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

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

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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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