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Mathematics LibreTexts

0.2 Introduction to Proofs/Contradiction

  • Page ID
    10958
  • [ "stage:draft", "article:topic", "license:ccbyncsa", "showtoc:yes" ]

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    In this section, we will explore different techniques of proving a mathematical statement  "If \(p\) then \(q\)". (\(p \to q\)).

     

    Direct Proof

    In this technique, we shall assume \(p\) and show that \(q\) is true.

    Theorem \(\PageIndex{1}\)

    Let \(n\) be an integer. If \(n\) is even then \(n^2\) is even.

    Proof

    Assume that \(n\) is even. Then  \(n=2m\) for some integer \(m).

    Consider \(n^2=(2m)^2=4m^2=2(2m^2).\)

    Thus \(n^2\) is even.

    Exercise \(\PageIndex{1}\)

    Show that  for all integers \( n\),  if \(n\) is odd then \(n^2\) is odd.

    Answer

     

    Assume that \(n\) is odd. Then  \(n=2m+1\) for some integer \(m).

    Consider \(n^2=(2m+1)^2=4m^2+4m+1=2(2m^2+2m)+1.\)

    Thus \(n^2\) is odd.

    Proof by Contrapositive

    In this technique, we shall assume \(\not p\) and show that \( \not q\) is true.

    Theorem \(\PageIndex{2}\)

    Let \(n\) be an integer. If \(n^2\) is even then \(n\) is even.

    Proof

    We shall prove this statement by assuming  \(n\) is odd. Then  \(n=2m+1\) for some integer \(m).

    Consider \(n^2=(2m+1)^2=4m^2+4m+1=2(2m^2+2m)+1.\)

    Thus \(n^2\) is odd.

    Exercise \(\PageIndex{2}\)

    Show that  for all integers \( n\),  if \(n^2\) is odd then \(n\) is odd.

    Answer

     

    We shall prove this statement by assuming \(n\) is even. Then  \(n=2m\) for some integer \(m).

    Consider \(n^2=(2m)^2=4m^2=2(2m^2).\)

    Thus \(n^2\) is even.

    Proof by Contradiction

    In this technique, we shall assume \( p\) and   \( \not  q\)  are true,  come to a contradiction.

     

    Proof by Counterexample

    Example \(\PageIndex{1}\):

    Decide whether the statement is true or false and justify your answer:

    For all integers \(a,b,u,v\), and \(u\ne 0, v \ne 0\), if \(au+bv =0\) then \(a=b=0.\)

    Solution:  The statement is false. 

    Counterexample:  Choose \(a=1,b=-1, u=2,v=2\), then  \(au+bv =0\) , but  \(a\ne 0. b \ne 0, a \ne b.\)

    Proof by induction 

     

    In mathematics, we use induction to prove mathematical statements involving integers. There are two types of induction: regular and strong. The steps start the same but vary at the end. Here are the steps. In mathematics, we start with a statement of our assumptions and intent:

    Let \(p(n) \forall n \geq n_0, \, n, \, n_0 \in \mathbb{Z}\) be a statement. We would show that p(n) is true for all possible values of n.

    1. Show that p(n) is true for the smallest possible value of n: In our case \(p(n_0)\). AND
    2. For regular Induction: Assume that the statement is true for \(n = k,\) for some integer \(k \geq n_0\). Show that the statement is true for n = k + 1.

    OR

    For Strong Induction: Assume that the statement p(r) is true for all integers r, where \(n_0 ≤ r ≤ k \) for some \(k ≥ n_0\). Show that p(k+1) is true.

    If these steps are completed and the statement holds, we are saying that, by mathematical induction, we can conclude that the statement is true for all values of \(n \geq n_0.\)

    Example \(\PageIndex{2}\):

    Prove that \(1 + 2 + ... + n = \displaystyle \frac{n(n + 1)}{2}, \, \forall n \in \mathbb{Z}\).

    Solution:

     

    Base step: Choose \(n = 1\). Then L.H.S =\(1\). and  R.H.S \( = \frac{(1)(1 + 1)}{2}=1\)

    Induction Assumption: Assume that \( 1 + 2 + ... +k= \displaystyle\frac{k(k + 1)}{2}\) , for \(k \in \mathbb{Z}\).

    We shall show that \(1 + 2 + ... + k + (k + 1) = \displaystyle\frac{(k + 1)[(k + 1) + 1]}{2} = \frac{(k + 1)(k + 2)}{2}\)

    Consider \(1 + 2 + ... + k + (k + 1) \)

    \(= \displaystyle \frac{k(k + 1)}{2} + (k + 1)\)

     \(= (k + 1) \left( \displaystyle\frac{k}{2} + \displaystyle\frac{1}{1}\right)\)

     \(= (k + 1) \left( \displaystyle\frac{k + 2}{2}\right)\)

    \(= \displaystyle \frac{(k + 1)(k + 2)}{2}\).

    Thus, by induction we have \(1 + 2 + ... + n = \displaystyle\frac{n(n + 1)}{2}, \, \forall n \in \mathbb{Z}\).

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