GVSU MTH 225 Lecture-Module #13: Induction

MTH 225 Lecture-Module #13:
Induction

induction is a further way to prove that a formula which is a universal quantification (∀ν...) is always true
but induction applies only to nonnegative integers
i.e. you can use induction only when the domain of discourse is ℕ

some sample formulas for which we need to use induction to prove that they are always true are as follows:

* (∀n∈ℕ)[n ≥ 10  →  n³ < 2ⁿ]
* (∀n∈ℕ)[10ⁿ mod 3 := 1]
* (∀n∈ℕ)[  ∑ i  := n(n+1)/2 ]
           1≤i≤n
* the binomial theorem

and do we care about facts such as these?

to invoke induction for such a goal:
(∀v∈ℕ)[φ<v>]
get two new goals as follows:
φ<0>  and  ∀v[φ<v> → φ<v+1>]
e.g. if our original goal was specifically    (∀x∈ℕ)[x ≤ 2·x]
then to prove it by induction we would work on the following goals (instead):
[0 ≤ 2·0]  and  ∀x[x ≤ 2·x  →  x+1 ≤ 2·(x+1)]
then would proceed as usual for proofs

now, why would proving  φ<0> and  ∀v[φ<v> → φ<v+1>] show
that original goal  (∀v∈ℕ)[φ<v>] would be true?
well, would proving those things establish that  φ<> is true for the specific integer 0,
i.e. that  φ<0> is true?

and would proving those things establish  φ<> is true for the specific integer 1?

and would it establish  φ<2> true?

and would it establish  φ<3> true?

thus  φ<> would be true for every nonnegative integer
i.e.  (∀v∈ℕ)[φ<v>] would be true

e.g. ~~Example 2 pages 268-269 of our textbook by Rosen:
what do you get if you sum odd numbers?

odd numbers | their sum
------------+-----------

1,3         | 
1,3,5       | 
1,3,5,7     | 
1,3,5,7,9   |

click here to see a proof for that
done in ProofBuilder

alternatively:

Proposition:

(∀n∈ℕ)[  ∑ (2i+1)  := (n+1)² ]
         0≤i≤n

(slightly different from book)

Proof by induction:

basis step:
need to prove that the following is true:

  ∑ (2i+1)  :=?= (0+1)²
0≤i≤0

Well, according to rules for summations,

  ∑ (2i+1)  :=  2·0 + 1
0≤i≤0

so:

  ∑ (2i+1)  :=?= (0+1)²
0≤i≤0
    ≡
2·0 + 1   :=?= 1²
    ≡
        :=?=

finished basis step

inductive step:
need to prove the following:

∀n( [  ∑ (2i+1)  := (n+1)² ]  →  [  ∑ (2i+1)  := ([n+1]+1)² ]
      0≤i≤n                        0≤i≤[n+1]

Well, to prove that this universally quantified formula is true,
we'll take k as an arbitrary value and try to prove the following:

[  ∑ (2i+1)  := (k+1)² ]  →  [  ∑ (2i+1)  := ([k+1]+1)² ]
 0≤i≤k                         0≤i≤[k+1]

Well, to prove that this implication is true,
we'll assume the antecedent and try to prove the consequent:
hypothesis:

  ∑ (2i+1)  := (k+1)²
0≤i≤k

~~real inductive step:

  ∑ (2i+1)  :=?= ([k+1]+1)²
0≤i≤[k+1]

Well,

  ∑ (2i+1)
0≤i≤[k+1]
    :=  // by rules for summations
  ∑ (2i+1)  +  
0≤i≤
    :=  // by 
  (k+1)²    +  
    :=  // by algebra
 + 2k + 
    :=  // by algebra

    :=  // by algebra

    :=  // by algebra
  which is what we wanted

finished inductive step

thus finished that proof by induction

for a version more like our textbook's Example 2, click here

for our textbook's Example 6, click here