Difference between revisions of "1959 IMO Problems/Problem 1"
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~ Solution by '''''z6z61im4s99''''' | ~ Solution by '''''z6z61im4s99''''' | ||
+ | <center> | ||
+ | '''SIXTH SOLUTION''' | ||
+ | </center> | ||
+ | |||
+ | we can use PRINCIPLE OF MATHEMATICAL INDUCTION TO EASILY PROVE THE ABOVE STATEMENT. | ||
+ | |||
+ | First, we show that the statement is true for n=1. For n=1, 25/17 is irreducible because gcd(25,17) =1. | ||
+ | |||
+ | |||
+ | Now, we show that '''If the statement is true for the natural number 'n', then it is also true for natural number 'n+1'.''' | ||
+ | |||
+ | |||
+ | PROOF: | ||
+ | |||
+ | Let the given statement be true for the natural number 'n', which means gcd (21n+4, 14n+3) =1 | ||
+ | |||
+ | Now we need to show that the statement is true for natural number 'n+1', which means we need to show gcd (21n+25, 14n+17) =1 | ||
+ | |||
+ | let us take a look at that, | ||
+ | |||
+ | gcd (21n+25, 14n+17) =1 | ||
+ | |||
+ | -> gcd [21n+4+(7*3), 14n+3+(7*2)] = 1, which is true since gcd (21n+25, 14n+17) =1 and adding some multiple of some number 'k=7' to both the numbers doesn't change the gcd. | ||
+ | |||
+ | Hence, the statement is true for natural number 'n+1'. | ||
+ | |||
+ | Hence, the statement is true for each natural number. | ||
+ | |||
Revision as of 16:35, 23 December 2013
Contents
[hide]Problem
Prove that the fraction is irreducible for every natural number .
Solutions
First Solution
We observe that
Since a multiple of differs from a multiple of by 1, we cannot have any postive integer greater than 1 simultaneously divide and . Hence the greatest common divisor of the fraction's numerator and denominator is 1, so the fraction is irreducible. Q.E.D.
Second Solution
Denoting the greatest common divisor of as , we use the Euclidean algorithm as follows:
As in the first solution, it follows that is irreducible. Q.E.D.
Third Solution
Let's assume that is a reducible fraction where is a divisor of both the numerator and the denominator:
Subtracting the second equation from the first equation we get which is clearly absurd.
Hence is irreducible. Q.E.D.
Fourth Solution
Let . Then where . Thus, . Note: This solution, in hindsight, is just the first solution above in a slightly different notation.
Fifth Solution
We notice that:
So it follows that and must be coprime for every natural number for the fraction to be irreducible. Now the problem simplifies to proving irreducible. We re-write this fraction as:
Since the denominator differs from a multiple of the numerator by 1, then the numerator and the denominator must be relatively prime natural numbers. Hence it follows that is irreducible.
Q.E.D
~ Solution by z6z61im4s99
SIXTH SOLUTION
we can use PRINCIPLE OF MATHEMATICAL INDUCTION TO EASILY PROVE THE ABOVE STATEMENT.
First, we show that the statement is true for n=1. For n=1, 25/17 is irreducible because gcd(25,17) =1.
Now, we show that If the statement is true for the natural number 'n', then it is also true for natural number 'n+1'.
PROOF:
Let the given statement be true for the natural number 'n', which means gcd (21n+4, 14n+3) =1
Now we need to show that the statement is true for natural number 'n+1', which means we need to show gcd (21n+25, 14n+17) =1
let us take a look at that,
gcd (21n+25, 14n+17) =1
-> gcd [21n+4+(7*3), 14n+3+(7*2)] = 1, which is true since gcd (21n+25, 14n+17) =1 and adding some multiple of some number 'k=7' to both the numbers doesn't change the gcd.
Hence, the statement is true for natural number 'n+1'.
Hence, the statement is true for each natural number.
For this fraction to be irreducible, Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1959 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |