Proof by contradiction is a powerful mathematical technique used to establish the truth of a statement by assuming the opposite is true. In this method, if the assumption leads to a contradiction, then the original statement must be true. This approach is valuable in various branches of mathematics and logic.
This example illustrates that the square root of 2 cannot be expressed as a fraction, showcasing the existence of irrational numbers.
Assume that \\( \sqrt{2} \) is rational, meaning it can be expressed as a fraction \\( \frac{a}{b} \) where \\( a \) and \\( b \) are integers with no common factors (i.e., in simplest form). Therefore, we can write:
\\[ \sqrt{2} = \frac{a}{b} \]
Squaring both sides gives:
\\[ 2 = \frac{a^2}{b^2} \]
This can be rearranged to:
\\[ a^2 = 2b^2 \]
This implies that \\( a^2 \) is even (since it is equal to 2 times another integer, \\( b^2 \)). Consequently, \\( a \) must also be even (as the square of an odd number is odd). Therefore, we can express \\( a \) as \\( a = 2k \) for some integer \\( k \).
Substituting back, we have:
\\[ (2k)^2 = 2b^2 \]
This simplifies to:
\\[ 4k^2 = 2b^2 \]
or:
\\[ b^2 = 2k^2 \]
Thus, \\( b^2 \) is also even, which means \\( b \) is even as well. Since both \\( a \) and \\( b \) are even, they share a common factor of 2, contradicting our original assumption that \\( a \) and \\( b \) have no common factors. Therefore, \\( \sqrt{2} \) is irrational.
This example demonstrates the infinitude of prime numbers, a fundamental theorem in number theory.
Assume there are only finitely many prime numbers, and let’s denote them as \\( p_1, p_2, \, \ldots, p_n \. \) Now, we can construct a new number:
\\[ N = p_1 \times p_2 \times \ldots \times p_n + 1 \]
This number \\( N \) is greater than 1 and is not divisible by any of the primes in our list (as it leaves a remainder of 1 when divided by any \\( p_i \)). Therefore, \\( N \) must either be prime itself or have prime factors that are not in our original list. In either case, we reach a contradiction of our assumption that there are only finitely many primes. Thus, there must be infinitely many prime numbers.
This example shows the property of odd and even numbers through proof by contradiction.
Assume that the sum of two odd numbers results in an odd number. Let’s denote two odd numbers as \\( a \) and \\( b \,\) which can be expressed in the form:
\\[ a = 2m + 1 \]
\\[ b = 2n + 1 \]
where \\( m \) and \\( n \) are integers. Now, we calculate their sum:
\\[ a + b = (2m + 1) + (2n + 1) \]
This simplifies to:
\\[ a + b = 2m + 2n + 2 = 2(m + n + 1) \]
Here, \\( 2(m + n + 1) \) is clearly even because it is 2 times an integer. This contradicts our assumption that the sum of two odd numbers is odd. Hence, the sum of two odd numbers must always be even.