Prove log2(3) Is Irrational (Math Proof Guide)

Assume (\log_2 3) were rational, so it could be written as (p/q) in lowest terms, with integers (p,q) and (q>0). Since (2^{p/q}=3), raising both sides to (q) gives (2^p=3^q). But the two sides have different prime factors, so they cannot be equal. The assumption fails; therefore (\log_2 3) is irrational.

A common mistake in this proof is to rely on a calculator’s decimal display. However many digits it shows, a finite decimal cannot prove that a number is irrational. The proof needs an exact contradiction, and the key is to turn the logarithm statement into an equation about whole-number powers.

Problem Overview & Proof Framework

This proof shows that (\log_2 3) is not a rational number. It uses proof by contradiction: assume the logarithm is rational, rewrite that assumption as an equation between integer powers, and show that the equation conflicts with unique prime factorization.

The notation (\log_2 3) means “the exponent to which 2 must be raised to get 3.” The claim is that this exponent cannot be written as a ratio of integers. We will assume the opposite and test what that would require.

What the logarithm statement means

A logarithm is the inverse of exponentiation. In general, if (b>0), (b\ne1), and (x>0), then [ \log_b x=y \quad\text{means}\quad b^y=x. ] Here the base is (2) and the argument is (3). Both are positive, and (2\ne1), so (\log_2 3) is defined.

Because (2>1), its powers increase as the exponent increases. Since (3>1), the exponent needed to produce 3 from 2 is positive. Thus (\log_2 3>0). This fact will let us conclude that the numerator (p) in our assumed fraction is positive.

Why use contradiction?

A proof by contradiction begins by assuming the opposite of the claim. If that assumption forces a statement that cannot be true, then the assumption must be false. Here, the impossible statement will be (2^p=3^q) for positive integers (p) and (q).

This approach is useful because it turns a question about a logarithm into one about integers, where prime factorization gives a clear test. The main task is to make sure the fraction and the powers meet the conditions needed for that test.

Definitions, Assumptions & Proof Requirements

The proof depends on three basic ideas: rational numbers can be written as fractions of integers, logarithms undo exponentiation, and every integer greater than 1 has a unique prime factorization. Keeping these ideas explicit prevents gaps, especially when we raise both sides of an equation to a power.

Rational numbers and lowest terms

A rational number is any number that can be written as (p/q), where (p) and (q) are integers and (q\ne0). Any rational number can be expressed in lowest terms, meaning its numerator and denominator have no common positive factor other than 1.

If (\log_2 3) were rational, we could therefore write [ \log_2 3=\frac pq, ] where (p,q\in\mathbb Z), (q>0), and (\gcd(p,q)=1). The condition (q>0) is a convenient choice of sign; it does not change the value of the fraction. Since the logarithm is positive and (q) is positive, (p) must also be positive.

Unique prime factorization

Unique prime factorization says that each whole number greater than 1 can be written as a product of prime numbers in only one way, apart from the order of the factors. For example, (12=2^2\cdot3), and no different collection of primes gives 12.

For this proof, the important point is that (2^p) contains only the prime factor 2, while (3^q) contains only the prime factor 3, provided (p,q) are positive integers. Such numbers cannot be equal: equal positive integers must have the same prime factors and the same exponents on those factors.

Conditions needed before raising to a power

The integers and sign conditions are not decoration. We need (q) to be a positive integer so that raising both sides to (q) produces an integer power, and we need (p) to be a positive integer for the prime-factorization comparison.

If we merely wrote an unspecified expression (p/q), the argument would be incomplete. Establishing (p,q\in\mathbb Z), (q>0), and (p>0) gives us exactly the form the contradiction requires.

Proof Execution, Common Errors & Verification

The proof now follows a short chain: assume rationality, use the definition of the logarithm, raise both sides to an integer power, and compare prime factors. Each step follows from the previous one, so there is no need to estimate the logarithm or rely on a decimal.

Step-by-step proof

Assume, for contradiction, that (\log_2 3) is rational. Write it in lowest terms: [ \log_2 3=\frac pq, \qquad p,q\in\mathbb Z,\quad q>0,\quad \gcd(p,q)=1. ] Since (3>1) and the base (2>1), (\log_2 3>0), so (p>0).

By the definition of a logarithm, [ 2^{\log_2 3}=3. ] Substitute (\log_2 3=p/q): [ 2^{p/q}=3. ] Now raise both sides to the positive integer power (q): [ \left(2^{p/q}\right)^q=3^q, ] so [ 2^p=3^q. ]

But (p) and (q) are positive integers. Thus (2^p) has only 2 as a prime factor, while (3^q) has only 3 as a prime factor. Unique prime factorization says these cannot be the same integer. This contradicts (2^p=3^q).

The original assumption must therefore be false. Hence [ \boxed{\log_2 3\text{ is irrational}.} ]

A quick check of each step

Common errors to avoid

One tempting shortcut is to say that (2^p) is even and (3^q) is odd. That observation works only after establishing that (p) is a positive integer. Without that step, the expressions have not yet been shown to be positive integer powers. The prime-factorization argument is direct and makes the same requirement visible.

Another error is to treat a decimal approximation as proof. A calculator may display (\log_2 3\approx1.58496), but this is only an approximation. A finite decimal is rational, and a displayed string of digits does not tell us whether the exact value has a repeating or nonterminating decimal expansion. The contradiction above establishes the claim exactly.

It is also important not to raise both sides to (q) before stating that (q) is an integer. The step works here because the assumed rational value can be represented with an integer denominator. That condition is what turns the equation into one about whole-number powers.

Mathematical Significance & Related Results

The proof illustrates a useful method: if a rational exponent would force two different primes to have equal positive powers, unique prime factorization can rule out that exponent. The same reasoning applies to logarithms whose bases and arguments are distinct primes.

This is not a claim that every logarithm is irrational. For example, (\log_2 8=3), which is rational, because (2^3=8). The important feature here is that no positive integer powers of 2 and 3 can match. The proof works because the assumed fraction turns the logarithm into exactly such a proposed match.

Why the decimal does not settle the question

A decimal approximation is useful for estimating a value, but it cannot replace an exact proof of irrationality. Any finite decimal, such as (1.58), can be written as a fraction: (1.58=158/100). Even a long calculator display gives only finitely many digits.

To prove irrationality, one must rule out every possible ratio of integers, not just a few simple fractions. The contradiction does that: it shows that any assumed fraction would force an impossible integer equation.

A related general rule

Suppose (a) and (b) are distinct prime numbers. Then (\log_a b) is irrational. If it were (p/q) in lowest terms, with (q>0), then (a^{p/q}=b), so raising to (q) would give (a^p=b^q). For positive (p,q), the two sides have different prime factors, which is impossible.

The same reasoning applies to (\log_2 3) because 2 and 3 are distinct primes. The rule does not apply in that form when the argument is a power of the base, as with (\log_2 8=3). The prime factors then agree, so this contradiction does not arise.

Conclusion

The essential move is to assume (\log_2 3=p/q) with integers (p,q) and (q>0), then use the logarithm definition to derive (2^p=3^q). Since (p>0), unique prime factorization rules out that equation. Therefore, (\log_2 3) is irrational.

Frequently Asked Questions

These answers review the key steps and clarify what the proof does—and does not—rely on. Use them to check whether a written solution has stated its assumptions, justified its integer powers, and reached an exact contradiction.

1. What does it mean for (\log_2 3) to be irrational?
It means there are no integers (p) and (q), with (q\ne0), such that (\log_2 3=p/q).

2. Why can we assume the fraction is in lowest terms?
Every rational number can be written in lowest terms by canceling common factors. This gives (\gcd(p,q)=1), although the prime-factor contradiction does not depend on that condition.

3. Why must (q) be positive?
A fraction’s sign can be placed in its numerator or denominator. We choose (q>0) as a standard form, which makes it easier to infer that (p>0) from the positive logarithm.

4. Why is (\log_2 3) positive?
Since (2^0=1) and (2^1=2), the exponent that gives 3 is greater than 1, so (\log_2 3>0). In particular, the numerator (p) must be positive when (q>0).

5. Why raise both sides to (q)?
The equation (2^{p/q}=3) has a fractional exponent. Raising both sides to the positive integer (q) gives the whole-number power equation (2^p=3^q), which can be checked using prime factors.

6. Why can’t (2^p) equal (3^q)?
For positive integers (p) and (q), the first number has only prime factor 2 and the second has only prime factor 3. Unique prime factorization rules out their equality.

7. Is it enough to say one number is even and the other is odd?
It can be a valid contradiction after showing (p>0) and (q>0) are integers. The prime-factor argument makes the reason more direct and emphasizes the role of unique prime factorization.

8. Can a calculator prove the logarithm is irrational?
No. A calculator gives an approximation with finitely many digits. Irrationality requires ruling out every ratio of integers, which this exact contradiction does.

9. Does this prove that every logarithm is irrational?
No. For example, (\log_2 8=3), which is rational. The proof applies here because assuming rationality would force equal positive powers of the distinct primes 2 and 3.

(This article was written by one of our staff writers, Alan Westbrook. Visit our Meet the Team page.)