Mathematical Reasoning: Induction, Successor Functions, and Order

This assignment delves into the core concepts of mathematical reasoning, concentrating on induction, successor functions, and order relations. It begins by exploring the definition of addition through recursion and the use of successor functions, providing a detailed proof using induction. The assignment then examines the properties of order relations, specifically irreflexivity, asymmetry, and transitivity, demonstrating how these properties apply within the context of successor sets. Through a series of proofs, the assignment establishes the validity of these properties, providing a clear understanding of how mathematical principles are applied and proven. This assignment is a comprehensive exploration of these fundamental concepts.

Running head: MATHEMATICAL REASONING
Mathematical Reasoning
Student
Institutional Affiliation

2
MATHEMATICAL REASONING
QUESTION 1
With regards to the usual arithmetic operations in conjuction with the property of induction on
nutural numbers, addition by reccursion from successor fuction is defined as
n+ 0=n ∀ n ∈ N
n+ ( k +1 )=(n+k )+1
Our task is to prove that n+ k=k + n by inducction on n
It clearly follows that 0+k=k=k +0 since the set of natural numbers is the closure of the empty
set under succesor.
NB: The empty set is 0 that is 0=ϕwhich has no predecessor
Now suppose n> 0be the successor of ksuch that n=k +1
Then it follows that
1+n=1+(k + 1)
¿( 1+ k )+1 by definition of addition property
¿ n+1 …………….*
{Remember 1+n=S ( 0 ) +n=S ( 0+ n )=S ( n ) }
Now with assuption that n+ k=k + n for some k in the set of natural numbers (k ∈ N ), it can
clearly be deduced that
n+ ( k +1 ) = ( n+ k ) +1 by definition of additive operation
( k +n ) +1with regards to the assumption above
¿ k + ( n+1 ) by definition of +¿ ………………**
Hence the proof

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MATHEMATICAL REASONING
Clarification:
From * the statement holds for S ( n )
That is 1+n=n+1 but 1=S ( 0 ) → n+S ( 0 ) =S ( 0+n ) =S ( n )
With the assumption that n+k=k + n ,we’ve also seen that the statement holds for S( n+ k ) i.e.
n+ ( k +1 )=n+ S ( k )=S ( n+k ) since k +1=S (k ) by definition
QUESTION 2
A relation < on the succesor set S(x ) is a strict order if
a) Irreflexive: S ( n ) < S( n) does not hold for any S ( n ) a predicessor set of S (x )
b) Asymmetric: if S ( k ) < S(n) then S ( n ) <S(k) does not hold for any
S ( k ) , S ( n ) predecessors of S ( x)
c) Transitive: S ( m ) < S (k ) and S ( k ) <S(n) implies S ( m ) <S ( n )for any S(m), S(k), and
S(n) predecessors of S(x)
QUESTION 3
The irreflexive property follows clearly by defition of the succesor function on the empty set,
that is
0=∅
taking the same trend with each natural number as the set of all its predicessors, it is clear that
S ( n ) < S(n) does not hold for any S ( n ) as a predicessor set
For transitivity:
If S ( m ) , S(k) and S ( n ) as predecessors of S ( x ) with k =m+1 and n=k +1

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MATHEMATICAL REASONING
For the first case, S ( m ) <S (k )
Since k =m+1 , it implies S ( m ) < S ( k ) =S ( m+1 ) =S ( 1+m ) =1+ S(m) hence the prove
For the second case, S ( k ) <S(n)
Since n=k +1 , it implies S ( k ) < S ( n ) =S ( k +1 ) =S ( 1+k ) =1+S (k ) hence the prove
The two cases clearly implies that S ( m ) <S (n) that is
S ( m )< S ( n )=S ( k +1 ) =S (1+ k )=1+ S ( k ) =1+ S ( m+1 ) =1+S ( 1+m )=1+ ( 1+S ( m ) ) =1=1+S(m)
hence the prove for transitivity
Combining irreflexive and transitive properties above it canclearly be deduced that if S ( k ) < S(n)
then S ( n ) < S( k) by the above definition of the relationship between n and k (assymetric
property), tha is to say since k < n holds, then n< k and k =n cannot hold