Proof that the Square Root of 2 is Irrational.

[Calu]

I am trying to prove that √2 is irrational using proof by contradiction. Here is my work so far:

√2 = p/q where p & q are in their lowest terms. Where q is non-zero.

2=p²/q²

2q² = p²

Which tells me that p² is an even number, using the definition of an even number. We can use this definition because we know that q is an integer and an integer squared is also an integer. Hence 2q² is even and as 2q=p² we see that p² is also even. Can we use the knowledge that p is an integer to say that p is also an even number? Or is there another rule that I'm not aware of?

If we can use this knowledge, then we can say that p=2n, where n is some integer, using the definition of an even number.

2q²=(2n)²
2q²=4n²
q²=2n² where n² is an integer, as n was an integer from the definition of an even number.

∴ q² is also an even number. If both p and q are even numbers then the fraction p/q can be simplified further and p/q is not in its lowest terms. This shows that √2 is in fact not a rational number and proves that it is irrational.

 

[Dick]

I am trying to prove that √2 is irrational using proof by contradiction. Here is my work so far:

√2 = p/q where p & q are in their lowest terms. Where q is non-zero.

2=p²/q²

2q² = p²

Which tells me that p² is an even number, using the definition of an even number. We can use this definition because we know that q is an integer and an integer squared is also an integer. Hence 2q² is even and as 2q=p² we see that p² is also even. Can we use the knowledge that p is an integer to say that p is also an even number? Or is there another rule that I'm not aware of?

If we can use this knowledge, then we can say that p=2n, where n is some integer, using the definition of an even number.

2q²=(2n)²
2q²=4n²
q²=2n² where n² is an integer, as n was an integer from the definition of an even number.

∴ q² is also an even number. If both p and q are even numbers then the fraction p/q can be simplified further and p/q is not in its lowest terms. This shows that √2 is in fact not a rational number and proves that it is irrational.

All you need is that the square of an even number is even and the square of an odd number is odd. You can prove that. So if p^2 is even, p must be even since it can't be odd.

 

[Calu]

All you need is that the square of an even number is even and the square of an odd number is odd. You can prove that. So if p^2 is even, p must be even since it can't be odd.

If I take p² = 6, p = √6 which is why I thought the definition of p as an integer was important, as √6 isn't even.

 

[Dick]

If I take p² = 6, p = √6 which is why I thought the definition of p as an integer was important, as √6 isn't even.

Of course, it's important that p is an integer. Now I'm not sure what your question actually is.

 

[Calu]

Of course, it's important that p is an integer. Now I'm not sure what your question actually is.

It was really just whether knowing p was an integer was important. Now I know it is, so thank you.