Difference between revisions of "1959 IMO Problems/Problem 1"
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Prove that the fraction <math>\frac{21n+4}{14n+3}</math> is irreducible for every natural number <math>n</math>. | Prove that the fraction <math>\frac{21n+4}{14n+3}</math> is irreducible for every natural number <math>n</math>. | ||
+ | == Solution 1 == | ||
− | + | Denoting the greatest common divisor of <math>a, b </math> as <math>(a,b) </math>, we use the [[Euclidean algorithm]]: | |
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− | + | <cmath>(21n+4, 14n+3) = (7n+1, 14n+3) = (7n+1, 1) = 1</cmath> | |
− | + | Their greatest common divisor is 1, so <math>\frac{21n+4}{14n+3}</math> is irreducible. Q.E.D. | |
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− | + | == Solution 2 == | |
[[Proof by contradiction]]: | [[Proof by contradiction]]: | ||
− | Assume that <math>\dfrac{14n+3}{21n+4}</math> is a [[reducible fraction]] where <math>p</math> is | + | Assume that <math>\dfrac{14n+3}{21n+4}</math> is a [[reducible fraction]] where <math>p</math> is the [[greatest common factor]] of <math>14n+3</math> and <math>21n + 4</math>. Thus, |
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<center> | <center> | ||
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<math>21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}</math> | <math>21n+4\equiv 0\pmod{p} \implies 42n+8\equiv 0\pmod{p}</math> | ||
</center> | </center> | ||
− | Subtracting the second [[equation]] from the first | + | Subtracting the second [[equation]] from the first equation we get <math>1\equiv 0\pmod{p}</math> which is clearly absurd. |
Hence <math>\frac{21n+4}{14n+3}</math> is irreducible. Q.E.D. | Hence <math>\frac{21n+4}{14n+3}</math> is irreducible. Q.E.D. | ||
<!--Solution by tonypr--> | <!--Solution by tonypr--> | ||
− | + | == Solution 3 == | |
[[Proof by contradiction]]: | [[Proof by contradiction]]: | ||
+ | |||
+ | Assume that <math>\dfrac{14n+3}{21n+4}</math> is a [[reducible fraction]]. | ||
If a certain fraction <math>\dfrac{a}{b}</math> is reducible, then the fraction <math>\dfrac{2a}{3b}</math> is reducible, too. | If a certain fraction <math>\dfrac{a}{b}</math> is reducible, then the fraction <math>\dfrac{2a}{3b}</math> is reducible, too. | ||
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− | + | == Solution 4 == | |
We notice that: | We notice that: | ||
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− | + | == Solution 5 == | |
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. | ||
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+ | == Solution 7 == | ||
+ | Let <math>\gcd (21n+4,14n+3)=a.</math> So for some co-prime positive integers <math>x,y</math> we have <cmath>21n+4 = ax \qquad (1)</cmath> <cmath>14n+3 = ay \qquad (2)</cmath> Multiplying <math>(1)</math> by <math>2</math> and <math>(2)</math> by <math>3</math> and then subtracting <math>(1)</math> from <math>(2)</math> we get <cmath>42n+9-(42n+8)=3ay-2ax</cmath> <cmath>\implies a(3y-2x)=1.</cmath> We must have <math>a=1</math>, since <math>a</math> is an positive integer. Thus, <math>\gcd(21n+4,14n+3)=1</math> which means the fraction is irreducible, as needed. | ||
+ | |||
+ | == Video Solution by OmegaLearn== | ||
+ | https://youtu.be/zfChnbMGLVQ?t=1266 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | https://www.youtube.com/watch?v=IpUh5cLP454&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=1 | ||
+ | - AMBRIGGS | ||
+ | |||
+ | https://youtu.be/Pfg4MVUML64 | ||
+ | |||
+ | - little fermat | ||
{{alternate solutions}} | {{alternate solutions}} |
Latest revision as of 09:59, 23 July 2023
Contents
[hide]Problem
Prove that the fraction is irreducible for every natural number .
Solution 1
Denoting the greatest common divisor of as , we use the Euclidean algorithm:
Their greatest common divisor is 1, so is irreducible. Q.E.D.
Solution 2
Assume that is a reducible fraction where is the greatest common factor of and . Thus,
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.
Solution 7
Let So for some co-prime positive integers we have Multiplying by and by and then subtracting from we get We must have , since is an positive integer. Thus, which means the fraction is irreducible, as needed.
Video Solution by OmegaLearn
https://youtu.be/zfChnbMGLVQ?t=1266
~ pi_is_3.14
https://www.youtube.com/watch?v=IpUh5cLP454&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=1 - AMBRIGGS
- little fermat
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 |