In the realm of mathematics, there is a powerful technique known as disproof by counterexample. This method is used to prove a conjecture or statement to be false, rather than proving it to be true like other proofs. This approach requires finding just one counterexample, making it a quick and effective way to disprove a statement.
As an example, consider the following statement about student test scores: "All students scored more than 80 on the test." To prove this statement true, we would need to collect and verify the scores of all students to ensure they are above 80. On the other hand, to disprove the statement, we only need to find one student with a score of 80 or below. This lone student's score would serve as our counterexample.
When encountering a conjecture, it is essential to understand the conditions that would disprove it by counterexample. This can provide valuable insight into why the statement is false. While randomly trying different values may work, it is more effective to approach this method in a systematic manner.
The first step in using disproof by counterexample is to understand precisely what the statement is saying. This helps us determine what criteria will make the conjecture fail.
This step is crucial in finding a counterexample. By determining the conditions in which the conjecture fails, we can begin to search for a suitable counterexample.
Next, we use the identified criteria to determine the value or statement of the counterexample.
Finally, we check that the chosen value does, in fact, disprove the conjecture and falls within any limitations on values. If all checks are passed, we have successfully disproven the conjecture by counterexample.
Example 1:
Proving the statement that the product of two irrational numbers is always irrational to be false.
Let our two numbers be √2 and √2. √2 · √2 = 2. 2 is a rational number, disproving the statement by counterexample.
Example 2:
Disproving the statement that the equation is true if and only if p = q, where p and q are real numbers.
Let p = 1. Then , meaning we now must find a value for q such that . By inspection, we can see that q = -1 would work. However, in this case, p ≠ q, making this a counterexample.
Note: This can be disproven generally by considering that, if , then . By the difference of two squares, we get (p²-q²) (p² + q²) = 0. Simplifying further, we get (p - q) (p + q) (p² + q²) = 0. This shows that p = q is the only condition that satisfies the original statement and thus disproves it.
Example 3:
Showing that the statement “ is not a perfect square for all values of n, where n belongs to the natural numbers” is false.
First, we must understand the statement, which means that when we substitute any natural number into the expression, it will not result in a perfect square. To disprove this, we need to find one value of n for which n²-n + 5 is indeed a perfect square. The key is to notice that if -n + 5 = 0, then we are left with only n², making it a perfect square. By setting n = 5, we have a counterexample, proving the statement to be false.
To sum it up, disproof by counterexample is a powerful tool in mathematics for quickly and effectively disproving a statement. Using a systematic approach and considering criteria, we can find a counterexample and prove a conjecture to be false with just one example. This method also provides valuable insight into why a statement may be incorrect, making it a useful tool for further learning and exploration. Next time you come across a challenging conjecture, give disproof by counterexample a try.
