Some important mathematical proofs
Introduction :
A proof is a valid argument that establishes the truth of a mathematical statement. A proof can use the hypothesis of the theorem, if any , axioms assumed to be true, and previously proven theorems. Using these ingredients and rules of inference, the final step of the proof establishes the truth of the statement being proved. But here we will mainly focus on more informal proof.
Methods of proving Theorems :
To prove a theorem of the form ∀x ( P(x) –> Q(x) ), our goal is to show that P(c) –> Q(c) is true, where c is an arbitary element of the domain, and then apply a universal generalization.
1. Direct Proof Method –
Let, ∀x ( P(x) –> Q(x) ), D be the domain. We start choosing an arbitrary member say a ∈ D. Then, for that a ,we can show that,
P(a) –> Q(a) is true .
P(a) is true.
Then, by Modus Ponens, Q(a) is true.
Then, by rule of universal generalization (u , G), it follows that , ∀x ( P(x) –> Q(x) ) is true.
Examples –
- Give a direct proof of the theorem “If n is an odd integer, then n^{2} is odd.”
Solution –
Note that this theorem states that ∀n (P(n) –> Q(n)) , where P(n) is ” n is an odd integer” and Q(n) is “n^{2} is odd.” By the definition of an odd integer, it follows that n=2k+1, where k is some integer. We want to show that n2 is odd. We can square both sides of the equation n=2k+1 to obtain a new equation that expresses n^{2} . When we do this, we find that n^{2} =(2k+1)^{2} =4k^{2}+4k+1 = 2(2k^{2} + 2k) + 1. By the definition of an odd integer, we can conclude that n2 is an odd integer (it is one more than twice an integer ). Consequently, we have proved that if n is an odd integer, then n^{2} is an odd integer. - Give direct proof that if m and n are both perfect squares, then nm is also a perfect square.
Solution –
Assume that m and n are odd integers. Then, by definition, m = 2k + 1 for some integer k and n = 2l + 1 for some integer l. Again, note that we have used different integers k and l in the definitions of m and n. We will now use this to show that mn is also an odd integer.
mn = (2k + 1)(2l + 1) by our definitions of m and n
= 4kl + 2k + 2l + 1 by expanding the brackets
= 2(2kl + k + l) + 1, since 2 is a common factor.
Hence, we have shown that mn has the form of an odd integer, since 2kl + k + l is an integer.
2. Indirect Proof Method –
Consider the implication p –> q . It is equivalent to ~q –> ~p.
In order to show p –> q is true, you can show that ~q –> ~p. It is called indirect proof or Proof by Contraposition.
Example –
- Prove that if n is an integer and 3n+2 is odd, then n is odd.
Solution –
The first step in a proof by contraposition is to assume that the conclusion of the conditional statement “If 3n+2 is odd, then n is odd.” is false; namely, assume that n is even. Then, by the definition of an even integer, n=2k for some integer k. Substituting 2k for n, we find that 3n+2 = 3(2k) + 2 = 6k + 2 = 2(3k+1). This tells us that 3n+2 is even (because it is a multiple of 2), and therefore not odd. This is a negation of the hypothesis of the theorem. Because the negation of the conditional statement implies that the hypothesis is false, the original conditional statement is true. Our proof by contraposition succeeded; we have proved the theorem “If 3n+2 is odd, then n is odd.” - Prove that if n = ab, where a and b are positive integers, then a≤√n or b≤√n .
Solution –
Assume b > √n and a > √n.
ab > (√n) . (√n) = n
So, n ≠ ab.
By contraposition, if n=ab, then a≤√n or b≤√n .
3. Proof by Contradiction :
In this case, we assume that the conclusion is not true. Then we arrive at some contradiction.
Example –
- Prove that √2 is irrational by giving proof by contradiction.
Solution –
Let’s suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero.
We additionally assume that this a/b is simplified to the lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in the simplest terms, both a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.
From equality √2 = a/b it follows that 2 = a^{2}/b^{2}, or a^{2} = 2 · b^{2}. So the square of a is an even number, since it is two times something.
From this we know that a itself is also an even number. Why? Because it can’t be odd; if a itself was odd, then a · a would be odd too. The odd number of times an odd number is always odd.
Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is the other number. We don’t need to know what k is; it won’t matter. Soon comes the contradiction.
If we substitute a = 2k into the original equation 2 = a^{2}/b^{2}, this is what we get:
2 = (2k)^{2}/b^{2}
2 = 4k^{2}/b^{2}
2*b^{2} = 4k^{2}
b^{2} = 2k^{2}
This means that b^{2} is even, from which follows again that b itself is even. And that is a contradiction!!!
WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lower terms, and now it turns out that a and b will both be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore, √2 cannot be rational. - Give a Proof by contradiction of the theorem “If 3n+2 is odd, then n is odd”.
Solution –
Let 3n + 2 is odd, but n is not odd. Then n is even and can be written in the form of 2m for some integer m
3n+2
=3(2m)+2
=6m+2
=2(3m+1)
⟹3n+2is even. Which is a contradiction to our assumptions.
Hence, n must be odd.
Conversely, let n be odd, then n can be written in the form of 2m + 1 for some integer m. We assume that 3n + 2 is even.
3n+2=3(2m+1)+2
=6m+5=6m+4+1
=2(3m+2)+1
⟹3 n + 2 is odd. Which is a contradiction to our assumptions.
Hence, if n is odd, 3n + 2 must be odd.
Mistakes in Proofs :
There are many common errors made in constructing mathematical proofs. We will briefly describe some of these here.
Example –
1. What is wrong with this famous supposed “proof” that 1 = 2?
“Proof:” We use these steps, where a and b are two equal positive integers.
S No. | Step | Reason |
1. | a = b | Given |
2. | a^{2} = ab | Multiply both sides of (1) by a |
3. | a^{2} – b^{2} = ab – b^{2} | Subtract b^{2} from both sides of (2) |
4 | (a-b) (a + b) = b(a-b) | Factor both sides of (3) |
5. | a + b = b | Divide both sides of (4) by a – b |
6. | 2b = b | Replace a by b in (5) because a = b and simplify |
7. | 2 = 1 | Divide both sides of (6) by b |
Solution –
Every step is valid except for one , step 5 where we divided both sides by (a – b) . The error is that (a – b) equals zero; division of both sides of an equation by the same quantity is valid as long as this quantity is not zero.
2. Suppose you want to prove the assertion –
Let a, b, ∈ Z where a = 1 mod 3 and b = 2 mod 3. Then (a + b) = 0 mod 3.
Incorrect Proof :
Since a = 1 mod 3 there is an integer k in Z such that a = 3k + 1. Since b = 2 mod 3, we can write b = 3k + 2. Thus a + b = (3k + 1) + (3k + 2) = 6k + 3 = 3(2k + 1), so (a + b) = 0 mod 3.
The error in the proof :
The attempted proof assumes that b = a + 1 since b − a = (3k + 2) − (3k + 1) = 1. So the proof is only valid for that limited set of choices for a and b.
A correct proof :
Since a = 1 mod 3 there is an integer k in Z such that a = 3k+1. Since b = 2 mod 3, there is an integer n in Z such that b = 3n + 2. Therefore a + b = (3k + 1) + (3n + 2) = 3k + 3n + 3 = 3(k + n + 1), so (a + b) = 0 mod 3
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