Prove the square root of 6 is not rational?

Answer 1

Assume that sqrt(6) is rational.

Then sqrt(6) = p/q where p and q are coprime integers.

sqrt(6)^2 = 6 = p²/q²

p² = 6q²

therefore p² is an even number since an even number multiplied by any other integer is also an even number. If p² is even then p must also be even since if p were odd, an odd number multiplied by an odd number would also be odd.

So we can replace p with 2k where k is an integer.

(2k)² = 6q²
4k² = 6q²
2k² = 3q²

Now we see that 3q² is even. For 3q² to be even, q² must be even since 3 is odd and an odd times an even number is even. And by the same argument above, if q² is even then q is even.

So both p and q are even which means both are divisible by 2. But that means they are not coprime, contradicting our assumption so sqrt(6) is not rational.

Answer 2

Proof by contradiction:
root[6]=a/b
since it's rational it's got a form with a/b where they are both rational, integers and in their most reduced form.

6=aa/bb
2x3=aa/bb
3x2xbb=aa
so aa=2bb so aa is even so a is even,
let a=2c
6=4cc/bb
6bb=4cc
3bb=2cc
(it doesn't matter that it's 3bb because 3 isn't a multiple of 2 so can;t contribute to 3bb's evenness)
therefore bb is even and b is even, but if a/b is in it's most reduced form and a and b can't be even so there is a contradiction and root 6 is irrational.

This is an adaption of the common root 2 is irrational proof. You could show that they are both multiplies of 3 doesn't matter.

Answer 3

Let a/b = √6, where a and b are integers with no common divisors. Then a² = 6b². Since 6 is divisible by 3, then a must be divisible by 3. But a² must be divisible by 3², which means that b must be divisible by 3, contradicting the original statement. QED.

Answer 4

I'll just sketch it. it goes like this...by way of contradiction,

Suppose p/q = sqr(6) where, p/q is a reduced fraction

Then p^2/q^2 = 6

p^2 = 6q^2

This means p^2 is even.

So p must also be even.

So let p = 2k

(2k)^2 =6q^2

4k^2 = 6q^2

2k^2 = 3q^2

This means q^2 is even. So then this would mean q is even. So q = 2j

But this would be a contradiction because....

figure out the punchline...hope this helps.

Answer 5

the square root of 6 = 2.449489743
if it were rational there would be some sort of repetition, which there clearly is not.