Proving $\sqrt2$ is irrational

Question

I used the method of contradiction by assuming that $\sqrt 2$ is a rational number. Then, by the definition of rational number, there exist two integers $p$ and $q$ whose ratio equals $\sqrt 2$. Thus, $$\frac pq = \sqrt2\tag{x}$$

Squaring both sides, $$p^2/q^2 = 2\tag{a}$$

or $$p^2 = 2q^2\tag{b}$$ This means that $p^2$ is an even number, implying that $p$ is even.

Now, any even integer can be written as $2^kf$ where $f$ is any odd integer and $k$ is some positive integer (the minimum value of $k$ and $f$ is $1$ since even numbers start from $2 = 2^1\cdot1$). For odd numbers, $k=0$.

For example, $8 = 2^3 \cdot 1$, $4= 2^2 \cdot 1$, $18= 2^1 \cdot 9$, $24 = 2^3 \cdot 3$, $-12= 2^2 \cdot(-3)$, etc.

Now, from $(b)$, $q^2$ can be even or odd.

Case 1: $q^2$ is even (thus meaning $q$ is even). Then $p= 2^{k_1} \cdot f_1$ (say) and $q = 2^{k_2} * f_2$ (say). Note here in this case, both conditions $k_1=k_2$ and $f_1=f_2$ can't hold simultaneously since that would mean $p/q =1$ and here $p/q =\sqrt2$ (from $(x)$)). Let's consider the condition when $k_1=k_2=k$ but $f_1\ne f_2$. Then $$\frac{p^2}{q^2}=\frac{(2^kf_1)^2}{(2^kf_2)^2} = \frac{f_1^2}{f_2^2} =2$$ (from $(a)$), i.e.: odd/odd ($f_1$ and $f_2$ are odd) can never equal $2$.

Now lets consider $f_1=f_2=f$ but $k_1\ne k_2$, thus $(2^{k_1}f)^2/(2^{k_2}f)^2= 2^{2(k_1-k_2)}= 2$ meaning $k_1 - k_2 = 0.5$ but $k_1,k_2$ are integers, so their difference can't be $0.5$. (Also here, $k_1-k_2$ must be greater or equal to $1$ since $2^{k_1-k_2}=\sqrt 2$, $k_1-k_2$ can't be negative since $\sqrt 2>1$, but $k_1-k_2\ge1$ can't satisfy $(a)$ since the minimum value of $p/q$ in this case will be $2$ which is surely greater than $\sqrt2$.)

Now lets consider $f_1\ne f_2$ and $k_1\ne k_2$, thus $\frac{(2^{k_1} f_1)^2}{(2^{k_2} f_2)^2}= 2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal $2$ since $\frac{f_1^2}{f_2^2}$ is either odd or "odd/odd" ie: it doesn't contain 2 as a factor and $2^{2(k_1-k_2)}$ is a power of $2$. So in $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$, there's no chance of cancellation of the odd factor $\frac{f_1^2}{f_2^2}$.

Case 2: $q^2$ is odd (thus meaning $q$ is odd). $q=f'$ (say, here $k=0$ for $q$ but not for $p$), thus from $(a)$ $2^{2k}\frac{f^2}{f'^2} = 2$ but this isn't possible since a power of $2$ multiplied with odd factor can't equal $2$.

Thus, both case 1 and 2 suggest that for any possible combination of $k_1,k_2,f_1$ and $f_2$, $p/q\ne\sqrt2$, i.e.: for no value of integers $p$ and $q$, $p/q = \sqrt2$. Thus, this contradicts our assumption that $\sqrt2$ is rational. Therefore it must be irrational.

PLS NOTE that i would like to clarify that $f_1^2$/$f_2^2$ is either an odd integer or a fraction of odd/odd or 1/odd form, thus not containing 2 as a factor,for any $2^n$, if it has to be reduced to 2 , must be multiplied with $1/ 2^(n-1)$ but $f_1^2$/$f_2^2$ can't be of $1/2^[n-1]$ form since for odd nos, in $2^k$*f notation, k is 0 , so 2 vanishes as $2^k$ becomes 1 in this case thus f1/f2 can't be represented as 1/ $2^(n-1)$ in which n has to be existent. therefore $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal 2 as to further explain the 3rd condition of case 1

Difficulty: is my approach correct? This proof which I thought is different from proofs found on the internet or books since I have used $p$ and $q$ as any integers, which may have a common factor. So I am not sure if I am on right track. Will someone please check this out and make kind comments?

Answer 1

I'm sorry, but try to understand this proof, which is the traditional one.

$\sqrt{2} = \dfrac{m}{n}\ $ where $\textrm{gcd}(m,n) = 1$

$2n^2 = m^2 \Rightarrow 2 \mid m^2 \Rightarrow 2\mid m \Rightarrow m^2 = 4L^2$

$2n^2 = 4L^2 \Rightarrow 2 \mid n^2 \Rightarrow 2 \mid n \Rightarrow \textrm{gcd}(m,n) \not = 1$ which is a contradiction.

Edit: $\sqrt{2} = \frac{m}{n}$.

If $\textrm{gcd}(m,n) = d>1$ then we can write $m = dx$ and $n = dy \Rightarrow \sqrt{2} = \dfrac{dx}{dy} = \dfrac{x}{y}$.

Then $\textrm{gcd}(x,y) = 1$. You can prove this by assuming for the sake of contradiction that it is not and showing that you get a divisior of $m,n$ which is greater than $d$.

Answer 2

Your approach is basically correct, but far too complicated, and it leads you into some trouble. For example, you write that $\frac{f_1^2}{f_2^2}$ is odd, but it may not be an integer, and you don't say what it means for a fraction to be odd. These problems can be repaired, but I recommend taking a simpler approach altogether.

You made the most important observation: we can write an integer as the product of an odd number and a power of $2$. So $p = 2^k a$, $q=2^l b$, where $a,b$ are odd integers, and $k,l$ are nonnegative integers.

Then $p^2 = 2q^2 \iff 2^{2k}a^2 = 2^{2l+1} b^2$. Since the number of powers of $2$ dividing an integer is unique, we have $2k=2l+1$. But an even integer cannot equal an odd integer, contradiction.

This is not any different from your proof—the core idea and most of the logic is nearly identical—but it avoids a lot of troubles by not breaking into cases.

Answer 3

Conceptually the proof compares the parity of the $\it\color{#90f}{unique}$ powers of $\,2\,$ on each side of $\,P^2 = 2Q^2.\,$ We explain how to view it in terms of unique factorization - but only using the single prime $\,p = 2$.

A simple induction shows every natural $> 0$ can be written $\it\color{#90f}{uniquely}$ in the form $\, 2^a n\,$ for odd $\,n.\,$ Thus $\, P = 2^a p,\ Q = 2^b q\,$ for $\,p,q\,$ odd, so $\, P^2 = 2 Q^2\,$ $\Rightarrow$ $\,2^{\color{#c00}{2a}} p^2 = 2^{\color{#0a0}{1+2b}} q^2$ for $\,p^2,q^2\,$ odd, contradicting said $\it\color{#90f}{uniqueness}$, because the LHS has $\rm\color{#c00}{even}$ power of $2$ vs. $\rm\color{#0a0}{odd}$ power on RHS.

Remark $ $ The above employed unique factorization $\,2^a n\,$ for a single prime $(p=2),\,$ a special case of the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations) whose proof is much easier than the general case involving arbitrary primes.

Answer 4

So continuing your way i.e not assuming gcd(p,q) = 1,

$2 = \frac{p^2}{q^2}$

This implies, $p^2=2q^2$.

As power of 2 in $q^2$ is even, power of 2 in $p^2$ will be odd. This is not possible as power of every prime in square of an integer is even.