How do you prove that the square root of 14 is irrational?

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How do you prove that the square root of 14 is irrational?

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Answer 1 · 2015-10-10T19:05:39

A rational number is expressed by ratio of integers.

Explanation:

The only square roots that are rational numbers are those who are perfect squares. $\sqrt{16}$ $sqrt16$ for example is a rational number because it equals 4 and 4 is an integer. $\sqrt{14}$ $sqrt14$ =3.74, which is not an integer and therefore is an irrational number.

Answer 2 · 2015-10-10T20:34:34

Use proof by contradiction...

Explanation:

Suppose $\sqrt{14}$ $sqrt(14)$ is rational.

Then $\sqrt{14} = \frac{p}{q}$ $sqrt(14) = p/q$ for some positive integers $p, q$ $p, q$ with $q \neq 0$ $q != 0$ .

Without loss of generality, we can suppose that $p$ $p$ and $q$ $q$ are the smallest such pair of integers.

${(\frac{p}{q})}^{2} = 14$ $(p/q)^2 = 14$

So:

$p^{2} = 14 q^{2}$ $p^2 = 14 q^2$

In particular, $p^{2}$ $p^2$ is even.

If $p^{2}$ $p^2$ is even, then $p$ $p$ must be even too, so $p = 2 k$ $p = 2k$ for some positive integer $k$ $k$ .

So:

$14 q^{2} = {(2 k)}^{2} = 4 k^{2}$ $14 q^2 = (2k)^2 = 4 k^2$

Dividing both sides by $2$ $2$ , we get:

$7 q^{2} = 2 k^{2}$ $7 q^2 = 2 k^2$

So $k^{2}$ $k^2$ must be divisible by $7$ $7$ . So $k$ $k$ must be divisible by $7$ $7$ too. So $k = 7 m$ $k = 7m$ for some positive integer $m$ $m$ .

$7 q^{2} = 2 {(7 m)}^{2} = 7 \cdot 14 m^{2}$ $7 q^2 = 2 (7m)^2 = 7*14m^2$

Divide both sides by $7$ $7$ to find:

$q^{2} = 14 m^{2}$ $q^2 = 14 m^2$

So $14 = \frac{q^{2}}{m^{2}} = {(\frac{q}{m})}^{2}$ $14 = q^2/m^2 = (q/m)^2$

So $\sqrt{14} = \frac{q}{m}$ $sqrt(14) = q/m$

Now $m < q < p$ $m < q < p$ , contradicting our supposition that $p$ $p$ and $q$ $q$ are the smallest pair of positive integers such that $\sqrt{14} = \frac{p}{q}$ $sqrt(14) = p/q$ .

So our supposition is false and therefore our hypothesis that $\sqrt{14}$ $sqrt(14)$ is rational is also false.

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How do you prove that the square root of 14 is irrational?

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