Difference between revisions of "1959 IMO Problems/Problem 1"
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By [[Bezout's Lemma]], <math>3 \cdot (14n+3) - 2 \cdot (21n + 4) = 1</math>, so the GCD of the numerator and denominator is <math>1</math> and the fraction is irreducible. | By [[Bezout's Lemma]], <math>3 \cdot (14n+3) - 2 \cdot (21n + 4) = 1</math>, so the GCD of the numerator and denominator is <math>1</math> and the fraction is irreducible. | ||
+ | ===Solution 6=== | ||
+ | To understand why its irreducible, let's take a closer look at the fraction itself. If we were to separate both fractions, we end up with <math> \frac{21n}{14n} </math> + <math> \frac{4}{3} </math>. | ||
+ | Simplifying the fraction, we end up with. <math> \frac{3}{2} </math>. Now combining these fractions with addition as shown before in the problem, we end up with <math> \frac{17}{6} </math>. It's important to note that 17 is 1 off from being divisible by 6, and you'll see why later down this explanation. Now we expirement with some numbers. Plugging 1 in gives us | ||
+ | <math> \frac{25}{17} </math>. Notice how both numbers are one off from being divisible by 8(25 is next to 24, 17 is next to 16). | ||
+ | Trying 2, we end up with <math> \frac{46}{31} </math>. Again, both results are 1 off from being multiples of 15. | ||
+ | Trying 3, we end up with <math> \frac{67}{45} </math>. Again, both end up 1 away from being multiples of 22. This is where the realization comes in that two scenarios keep reoccurring: | ||
+ | |||
+ | 1. Both Numerator and Denominator keep ending up 1 value away from being multiples of the same number, but | ||
+ | |||
+ | 2. One always ends up being a prime number. Knowing that prime numbers only factors are 1 and itself, the fraction ends up being a paradoxical expression where one prime number is always being produced, and even with larger values, like say 15, implementing it in gives us <math> \frac{319}{199} </math>, we keep ending up with results that at the end will only have a common divisor of 1. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 19:11, 11 January 2021
Contents
Problem
Prove that the fraction is irreducible for every natural number
.
Solutions
Solution
Denoting the greatest common divisor of as
, we use the Euclidean algorithm:
It follows that is irreducible. Q.E.D.
Solution 2
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.
Solution 3
Assume that is a reducible fraction.
If a certain fraction is reducible, then the fraction
is reducible, too.
In this case,
.
This fraction consists of two consecutives numbers, which never share any factor. So in this case, is irreducible, which is absurd.
Hence is irreducible. Q.E.D.
Solution 4
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, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that
is irreducible.
Q.E.D
Solution 5
By Bezout's Lemma, , so the GCD of the numerator and denominator is
and the fraction is irreducible.
Solution 6
To understand why its irreducible, let's take a closer look at the fraction itself. If we were to separate both fractions, we end up with +
.
Simplifying the fraction, we end up with.
. Now combining these fractions with addition as shown before in the problem, we end up with
. It's important to note that 17 is 1 off from being divisible by 6, and you'll see why later down this explanation. Now we expirement with some numbers. Plugging 1 in gives us
. Notice how both numbers are one off from being divisible by 8(25 is next to 24, 17 is next to 16).
Trying 2, we end up with
. Again, both results are 1 off from being multiples of 15.
Trying 3, we end up with
. Again, both end up 1 away from being multiples of 22. This is where the realization comes in that two scenarios keep reoccurring:
1. Both Numerator and Denominator keep ending up 1 value away from being multiples of the same number, but
2. One always ends up being a prime number. Knowing that prime numbers only factors are 1 and itself, the fraction ends up being a paradoxical expression where one prime number is always being produced, and even with larger values, like say 15, implementing it in gives us , we keep ending up with results that at the end will only have a common divisor of 1.
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 |