# arrow_back Prove that $\log _{2} 3$ and $\sqrt{2}$ are irrational.

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Prove that $\log _{2} 3$ and $\sqrt{2}$ are irrational.

Definition: A rational number is any number that can be expressed as the quotient or fraction a/b of two integers, with the denominator $\mathrm{b}$ not equal to zero. A real number that is not rational is called irrational.

Proof: (a) Let's assume $\log _{2} 3$ is rational, then there exist integers $\mathrm{p}$ and $\mathrm{q}$, such that

$$\log _{2} 3=\frac{p}{q}$$

Since $\log _{2} 3$ is positive, we can assume both $\mathrm{p}$ and qto be positive integers.

Taking the power of 2 on both sides of (11.1), we get

\begin{aligned} &2^{\log _{2} 3}=2^{\frac{p}{q}} \\ &3=2^{\frac{p}{q}} \end{aligned}

Raising both sides by $q$ -th power, we have

$$3^{q}=2^{p}$$

However, $3^{\mathrm{q}}$ is an integer not divisible by 2, so it can not be a positive power of 2 which makes the equality $3^{\mathrm{q}}=2^{\mathrm{p}}$ impossible. This contradiction proves that the original assumption that $\log _{2} 3$ is rational is incorrect. So $\log _{2} 3$ is irrational.

(b) Assume $\sqrt{2}$ is rational, then there exist integers $\mathrm{p}$ and $\mathrm{q}$, such that

$$\sqrt{2}=\frac{\mathrm{p}}{\mathrm{q}}$$

We can also assume that $\mathrm{p}$ and $\mathrm{q}$ have no common divisors because then we can reduce the fraction $\frac{\mathrm{p}}{\mathrm{q}}$ to an irreducible one dividing both $\mathrm{p}$ and $\mathrm{q}$ by the largest common divisor.

Taking the square of both sides of $(11.2)$, we get

\begin{aligned} &2=\frac{p^{2}}{q^{2}} \\ &2 q^{2}=p^{2} \end{aligned}

If $\mathrm{p}$ is odd then $p^{2}$ is also odd but $p^{2}$ is equal to $2 q^{2}$ therefore $p^{2}$ is divisible by 2 and so $\mathrm{p}$ must be even, which means that $p=2 k$ for some integer $k$.

Substituting $p=2 k$ into (11.3), we get

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

Dividing by 2 , we get

$$\mathrm{q}^{2}=2 \mathrm{k}^{2}$$

which implies that $\mathrm{q}$ must also be even which leads us to a contradiction with the assumption that $\frac{\mathrm{p}}{\mathrm{q}}$ is irreducible. This means that the original assumption of rationality of $\sqrt{2}$ is also incorrect and $\sqrt{2}$ is irrational.
by Platinum
(119,120 points)

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