# Sqrt of 4 is irrational

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• Jun 24th 2010, 08:13 AM
Mukilab
Sqrt of 4 is irrational
Hello please could you tell me if my proof that the sqrt of 4 is irrational is correct.

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

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

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

Equation 1.2 contradicts equation 1.1. This shows that a and b cannot be whole positive integers.
• Jun 24th 2010, 08:16 AM
Ackbeet
Going from $\displaystyle (a/2)^{2}=b^{2}$ to $\displaystyle a^{2}=2b^{2}$ is incorrect. You forgot to square the 2.
• Jun 24th 2010, 08:34 AM
Mukilab
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 :)
• Jun 24th 2010, 08:52 AM
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
• Jun 24th 2010, 08:57 AM
undefined
Quote:

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 $\displaystyle \sqrt{4}=2$ which is rational. No need for fancy proof.
• Jun 24th 2010, 08:58 AM
Mukilab
you need to prove that it is a rational number. I know it's a perfect square... that was never the point
• Jun 24th 2010, 08:59 AM
Ackbeet
Here's my proof: $\displaystyle (\pm 2)^{2}=4$, therefore, by the definition of square root, $\displaystyle \sqrt{4}=\pm 2$. Now, $\displaystyle -2=\frac{-2}{1}$, and $\displaystyle 2=\frac{2}{1}$. In either case, by the definition of rational number, we have a rational number. Therefore, $\displaystyle \sqrt{4}$ is rational.
• Jun 24th 2010, 09:02 AM
Mukilab
much simpler :)
• Jun 24th 2010, 09:07 AM
undefined
Quote:

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

This is actually not correct. The $\displaystyle \sqrt{}$ sign is defined as nonnegative (for reals).

Quote:

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 $\displaystyle \sqrt{4}=2 \in \mathbb{Q} \therefore \sqrt{4}\ \text{is rational.}$ is a valid proof.
• Jun 24th 2010, 09:08 AM
Mukilab
Your proof feels like cheating :(
• Jun 24th 2010, 09:09 AM
undefined
Quote:

Originally Posted by Mukilab
Your proof feels like cheating :(

But it's not. Prove that 6 is composite. $\displaystyle 6=2\cdot3\therefore6\ \text{is composite}.$
• Jun 24th 2010, 09:10 AM
Ackbeet
I knew someone was going to flag me on that one. Just interpret it to mean this: either $\displaystyle \sqrt{4}=2$ or $\displaystyle \sqrt{4}=-2$, which is certainly a true statement. I should have written it out in more fullness.
• Jun 24th 2010, 09:13 AM
undefined
Quote:

Originally Posted by Ackbeet
I knew someone was going to flag me on that one. Just interpret it to mean this: either $\displaystyle \sqrt{4}=2$ or $\displaystyle \sqrt{4}=-2$, 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 $\displaystyle \forall x\in\mathbb{R}:x=\pm x$. 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:

$\displaystyle 5=\pm5$

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

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

Contrasted with

$\displaystyle \forall x,y\in\mathbb{R},y\ge0:x^2=y \Leftrightarrow x=\pm \sqrt{y}$
• Jun 24th 2010, 09:13 AM
Mukilab
Quote:

Originally Posted by undefined
But it's not.

Quote:

Originally Posted by Mukilab
feels like

.
• Jun 24th 2010, 09:14 AM
Ackbeet
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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