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Chapter 4: Problem 7

Prove that \(\sqrt{p}\) is irrational for any prime \(p\).

Short Answer

Expert verified
By contradiction, we proved that \(\sqrt{p}\) is irrational for any prime \(p\).

Step by step solution

01

Assumption

Assume to the contrary that \(\sqrt{p}\) is rational. That is, suppose \(\sqrt{p}\) can be written as a ratio of two integers. Express it as \(\frac{a}{b}\), where \(a\) and \(b\) are integers that have no common factors other than 1 (i.e., \(a\) and \(b\) are coprime), and \(b\) is not equal to 0.
02

Squaring both sides

Square both sides of the equation \(\sqrt{p} = \frac{a}{b}\). This gives \(p = \frac{a^2}{b^2}\), or equivalently, \(a^2 = p \cdot b^2\). This implies that \(a^2\) is a multiple of \(p\).
03

Applying Properties of Primes

Since \(a^2\) is a multiple of the prime \(p\), \(a\) is also a multiple of \(p\), due to a property of prime numbers that states if a prime divides a number's square, it must divide the number. Therefore \(a = p \cdot k\) for some integer \(k\).
04

Substitute

Substitute \(a\) in the equation from step 2 with \(p \cdot k\). This gives \(p \cdot k^2 = p \cdot b^2.\) Then, we can cancel out \(p\) from both sides to get \(k^2 = b^2\). This indicates \(b\) is also a multiple of \(p\).
05

Identifying the Contradiction

Here is the contradiction: we assumed \(a\) and \(b\) are coprime, but we have shown both \(a\) and \(b\) to be divisible by \(p\), meaning they have a common factor other than 1. This contradiction proves the initial assumption that the square root of a prime could be rational is false.
06

Conclusion

Since the assumption leads to a contradiction, we conclude that \(\sqrt{p}\) is irrational.

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Most popular questions from this chapter

Let \(a, b \in \mathbf{Z}^{*}\) where \(a \geq b\). Prove that \(\operatorname{gcd}(a, b)=\operatorname{gcd}(a-b, b)\).

Let \(a, b \in \mathbf{Z}^{+}\). If \(b \mid a\) and \(b \mid(a+2)\), prove that \(b=1\) or \(b=2\).

Let \(a, b, c, d\) be fixed positive integers. If \((a d-b c) \mid a\) and \((a d-b c) \mid c\), prove that \(\operatorname{gcd}(a n+b, c n+d)=1\) for any \(n \in \mathbf{Z}^{+}\).

If \(n \in \mathbf{Z}^{+}\), prove that \(\sum_{i=1}^{2 n} F_{i} F_{i-1}=F_{2 n}^{2}\).

A grocery store runs a weekly contest to promote sales. Each customer who purchases more than $$\$ 20$$ worth of groceries receives a game card with 12 numbers on it; if any of these numbers sum to exactly 500 , then that customer receives a $$\$ 500$$ shopping spree (at the grocery store). After purchasing $$\$ 22.83$$ worth of groceries at this store, Eleanor receives her game card on which are printed the following 12 numbers: \(144,336,30,66\), \(138,162,318,54,84,288,126\), and 456 . Has Eleanor won a $$\$ 500$$ shopping spree?
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