Simple properties of

[Aion]

How do we prove these two statements?

If $a>1$ and $\alpha < \beta$ then $a^{\alpha}< a^{\beta}$

If $0<a<b$ and $\alpha>0$ then $a^{\alpha}< b^{\alpha}$

I tried to prove the second statement with induction.

For the base case, let $\alpha = 1$, then clearly $a< b$. Suppose for induction that $a^k< b^k$. Then given that $0<a<b$ we multiply by $a^k$ and obtain $0<a^{k+1}<a^{k}b$. But by the induction hypothesis know that $a^k< b^k$, hence

$$0<a^{k+1}<a^{k}b<b^{k+1}$$
Therefore by induction principle, we can conclude that the statement holds for whole numbers $\alpha \geq1.$

Is this proof correct? I have no clue how to prove the general case, however.

 

[fresh_42]

The problem with induction (which you performed correctly) is, that it doesn't prove the general case.

How about taking the logarithm on both sides? In that case, you only have to find an argument that the logarithm preserves the inequality.

 

[Dr.Peterson]

How do we prove these two statements?

If $a>1$ and $\alpha < \beta$ then $a^{\alpha}< a^{\beta}$

If $0<a<b$ and $\alpha>0$ then $a^{\alpha}< b^{\alpha}$
...
Is this proof correct? I have no clue how to prove the general case, however.

What is the context of the question? How have exponential functions (as functions of a real number) been defined, and what theorems are available? Can you use the derivative? Is there a reason you put this under "Beginning Algebra"?

 

[Aion]

What is the context of the question? How have exponential functions (as functions of a real number) been defined, and what theorems are available? Can you use the derivative? Is there a reason you put this under "Beginning Algebra"?

I'm currently reading an introductory book on calculus and know very little about the topic. These two statements were given without proof in chapter one, and are then used to prove other properties for the power function.

What's known is that if $a>0$. Then for $q>0$ we can define $a^{\frac{1}{q}}$ as the unambiguous positive number that solves the equation $x^q=a.$ Now we can define $a^t$ for rational exponents $t=\frac{p}{q}, q>0$ by

$$a^{\frac{p}{q}}=\left(a^{\frac{1}{q}}\right)^p$$
Lastly, we can define $a^{\alpha}$ for arbitrary real number $\alpha$ and $a>0$ by approximating $\alpha$ as a rational number $\frac{p}{q}.$

Is it possible to prove these inequalities without using anything too advanced? If proving these inequalities requires properties of the exponential function that may rely on this inequality, then the book fails to provide a logical foundation for the topic in general.

 

[Dr.Peterson]

I'm currently reading an introductory book on calculus and know very little about the topic. These two statements were given without proof in chapter one, and are then used to prove other properties for the power function.

What's known is that if $a>0$. Then for $q>0$ we can define $a^{\frac{1}{q}}$ as the unambiguous positive number that solves the equation $x^q=a.$ Now we can define $a^t$ for rational exponents $t=\frac{p}{q}, q>0$ by

$$a^{\frac{p}{q}}=\left(a^{\frac{1}{q}}\right)^p$$
Lastly, we can define $a^{\alpha}$ for arbitrary real number $\alpha$ and $a>0$ by approximating $\alpha$ as a rational number $\frac{p}{q}.$

Is it possible to prove these inequalities without using anything too advanced? If proving these inequalities requires properties of the exponential function that may rely on this inequality, then the book fails to provide a logical foundation for the topic in general.

I'd like to see some of what it says, in a link or an image. But it sounds like they are either presenting ideas without formal proof (which is appropriate for an introductory book), or are developing the concept of the exponential function as a step-by-step extension from positive integer powers to negative and rational and finally real powers. A full explanation of the last step (including the proof you ask for) requires knowledge of calculus.

So at the start of calculus you may need to just recognize that $a^x$ has the graph they show you, which is always increasing (for $a>1$), and leave formal proofs for later. But again, if I saw how they are presenting this, I might have a different answer!

 

[Aion]

I'd like to see some of what it says, in a link or an image. But it sounds like they are either presenting ideas without formal proof (which is appropriate for an introductory book), or are developing the concept of the exponential function as a step-by-step extension from positive integer powers to negative and rational and finally real powers. A full explanation of the last step (including the proof you ask for) requires knowledge of calculus.

So at the start of calculus you may need to just recognize that $a^x$ has the graph they show you, which is always increasing (for $a>1$), and leave formal proofs for later. But again, if I saw how they are presenting this, I might have a different answer!

I could provide an image, but the book is written in Swedish so it might not tell you much. You're right, I should leave the problem for now and return to it later. It just felt like circular reasoning as the part afterwards on power functions relies on the above inequality.

 

[Dr.Peterson]

I could provide an image, but the book is written in Swedish so it might not tell you much. You're right, I should leave the problem for now and return to it later. It just felt like circular reasoning as the part afterwards on power functions relies on the above inequality.

You could leave it at that for now; but I could be interested in an image, to see if there's more I could say about it.

(Does the name Peterson suggest anything? I don't know much Swedish, but my great-grandfather was from there, and I have been there ... and have a dictionary ... and Google.)

 

[Aion]

You could leave it at that for now; but I could be interested in an image, to see if there's more I could say about it.

(Does the name Peterson suggest anything? I don't know much Swedish, but my great-grandfather was from there, and I have been there ... and have a dictionary ... and Google.)

Yes, Dr. Peterson, your name is very similar to a common Swedish surname, Petersson, which is widely used in Sweden . I will provide some photos of the section discussed above.

 

[Aion]

The section continues and discusses the elementary properties of the exponential function.

 

[blamocur]

The only Swedish I remember from my one week in the country decades ago is that all
street names end with ${\displaystyle }$gatan'', but here are my two öre's worth:
First, replace (26) with an equivalent statement, $$a > 1 \wedge \gamma > 0 \Longrightarrow a^\gamma > 1$$.
Now, prove for arbitrary natural numbers $p,q$
$$a > 1 \Longrightarrow a^p > 1$$$$a < 1 \Longrightarrow a^q < 1$$$$a > 1 \Longrightarrow a^{\frac{p}{q}} > 1$$Then use the definition of $a^\gamma$ for arbitrary $\gamma$'s through approximations of $\gamma$ with rational numbers.
Does this help?

 

[Dr.Peterson]

Thanks.

Looking through this with the help of Google, it appears that the key parts are:

The numbers a and α are called base and exponent, respectively. With modern pocket calculators, you can easily get very good approximate values for numbers of the form (20). But how does the calculator know which approximation of a^α it should specify? Apparently, the calculator must be equipped with a built-in instruction for calculating powers. To be able to make such a calculation algorithm, one must of course know exactly what it is to be calculated; one must therefore have a proper definition of numbers of the form a^α.​

​

Indeed, it turns out to be quite difficult to establish a tenable definition of a^α for general bases and exponents. In this course, we have to content ourselves with mastering the actual calculation of powers. However, a small sketch of how to define a^α is in order.​

​

...​

​

Officially, one defines a^α for arbitrary real numbers α and a>0 by approximating a with rational numbers and letting a be the driving "boundary value" of the already defined numbers a^p/q.​

​

That the laws of arithmetic remain valid at each step in this successive expansion of a^α to ever more general exponents naturally requires proof. However, we bypass these.​

This is essentially what I supposed:

But it sounds like they are either presenting ideas without formal proof (which is appropriate for an introductory book), or are developing the concept of the exponential function as a step-by-step extension from positive integer powers to negative and rational and finally real powers. A full explanation of the last step (including the proof you ask for) requires knowledge of calculus.

So at the start of calculus you may need to just recognize that $a^x$ has the graph they show you, which is always increasing (for $a>1$), and leave formal proofs for later. But again, if I saw how they are presenting this, I might have a different answer!

Now, looking for the specific statements you asked about (that the exponential function is monotonically increasing), they do indeed just state them without proof, as they had just said (in what I translated above):

​

To prove this properly requires a more complete definition for real numbers (as the unique continuous extension of the exponential function on rational numbers); it's not too hard to prove for rational numbers, as @blamocur has suggested.

Most sites I find in searching for a proof start with other definitions that require calculus even to read!

 

[Aion]

The only Swedish I remember from my one week in the country decades ago is that all
street names end with ${\displaystyle }$gatan'', but here are my two öre's worth:
First, replace (26) with an equivalent statement, $$a > 1 \wedge \gamma > 0 \Longrightarrow a^\gamma > 1$$.

My logic is a little rusty, I think your statement is equivalent, for if we substitute $\alpha=0$ and $\beta=\gamma$ in (26) then clearly $a^{ \gamma}>1$. For the second direction, let $a>1$ and assume $a^{\gamma}>1$ holds for $\gamma >0$. Now suppose $\alpha<\beta$, then $\beta-\alpha>0$ and by the properties of the exponents

$$a^{\beta}=a^{\alpha}\cdot a^{\beta-\alpha}$$Since $\beta-\alpha>0$, it follows from our assumption that $a^{\beta-\alpha}>1$ and therefore

$$a^{\beta}=a^{\alpha}\cdot a^{\beta-\alpha}>a^{\alpha}\cdot 1=a^{\alpha}$$
Hence they are logically equivalent statements.

Now, prove for arbitrary natural numbers $p,q$

$$a > 1 \Longrightarrow a^p > 1$$$$a < 1 \Longrightarrow a^q < 1$$

These two statements I managed to prove easily using induction

Suppose $a>1$. Then for $p=1$ we have $a^p=a$. Since $a>1$ the base case follows. Suppose for induction that $a^k>1$. Then $a^{k+1}=a^{k}\cdot a$. But we know by the induction hypothesis that $a^k>1$ and $a>1$, $a^{k+1}=a^{k}\cdot a >1 \cdot 1 =1$. Hence $a^{k+1}>1$.

I used a similar argument to show the second statement.

$$a > 1 \Longrightarrow a^{\frac{p}{q}} > 1$$

I found this part challenging and unsure if my argument was valid.

Suppose $a>1$and let $q>0$ be some whole-number. The rational exponent $\frac{p}{q}$ for base $a$ can be expressed as

$$a^{\frac{p}{q}}=\left( a^{p} \right)^{\frac{1}{q}}$$
Now $a^p>1$ as we have already proven this. Suppose $x=a^{p}$, then we need to prove that if $x>1$ then $x^{\frac{1}{q}}>1$. The $q$-th root is defined as the unique real number $y$ that solves the equation. $$y^q=x$$
If $x>1$ then $y^q=x>1$. And our goal is to show that $y>1$. Assume for contradiction that $y\leq1$. If $y=1, y^q=1^q=1$, which contradicts the assumption $y^q>1$, suppose $y<1$, since $q$ is a natural number, $y^q<1^q$ (by previous theorem) which also contradicts that $y^q=x>1$, therefore $y>1$.

This proves that if $a>1$, then $a^{\frac{p}{q}}>1$

Then use the definition of $a^\gamma$ for arbitrary $\gamma$'s through approximations of $\gamma$ with rational numbers.
Does this help?

Since any real number $\gamma$ can be approximated to any degree of accuracy with a fraction $p/q$, the property $a^{\frac{p}{q}}$ holds for $a>1$ and rational $p/q$, it must hold for $a>1$ and real $\gamma$ by the continuity of the exponential function.

What do you think about my attempt to use your suggestion blamocur?

 

[blamocur]

What do you think about my attempt to use your suggestion blamocur?

Exactly what I had in mind.
Someone might ask for more rigorous treatment of the last part. E.g., we can express $\lambda = \lim_{k\rightarrow\infty} r_k$, where all $r_k$'s are rational, but it does not follow that $r_k > 0$ for all $k$'s.

 

[Aion]

$$a < 1 \Longrightarrow a^q < 1$$

We make the implicit assumption that $0<a<1$ right? If $a<0$ then $a^q$ alternates between negative and positive values depending on whether $q$ is even or odd. So induction only works on $0<a<1$.

 

[Aion]

Exactly what I had in mind.
Someone might ask for more rigorous treatment of the last part. E.g., we can express $\lambda = \lim_{k\rightarrow\infty} r_k$, where all $r_k$'s are rational, but it does not follow that $r_k > 0$ for all $k$'s.

Sorry, but I don't understand what you mean by that limit of $r_k.$ What is $r$ in this context?

 

[blamocur]

We make the implicit assumption that $0<a<1$ right? If $a<0$ then $a^q$ alternates between negative and positive values depending on whether $q$ is even or odd. So induction only works on $0<a<1$.

That's correct. I implicitly assumed that $a>0$ because $a^\gamma$ is not defined for negative $a$ and arbitrary real $\gamma$'s, at least not in the domain of real numbers.

 

[blamocur]

Sorry, but I don't understand what you mean by that limit of $r_k.$ What is $r$ in this context?

I should have specified that $r = \frac{p}{q}$ -- sorry. And $r_k$ is a sequence of approximations for $\lambda$, i.e. $\lim_{k\rightarrow\infty} r_k = \lambda$. I was thinking of a more formal way to say that any real $\lambda$ can be approximated with any degree of accuracy with rational numbers.

 

[Aion]

I should have specified that $r = \frac{p}{q}$ -- sorry. And $r_k$ is a sequence of approximations for $\lambda$, i.e. $\lim_{k\rightarrow\infty} r_k = \lambda$. I was thinking of a more formal way to say that any real $\lambda$ can be approximated with any degree of accuracy with rational numbers.

I read about a theorem in Higher Algebra by Hall and Knight some time ago, and I think it might be applicable in this context.

Suppose that $\lambda_1$ and $\lambda_2$ are incommensurable. Divide $\lambda_2$ into $q$ equal parts each equal to $\beta$ such that $\lambda_2=q\beta$ where $q$ is a positive integer. Also, suppose $\beta$ is contained in $\lambda_1$ more than $p$ times and less than $p+1$ times. Then we have $\frac{\lambda_1}{\lambda_2}>\frac{p\beta}{q\beta}$ and $\frac{\lambda_1}{\lambda_2}<\frac{(p+1)\beta}{q\beta}$. Suppose we define $\lambda=\frac{\lambda_1}{\lambda_2}$. Then the following inequality holds

$$\frac{p}{q}<\lambda<\frac{p+1}{q}$$
So that $\lambda$ differs from $\frac{p}{q}$ by a quantity less than $\frac{1}{q}$. And since we can choose $\beta$ (our unit of measurement) as small as we please, $q$ can be made as great as we please. Hence $\frac{1}{q}$ can be made as small as we please, and the two integers $p$ and $q$ can be found whose ratio will express that of $\lambda$ to any degree of accuracy.

 

[blamocur]

Looks good, but can't we simplify it a little bit? Why do we need $\lambda_1$ and $\lambda_2$ ? I.e. given an arbitrary $\lambda$ and a natural $q$ pick $p = \lceil q\lambda\rceil$, then $$\frac{p}{q} \leq \lambda \leq \frac{p+1}{q}$$ which means that $$\left|\lambda-\frac{p}{q}\right|\leq \frac{1}{q}$$