math – update

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Day #:2 Craziness of the Square Roots of Primes!

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sq2-1

Q: Could you give a proof of the irrationality of the square root of Your favorite prime number p??…

A: I think we ALL should try this proof with our favorite prime! My – smallest – favorite prime is 3 and here is ONE proof:

The Irrationality of square root of three

Problem:
Prove that
square root of three is an irrational number.

Solution:
The number,
square root of three, is irrational, ie., it cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us assume that square root of three is rational so that we may write

square root of three = a/b
1.

for a and b = any two integers. We must then show that no two such integers can be found. We begin by squaring both sides of eq. 1:

3 = a2/b2
2.
or
3b2 = a2
2a.

If b is odd, then b2 is odd; in this case, a2 and a are also odd. Similarly, if b is even, then b2, a2, and a are even. Since any choice of even values of a and b leads to a ratio a/b that can be reduced by canceling a common factor of 2, we must assume that a and b are odd, and that the ratio a/b is already reduced to smallest possible terms. With a and b both odd, we may write

a = 2m + 1
3.
and
b = 2n +1
4.

where we require m and n to be integers (to ensure integer values of a and b). When these expressions are substituted into eq. 2a, we obtain

3(4n2 + 4n + 1) = 4m2 + 4m + 1
5.

Upon performing some algebra, we acquire the further expression

6n2 + 6n + 1 = 2(m2 + m)
6.

The Left Hand Side of eq. 6 is an odd integer. The Right Hand Side, on the other hand, is an even integer. There are no solutions for eq. 6. Therefore, integer values of a and b which satisfy the relationship square root of three = a/b cannot be found. We are forced to conclude that square root of three is irrational.

Source: Irrationality of square root of 3, Square root of 3 , Square root of 2,

 

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