Mathematical Reasoning


In mathematics we have to assume many things and Mathematical Reasoning is the part of assumption mathematics. In mathematical reasoning or math reasoning if we are performing some calculation like checking of inequalities, performing logical operations etc, we have to consider some previously defined reasons, that means we have to work with those defined logical reasons.

Introduction mathematical reasoning is all about providing some reasons or consider some math reasons in order to perform some mathematical operations. In mathematical reasoning we have to assume things like “set exists” and “objects belong to a set”. And also includes concepts like like”equality of sets” and “sub sets” this is because of math reasoning and will result in introduction of new concepts, creating new properties and relationships.
Here proofs are arguments or statements that are essentially needed in order to make those properties and relationships true.

Determination of proofs is like an art to solve a given problem in our own way, there is no single method defined to solve each and every reasoning problem and to conclude the proof. Some times we use the previously determined proofs to prove theorems.
Theorem 1:- P Í U,
Proof: - Here in order to prove above theorem we have to deal with the definition of 'Í', and show that "x [xÎ P ® x Î U ], here the given arbitrary x, [xÎ P ® xÎ U].
Is represented according to interference rule of “universal generalization”.
xÎ U is true.
xÎ P ® xÎU (here p®q is always said to be true if the defined domain of 'q' is true and doesn’t depend on the occurrence of 'p')
Thus according to defined domain of Universal Generalization "x [xÎ P ® x Î U], that is P Í U by definition of subset.
So from this proof that we derived we can say that p®q is trivially true if 'q' is true, and this type if proof is called trivial proof.

Theorem 2:- f Í P,
Proof:- Here according to the definition of 'Í', we can say "x [xÎ f ® x Î P ].
To prove this we have to show that for any arbitrary 'x', [xÎ f ® xÎ P] holds.
And after that just simply apply the Universal Generalization.
As "x xÏ f, and for any arbitrary x, xÎf is false as according to Universal instantiation.
Therefore [xÎf ® xÎP] is true for any arbitrary 'x'.
So by Universal Generalization "x [xÎ f ® x Î P] gives f Í P by definition of subset.
Here we can say that p®q is true if 'p' is false, and this kind of proof is called a vacuous proof.

Math Topics