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Chapter 5: Problem 14

Let \(x \in \mathbb{R}, x>0\). Show that if \(m, p \in \mathbb{Z}, n, q \in \mathbb{N}\), and \(m q=n p\), then \(\left(x^{1 / n}\right)^{m}=\left(x^{1 / q}\right)^{p}\).

Short Answer

Expert verified
It is true that \(\left(x^{1 / n}\right)^{m}=\left(x^{1 / q}\right)^{p}\) when \(mq=np\) for given conditions. This is proven by applying the rules of exponents and using the provided hypothesis \(mq=np\).

Step by step solution

01

Interpret the Question

Start by understanding what we are asked to prove: \(\left(x^{1 / n}\right)^{m}=\left(x^{1 / q}\right)^{p}\). This will involve some algebraic manipulation of exponents.
02

Apply Exponent Rules

We know that when we raise a power to a power, we multiply the exponents. Thus, we can simplify the expression to \(x^{m/n}=x^{p/q}\). Notice the exponents are fractions which can be interpreted as a division.
03

Apply the Hypothesis

Our hypothesis states that \(mq=np\). We can substitute this into our expression to verify \(x^{m/n}=x^{mq/np}\) which simplifies to \(x^{m/n}=x^{m/n}\).
04

Verify the Equality

We can see that both sides of the equation are equal, thus the proposed statement is true under the given conditions. That is, \(\left(x^{1 / n}\right)^{m}=\left(x^{1 / q}\right)^{p}\) whenever \(mq=np\).

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Most popular questions from this chapter

Let \(h: \mathbb{R} \rightarrow \mathbb{R}\) be continuous on \(\mathbb{R}\) satisfying \(h\left(m / 2^{n}\right)=0\) for all \(m \in \mathbb{Z}, n \in \mathbb{N} .\) Show that \(h(x)=0\) for all \(x \in \mathbb{R}\)

Let \(I:=[0,1]\) and let \(f: I \rightarrow \mathbb{R}\) be defined by \(f(x):=x\) for \(x\) rational, and \(f(x):=1-x\) for \(x\) irrational. Show that \(f\) is injective on \(I\) and that \(f(f(x))=x\) for all \(x \in I\). (Hence \(f\) is its own inverse function!) Show that \(f\) is continuous only at the point \(x=\frac{1}{2}\).

Give an example of functions \(f\) and \(g\) that are both discontinuous at a point \(c\) in \(\mathbb{R}\) such that (a) the sum \(f+g\) is continuous at \(c\), (b) the product \(f g\) is continuous at \(c\).

Suppose that \(f\) is a continuous additive function on \(\mathbb{R}\). If \(c:=f(1)\), show that we have \(f(x)=c x\) for all \(x \in \mathbb{R}\). [Hint: First show that if \(r\) is a rational number, then \(f(r)=c r\). ]

Show that both \(f(x):=x\) and \(g(x):=x-1\) are strictly increasing on \(I:=[0,1]\), but that their product \(f g\) is not increasing on \(I\).
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