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alfers101

  • 3 years ago

proof that square root of 2 is irrational

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  1. saifoo.khan
    • 3 years ago
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    Let's suppose √2 were a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero. We additionally make it so that this a/b is simplified to the lowest terms, since that can obviously be done with any fraction. It follows that 2 = a2/b2, or a2 = 2 * b2. So the square of a is an even number since it is two times something. From this we can 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. Odd number times odd number is always odd. Check if you don't believe that! Okay, if a itself is an even number, then a is 2 times some other whole number, or a = 2k where k is this other number. We don't need to know exactly what k is; it won't matter. Soon is coming the contradiction: If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get: 2 = (2k)2/b2 2 = 4k2/b2 2*b2 = 4k2 b2 = 2k2. This means b2 is even, from which follows again that b itself is an even number!!! WHY is that a contradiction? Because we started the whole process saying that a/b is simplified to the lowest terms, and now it turns out that a and b would both be even. So √2 cannot be rational.

  2. alfers101
    • 3 years ago
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    thank u so much for all your help :)

  3. saifoo.khan
    • 3 years ago
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    Np Man!

  4. joemath314159
    • 3 years ago
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    Theres another good proof using the Fundamental Theorem of Arithmetic, im gonna up load it in a sec. It starts out the same way though, assume its rational.

  5. alfers101
    • 3 years ago
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    ok

  6. joemath314159
    • 3 years ago
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    Alright, the goal is the same, force a contradiction. The Fund. Theorem of Arithmetic lets you prime factor a number. In this proof, you prime factor a and b, then count the number of primes on both sides of the equation. If the numbers were equal, they would have the same prime factorization. but thats impossible for this case.

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