# Help: Proof that square root of 2 is irrational

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Hello

As part of my AS Maths Summer work, i have been asked to 'Search either books or the internet to find a proof ( a clear explanation that would convince a critical reader) that the square root of 2 is an irrational number' - I really cannot find anything, i tried to make one up but they're not really solid enough.......does anyone have any ideas? Thanx in advance, Shane

As part of my AS Maths Summer work, i have been asked to 'Search either books or the internet to find a proof ( a clear explanation that would convince a critical reader) that the square root of 2 is an irrational number' - I really cannot find anything, i tried to make one up but they're not really solid enough.......does anyone have any ideas? Thanx in advance, Shane

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#2

The classic one goes:

Any rational number can be expressed in the form p/q where p and q are integers and the fraction can be reduced no more.

So assume rt2 is rational

Let rt2 = p/q

2 = p^2/q^2

p^2 = 2q^2

implies that p^2 is even, implies that p is even, since an even x even is even, and odd x odd is odd for all integers.

so p is even, let p = 2m where m is also an integer (since p is even)

4m^2 = 2q^2 implies that 2m^2 = q^2, which implies that q^2 is even and so q is even.

so let q = 2n

implies that rt2 = 2m/2n, but this cancels to m/n

our original premise said that the fraction shouldn't cancel any further, but it has done, so our original premise must be wrong and rt2 must be irrational.

Any rational number can be expressed in the form p/q where p and q are integers and the fraction can be reduced no more.

So assume rt2 is rational

Let rt2 = p/q

2 = p^2/q^2

p^2 = 2q^2

implies that p^2 is even, implies that p is even, since an even x even is even, and odd x odd is odd for all integers.

so p is even, let p = 2m where m is also an integer (since p is even)

4m^2 = 2q^2 implies that 2m^2 = q^2, which implies that q^2 is even and so q is even.

so let q = 2n

implies that rt2 = 2m/2n, but this cancels to m/n

our original premise said that the fraction shouldn't cancel any further, but it has done, so our original premise must be wrong and rt2 must be irrational.

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#3

proof by contradiction my fellow mathematician!

Assume that root 2 is rational. If it is rational, it can be represetented as p/q where p and q are integers and have no common factors (Simplest ratio). So 2 = p^2/q^2. So 2q^2 = p^2

So the square of p is even, so p itself must be even (Because all odd numbers squared are odd, so p is even).

So lets say p=2r (since p is even)

then p^2 = 4r^2

so 2q^2 = 4r^2

so q^2 = 2r^2

so q^2 must also be even, so q must be even

BUT HOLD ON. That means that both p and q are even, and so p/q is not in it's simplest form (since you can divide by 2) so Root 2 can not be rational!

Assume that root 2 is rational. If it is rational, it can be represetented as p/q where p and q are integers and have no common factors (Simplest ratio). So 2 = p^2/q^2. So 2q^2 = p^2

So the square of p is even, so p itself must be even (Because all odd numbers squared are odd, so p is even).

So lets say p=2r (since p is even)

then p^2 = 4r^2

so 2q^2 = 4r^2

so q^2 = 2r^2

so q^2 must also be even, so q must be even

BUT HOLD ON. That means that both p and q are even, and so p/q is not in it's simplest form (since you can divide by 2) so Root 2 can not be rational!

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#4

Alas, even easier

2p^2 = q^2. Consider factors of 2 on both sides, the LHS must have an even factor of two, whilst the right must be even. Contradiction.

2p^2 = q^2. Consider factors of 2 on both sides, the LHS must have an even factor of two, whilst the right must be even. Contradiction.

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Hey thanx, you all helped me loads - i didnt have a clue! Wow, people always reply so quickly for maths - the mathematicians must be the kindest lot on the board. Thanx again, x

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#6

The classic proof- and actually conception of the irrational number concept- is the triangle, with sides 1, 1 and hypotenuse square root 2.

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#7

(Original post by

The classic proof- and actually conception of the irrational number concept- is the triangle, with sides 1, 1 and hypotenuse square root 2.

**tomcoolinguk**)The classic proof- and actually conception of the irrational number concept- is the triangle, with sides 1, 1 and hypotenuse square root 2.

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#8

Right you have your right angled triangle with sides a, b and h.

a= 1

b=1

therefore by pythagoras

h2= a2+ b2

h2= 2

therefore h= sq. root 2

a= 1

b=1

therefore by pythagoras

h2= a2+ b2

h2= 2

therefore h= sq. root 2

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#10

(Original post by

How does that show that h is irrational?

**fishpaste**)How does that show that h is irrational?

When I click the Fraction button on my calculator, the calculator fails to convert the decimal number into a fraction, which it can do if the decimal number is rational.

e.g.) 4.5 ---> Click Fraction Button -----> Shows 4.5 in Fraction form.

Hence: Square root 2 is irrational.

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#11

(Original post by

Right you have your right angled triangle with sides a, b and h.

a= 1

b=1

therefore by pythagoras

h2= a2+ b2

h2= 2

therefore h= sq. root 2

**tomcoolinguk**)Right you have your right angled triangle with sides a, b and h.

a= 1

b=1

therefore by pythagoras

h2= a2+ b2

h2= 2

therefore h= sq. root 2

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#12

(Original post by

Square root 2 = 1.4142...

When I click the Fraction button on my calculator, the calculator fails to convert the decimal number into a fraction, which it can do if the decimal number is rational.

e.g.) 4.5 ---> Click Fraction Button -----> Shows 4.5 in Fraction form.

Hence: Square root 2 is irrational.

**Invisible**)Square root 2 = 1.4142...

When I click the Fraction button on my calculator, the calculator fails to convert the decimal number into a fraction, which it can do if the decimal number is rational.

e.g.) 4.5 ---> Click Fraction Button -----> Shows 4.5 in Fraction form.

Hence: Square root 2 is irrational.

[random observations] I think if you get an ugly enough rational number, it won't convert it, if the period of the digits is longer than the capacity of the calculator. And also your calculator uses a rational approximation to rt2 anyway. Probably from the binomial expansions.

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#13

Haha, great, now we can add to Proof by Induction, Proof by Contradiction a brand new proof, "Proof by my calculator says so, nyah!".

Add to that several proofs my lecturers like such as "Proof because I say so" and "Proof by vigourous hand-waving".

Add to that several proofs my lecturers like such as "Proof because I say so" and "Proof by vigourous hand-waving".

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