# Prove or disprove a claim regarding irrational numbers

I am trying to prove the following claim:

Let $$0\leq n \in \Bbb Z$$ and suppose that there exists a $$k \in \Bbb Z$$ such that $$n=4k+3$$. Prove or disprove: $$\sqrt n \notin \Bbb Q$$ .

The problem I am having is that I am trying to assume by contradiction that $$\sqrt n \in \Bbb Q$$ and then I say that there are $$a,b \in \Bbb Z$$ such that $$n=\sqrt {4k+3}=\frac ab$$. I finally get to a point where $$k=\frac {a^2-3b^2}{4b^2}$$. Yet I can't find any $$a,b \in \Bbb Z$$ that will help me show that the claim is false, nor show a contradiction that will cause the claim to be true. Any help will be welcomed.

• $k$ is an integer. So that, you don't need $b$. Thus, $b=1$. Oct 24 at 20:24
• Try to solve “If $a^2-3b^2$ is divisible by $4,$ then $a,b$ are both even.” Oct 24 at 20:25
• @lonestudent What? That’s a jump. Oct 24 at 20:27
• @ThomasAndrews Am I thinking wrong?.. Oct 24 at 20:29
• Well, $\frac A{4b^2}$ can be an integer, without $b=1.$ The question is what about $a^2-3b^2$ let’s us say this particular $b$ must be $1.$ Hence I called it “quite a jump” rather than “wrong.” @lonestudent Oct 24 at 20:39

Statement:

Let $$a,b,k\in\mathbb Z^{+}$$, where $$\gcd (a,b)=1$$ and if $$4k+3=\frac{a^2}{b^2}$$, then $$b^2=1$$ or $$b=1$$.

Thus we have,

$$\sqrt{4k+3}=a,\thinspace a\in\mathbb Z^{+}$$

and

$$k=\frac{a^2-4+1}{4}=\frac{a^2+1}{4}-1$$

This immediately implies,

$$a=2m-1, \thinspace m\in\mathbb Z^{+}$$

This means,

\begin{align}a^2+1&=4(m^2-m)+2\not\equiv 0\thinspace\thinspace\thinspace\text{(mod 4)}.&\end{align}

Conclusion:

We conclude that, there doesn't exist $$n=4k+3,\thinspace k\in\mathbb Z^{+}$$, such that $$\sqrt n\in\mathbb Q^{+}$$.

• I don't understand why $k \in \Bbb Z$ implies that $\sqrt {4k+3} \in \Bbb Z$ For example for $k=1$ we have that $a=\sqrt 7 \notin \Bbb Z$ Oct 24 at 20:39
• I understand but no one said that $\sqrt n$ is an integer, thus I don't know why you assume that ${4k+3}$ is a perfect square. Oct 24 at 20:57
• @Yuval I expanded my answer. Now, everything is clear. Oct 24 at 21:15

The square of an even integer is $$4k$$, the square of an odd integer is $$8k+1$$. $$4k+3$$ is never the square of an integer, neither is $$4k+2$$ nor $$8k+5$$. So the square root of $$n$$ is not an integer.

Now you need to remember the well known proof that the square root of 2 is irrational; that proof can be adapted to show that the square root of any integer is either an integer or irrational.

It is given that $$n=4k+3$$, so I think you mean to say that there exist $$a,b\in\Bbb{Z}$$ such that $$\sqrt{n}=\sqrt{4k+3}=\tfrac ab,$$ in the hopes of reaching a contradiction. Indeed some algebra then leads to $$k=\frac{a^2-3b^2}{4b^2},$$ which means that $$4b^2$$ should divide $$a^2-3b^2$$ because $$k$$ is an integer. In particular $$4$$ should divide $$a^2-3b^2$$. This implies that $$a$$ and $$b$$ are both even [prove this!], say $$a=2A$$ and $$b=2B$$. Plugging this in then gives $$k=\frac{(2A)^2-3(2B)^2}{4(2B)^2}=\frac{4A^2-12B^2}{16B^2}=\frac{A^2-3B^2}{4B^2},$$ and so by the same argument $$A$$ and $$B$$ are again both even. Can you see the contradiction from here?

• I understand that this process can go on and on and reach the same point every time, but how can I reach a contradiction from here? Oct 24 at 20:43
• There are two ways to proceed. One way is to conclude that both $a$ and $b$ are integers that can be divided by $2$ indefinitely, which implies $a=b=0$, a contradiction. The other way is to start with a reduced fraction $\tfrac{a}{b}$, so that $a$ and $b$ are coprime. Then the conclusion that both are even is already a contradiction. Oct 24 at 20:44

Note the if $$q$$ is rational but not integral then $$q^2$$ will also not be integral. So you only need to prove that $$n$$ is not a power of an integer.

Note that if $$m=2l$$ then $$m^2=4l^2$$, if $$m=4l+1$$ then $$m^2=16l^2+4l+1$$ and if $$m=4l+3$$ then $$m^2 = 16l^2+12l + 8 +1$$. So for any $$m$$ we have that $$m^2$$ is either of the form $$4r$$ or of the form $$4r+1$$.

In different words: Modulo $$4$$ we have $$0^2=0,1^2=1,2^2=0,3^2=1$$.

So any number of the form $$4k+3$$ is not a square of an integer.

$$\sqrt n=\sqrt{4k+3}$$ is a solution of the equation $$x^2 -(4k+3)=0$$

Now by Rational Zeros Theorem $$\biggr($$Which states that : $$x=p/q$$ is a Rational Number satisfying the polynomial equation $$\sum_{r=0}^n a_rx^r$$ then $$q(\neq0)|a_n$$ and $$p|a_0$$ and gcd(p,q)=1 $$\biggr)$$ , the only Rational roots of the equation are $$±1,±(4k+3),±n|(4k+3) \forall k\in \Bbb Z^+$$ where $$n \in \Bbb Z^+$$. But $$\sqrt{4k+3}$$ is neither of them . So $$\sqrt n=\sqrt{4k+3}$$ is irrational.