Logic, Mathematical Induction, pts, Natural number, induction step, Recursive definition

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