Proof that the square root of 2 is irrational

In summary, an irrational number is one that cannot be expressed as a ratio of two integers and has a non-terminating and non-repeating decimal representation. The square root of 2 is considered irrational because it cannot be simplified to a whole number or fraction. Its proof of irrationality, discovered by Pythagoras, is important as it led to a deeper understanding of numbers and their properties and has many real-life applications in mathematics, physics, and engineering.
  • #1
Mr Davis 97
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Quick question: In the proof that the square root of 2 is irrational, when we are arguing by contradiction, why are we allowed to assume that ##\displaystyle \frac{p}{q}## is in lowest terms? What if we assumed that they weren't in lowest terms, or what if we assumed that ##\operatorname{gcd} (p,q)=2##? Would the contradiction still follow?
 
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  • #2
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Mr Davis 97 said:
why are we allowed to assume that ##\frac{p}{q}## is in lowest terms?
We are using (without citing it) the fact that ##p,q## are positive integers and that the set of positive integers is well-ordered, meaning that every subset has a least element. That guarantees that there exists a unique representation of the fraction in lowest terms, because the set S of all numerators of representations of the fraction is a subset of the positive integers and so must have a least element m. We choose the subset of S that have numerators equal to m. Then we observe that if m/a and m/b are both representations of the fraction, we have m/a=m/b and neither of a nor b is zero, so that a=b, so there is only one representation with numerator m.

BTW, we don't need to assume p and q are on lowest terms. All we need to assume to get the proof to work is that 2 is not a common factor of both, ie that ##2\not| \ \gcd(p,q)##.

If we don't assume that, as would be the case if ##\gcd(p,q)=2##, we can't get our contradiction. We could probably recover it by then entering into a long induction, but there's no need to do that since we can just assume lowest terms, which implies ##2\not| \ \gcd(p,q)##.
 
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  • #3
This discussion may help:

http://mathforum.org/library/drmath/view/52612.html

Basically, we start by assuming that ##\sqrt{2}## is rational then we can conclude that there exists a fully reduced ratio of integers p/q that represent it. From there the proof goes on to show that p/q isn't fully reduced.

It's a key part of the proof. You could start with that notion and then state that there is a common factor to the top and bottom that can be factored out to get a reduced ratio but it is just a side show and it just detracts from the elegance of the proof.

But wait until other mentors reply to get their opinion.
 
  • #4
Mr Davis 97 said:
Quick question: In the proof that the square root of 2 is irrational, when we are arguing by contradiction, why are we allowed to assume that ##\displaystyle \frac{p}{q}## is in lowest terms? What if we assumed that they weren't in lowest terms, or what if we assumed that ##\operatorname{gcd} (p,q)=2##? Would the contradiction still follow?
You only need the fact that every integer can be written as a product of primes in a unique way (##\,##up to ##\pm1\,##). This way you get ##\sqrt{2}=\frac{n}{m} \Longrightarrow 2m^2=n^2## and you end up with one factor ##2## more on the left than on the right. Strictly speaking one also uses that ##\mathbb{Z}## has no zero divisors, because it allows to cancel out all primes, which are on the left and on the right.
 
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  • #5
Mr Davis 97 said:
Quick question: In the proof that the square root of 2 is irrational, when we are arguing by contradiction, why are we allowed to assume that ##\displaystyle \frac{p}{q}## is in lowest terms? What if we assumed that they weren't in lowest terms, or what if we assumed that ##\operatorname{gcd} (p,q)=2##? Would the contradiction still follow?

It would follow that ##gcd(p,q) = 2##, meaning both numerator and denumerator have a factor ##2##. You can cancel it. Repeat the process until no more common factors 2 are there (why is this possible?). Conclude that ##\sqrt{2}## is irrational.
 

Related to Proof that the square root of 2 is irrational

What does it mean for a number to be irrational?

A number is considered irrational if it cannot be expressed as a ratio of two integers (whole numbers). This means that the decimal representation of an irrational number is non-terminating and non-repeating.

Why is the square root of 2 considered an irrational number?

The square root of 2 is considered irrational because it cannot be expressed as a fraction of two integers. It has an infinite number of non-repeating decimal places and cannot be simplified to a whole number or fraction.

What is the proof that the square root of 2 is irrational?

The proof that the square root of 2 is irrational was first discovered by the ancient Greek mathematician, Pythagoras. It involves assuming that the square root of 2 can be expressed as a fraction and then using logical reasoning to show that this assumption leads to a contradiction.

Why is the proof of the square root of 2 being irrational important?

The proof of the square root of 2 being irrational is important because it is one of the first proofs of irrationality and led to a deeper understanding of numbers and their properties. It also paved the way for the development of other proofs of irrationality for different numbers.

How is the proof of the square root of 2 being irrational used in real life?

The proof of the square root of 2 being irrational has many applications in mathematics, physics, and engineering. It is used in calculations involving right triangles, Pythagorean triples, and other geometric shapes. It is also used in cryptography to ensure secure communication and in computer algorithms for efficient calculations.

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