Is the square root of two rational or irrational?
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Zoe Reyes
Studied at the University of Barcelona, Lives in Barcelona, Spain.
As a domain expert in mathematics, I specialize in the field of number theory, which deals with the properties and relationships of numbers, including rational and irrational numbers. One of the most fundamental and intriguing results in this area is the proof that the square root of two (\(\sqrt{2}\)) is an irrational number. This means that it cannot be expressed as a simple fraction of two integers. Let's delve into the proof and its implications.
The proof that \(\sqrt{2}\) is irrational is a classic example of a proof by contradiction, which is a common method in mathematics. The idea is to assume the opposite of what we want to prove and then show that this assumption leads to a contradiction. Here's a step-by-step breakdown of the proof:
1. Assumption of Rationality: We start by assuming that \(\sqrt{2}\) is rational. This means there exist two relatively prime integers \(a\) and \(b\) (where \(b \neq 0\)) such that \(\sqrt{2} = \frac{a}{b}\).
2. Squaring Both Sides: To work with this fraction, we square both sides of the equation to eliminate the square root, which gives us \(2 = \left(\frac{a}{b}\right)^2\), or \(2b^2 = a^2\).
3. Divisibility and Evenness: Since \(a^2\) is divisible by \(2b^2\), and \(2b^2\) is even (because it's a multiple of 2), it follows that \(a^2\) must also be even. An even number is one that can be expressed as \(2k\) for some integer \(k\). Therefore, we can write \(a = 2k\) for some integer \(k\).
4. Substitution and Simplification: Substituting \(a = 2k\) into the equation \(2b^2 = a^2\), we get \(2b^2 = (2k)^2\), which simplifies to \(b^2 = 2k^2\).
5. Evenness of \(b^2\): From the equation \(b^2 = 2k^2\), it's clear that \(b^2\) is also even because it is a multiple of 2. This means \(b\) must also be even, as the square of an odd number is odd.
6. Contradiction: We have now established that both \(a\) and \(b\) are even. However, this contradicts our initial assumption that \(a\) and \(b\) are relatively prime (which means they have no common factors other than 1). If both \(a\) and \(b\) were even, they would both be divisible by 2, which means they would have a common factor of 2, not just 1.
7.
Conclusion: Since our assumption that \(\sqrt{2}\) is rational led to a contradiction, we conclude that \(\sqrt{2}\) cannot be rational. It must be irrational.
This proof is significant because it was one of the first to demonstrate the existence of irrational numbers, which are numbers that cannot be expressed as a fraction of two integers. It also shows the power of proof by contradiction as a method for establishing mathematical truths.
Now, let's translate this explanation into Chinese.
The proof that \(\sqrt{2}\) is irrational is a classic example of a proof by contradiction, which is a common method in mathematics. The idea is to assume the opposite of what we want to prove and then show that this assumption leads to a contradiction. Here's a step-by-step breakdown of the proof:
1. Assumption of Rationality: We start by assuming that \(\sqrt{2}\) is rational. This means there exist two relatively prime integers \(a\) and \(b\) (where \(b \neq 0\)) such that \(\sqrt{2} = \frac{a}{b}\).
2. Squaring Both Sides: To work with this fraction, we square both sides of the equation to eliminate the square root, which gives us \(2 = \left(\frac{a}{b}\right)^2\), or \(2b^2 = a^2\).
3. Divisibility and Evenness: Since \(a^2\) is divisible by \(2b^2\), and \(2b^2\) is even (because it's a multiple of 2), it follows that \(a^2\) must also be even. An even number is one that can be expressed as \(2k\) for some integer \(k\). Therefore, we can write \(a = 2k\) for some integer \(k\).
4. Substitution and Simplification: Substituting \(a = 2k\) into the equation \(2b^2 = a^2\), we get \(2b^2 = (2k)^2\), which simplifies to \(b^2 = 2k^2\).
5. Evenness of \(b^2\): From the equation \(b^2 = 2k^2\), it's clear that \(b^2\) is also even because it is a multiple of 2. This means \(b\) must also be even, as the square of an odd number is odd.
6. Contradiction: We have now established that both \(a\) and \(b\) are even. However, this contradicts our initial assumption that \(a\) and \(b\) are relatively prime (which means they have no common factors other than 1). If both \(a\) and \(b\) were even, they would both be divisible by 2, which means they would have a common factor of 2, not just 1.
7.
Conclusion: Since our assumption that \(\sqrt{2}\) is rational led to a contradiction, we conclude that \(\sqrt{2}\) cannot be rational. It must be irrational.
This proof is significant because it was one of the first to demonstrate the existence of irrational numbers, which are numbers that cannot be expressed as a fraction of two integers. It also shows the power of proof by contradiction as a method for establishing mathematical truths.
Now, let's translate this explanation into Chinese.
2024-05-12 22:50:40
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Works at Apple, Lives in Cupertino, CA
Proof that the square root of 2 is irrational. Assume is rational, i.e. it can be expressed as a rational fraction of the form , where and are two relatively prime integers. Now, since , we have , or . Since is even, must be even, and since is even, so is .
2023-06-19 02:45:04
Amelia Lewis
QuesHub.com delivers expert answers and knowledge to you.
Proof that the square root of 2 is irrational. Assume is rational, i.e. it can be expressed as a rational fraction of the form , where and are two relatively prime integers. Now, since , we have , or . Since is even, must be even, and since is even, so is .
