# Thread: Sqrt of 4 is irrational

1. ## Sqrt of 4 is irrational

Hello please could you tell me if my proof that the sqrt of 4 is irrational is correct.

$\sqrt{4}=\frac{a}{b}$ therefore $4=(\frac{a}{b})^{2}$ so $4b^2=a^2$....1.1

This implies that a is greater than b and that a is even.

a is even so a=2k

put into 1.1 $4b^2=4k^2$

cancel by 4 $b^2=k^2$

this can be written as $(\frac{2k}{2})^{2}=b^2$ this implies that $(\frac{a}{2})^{2}=b^2$ which implies that $a^2=2b^2$....1.2

Equation 1.2 contradicts equation 1.1. This shows that a and b cannot be whole positive integers.

2. Going from $(a/2)^{2}=b^{2}$ to $a^{2}=2b^{2}$ is incorrect. You forgot to square the 2.

3. oh. Thanks for that, I didn't realise it. Please can nobody post a solution until I ask for it, I want to solve it on my own

4. I think I've proved it is rational.

a^2=4b^2 = 2b=a

Because we have proved a is an even number, b also must be even

Because a is greater than b and they are both integers we know it cannot be a decimal and we know that if a number can be written as a fraction then it is rational. Because they are both even it can be written as a fraction.

Q.E.D

5. Originally Posted by Mukilab
I think I've proved it is rational.

a^2=4b^2 = 2b=a

Because we have proved a is an even number, b also must be even

Because a is greater than b and they are both integers we know it cannot be a decimal and we know that if a number can be written as a fraction then it is rational. Because they are both even it can be written as a fraction.

Q.E.D
4 is a perfect square, and $\sqrt{4}=2$ which is rational. No need for fancy proof.

6. you need to prove that it is a rational number. I know it's a perfect square... that was never the point

7. Here's my proof: $(\pm 2)^{2}=4$, therefore, by the definition of square root, $\sqrt{4}=\pm 2$. Now, $-2=\frac{-2}{1}$, and $2=\frac{2}{1}$. In either case, by the definition of rational number, we have a rational number. Therefore, $\sqrt{4}$ is rational.

8. much simpler

9. Originally Posted by Ackbeet
Here's my proof: $(\pm 2)^{2}=4$, therefore, by the definition of square root, $\sqrt{4}=\pm 2$. Now, $-2=\frac{-2}{1}$, and $2=\frac{2}{1}$. In either case, by the definition of rational number, we have a rational number. Therefore, $\sqrt{4}$ is rational.
This is actually not correct. The $\sqrt{}$ sign is defined as nonnegative (for reals).

Originally Posted by Mukilab
you need to prove that it is a rational number. I know it's a perfect square... that was never the point
Writing $\sqrt{4}=2 \in \mathbb{Q} \therefore \sqrt{4}\ \text{is rational.}$ is a valid proof.

10. Your proof feels like cheating

11. Originally Posted by Mukilab
But it's not. Prove that 6 is composite. $6=2\cdot3\therefore6\ \text{is composite}.$

12. I knew someone was going to flag me on that one. Just interpret it to mean this: either $\sqrt{4}=2$ or $\sqrt{4}=-2$, which is certainly a true statement. I should have written it out in more fullness.

13. Originally Posted by Ackbeet
I knew someone was going to flag me on that one. Just interpret it to mean this: either $\sqrt{4}=2$ or $\sqrt{4}=-2$, which is certainly a true statement. I should have written it out in more fullness.
Ah, you're right, didn't think of that. In fact it's true that $\forall x\in\mathbb{R}:x=\pm x$. Misinterpretation on my part.

Edit: Actually, I'm wondering whether this is normal syntax for the "=" sign, or for equations containing an unknown.

Edit 2: I think the syntax is perfectly valid, after a little thought. I don't think anyone would have a problem with this statement:

$5=\pm5$

Then in the context of solving equations, we have, for example

$
\forall x,y\in\mathbb{R}:x=y \Rightarrow x=\pm y$

Contrasted with

$
\forall x,y\in\mathbb{R},y\ge0:x^2=y \Leftrightarrow x=\pm \sqrt{y}$

14. Originally Posted by undefined
But it's not.
Originally Posted by Mukilab
feels like
.

15. Well, more than that, I was just trying to get at all numbers that square to equal 4. Perhaps that's more even than is called for.

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