We say that an infinite sequence a0,a1,a2,a3,… of real numbers has the limit L if for...

We say that an infinite sequence a0,a1,a2,a3,… of real numbers has the limit L if for every strictly positive number ε, there is a natural number n such that all the elements an,an+1,an+2,… are within distance ε of the value L. In this case, we write lim a = L.

Express the condition that lim a = L as a formula of predicate logic. Your formula may use typical mathematical functions like + and absolute value and mathematical relations like > or ∈ . Check that your formula has exactly two free variables: a and L.
For any two infinite sequences of real numbers a = a0,a1,a2,… and b = b0,b1,b2,… we define their sum (written a+b) to be an infinite sequence c0,c1,c2,c3,… such that ci = ai + bi for every i≥0. Prove (in complete sentences of “mathematical English”) that if we have two infinite sequences with lim a = L and lim b = M then lim (a+b) = (L+M).
Identify 5 places in your Part B proof where you took a step corresponding to a natural deduction rule, explicitly or implicitly. Each place should correspond to a different natural deduction rule.
Your proof almost certainly used well-known mathematical “facts” about natural numbers or real numbers. These “facts” are just small theorems that are so obvious or well-known that they can be used in proofs without comment. Give predicate logic formulas to specify three theorems you used. (no proofs for these, but they should all be true formulas about numbers, with no free variables.)

Expert Solution

Feel free to ask if you don't understand any of the symbols.

5 Places where natural deduction rule is used are where the following occur :

1. If >0, then /2 >0.

2. (a+b)-(L+M) = a+b-L-M

3. a+b-L-M = a-L+b-M (commutative law)

4. k=largest of k1 and k2 that is k>k1 and k>k2.

5. /2+/2=.

Three theorems used :

1. Real numbers are commutative.

2. (a+b)-(L+M) = a+b-L-M

3. If a number is positive, the half of the number will also be positive,

Colby Messinger answered 2 years ago

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