# Number raised to power of irrational number

What is the consequence of raising a number to the power of irrational number?

Ex: $2^\pi , 5^\sqrt2$

1. Does this mathematically makes sense? (Are there any problems in physics world where we encounter such a calculation?)
2. How does one calculate or maybe estimate its value ? (I want to know if there is an infinite summation formula, instead of simply rounding $\pi$ to 3.14)
• Basically you can let sequence of rational number converge to that irrational number, and power with rational number makes perfect sense. – MonkeyKing May 8 '15 at 7:12
• For $a > 0$ you squeeze your irrational number $b$ with rationals, so if $\frac {m_1}{n_1} < b < \frac {m_2}{n_2}$, then $$a^{\frac {m_1}{n_1}} < a^b < a^{\frac {m_2}{n_2}}$$ – Kaster May 8 '15 at 7:14
• thanks @Kaster for pointing this property – Bharath May 8 '15 at 7:26
• $\{\pi\}=(3,4)\cap(3.1,3.2)\cap(3.14,3.15)\cap(3.141,3.142)\dotsb$, so we can define $\{2^\pi\}=(2^3,2^4)\cap(2^{3.1},2^{3.2})\cap\dotsb$. – Akiva Weinberger May 8 '15 at 13:27

Formally, we have $a^b = e^{b \ln(a)}$ and

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$

$$\ln x = \int_1^x \frac{dt}{t}$$

And for integer $n$, we define $x^n$ as

$$\prod^n_{i=1} x$$

This is needed because we don't want to define the powers in $e^x$ circulary.

Also note that since we use $\ln a$ in this definition, we must have $a>0$.

You can also just approximate the exponent with a rational number.

A good approximation for $\pi$ is $\frac{355}{113}$.

• thanks. oddly, 'e' is also irrational. – Bharath May 8 '15 at 7:28
• Is there any benefit to defining it via a Taylor series when when you can just define it as a limit of rational approximations...? – Mehrdad May 8 '15 at 8:13
• @Mehrdad Note that even $x^y$ for rational $x,y$ does not usually give a rational number, so if that is what you mean it is not a limit of rational approximations. The power series for $e^x$ certainly is, however. – Mario Carneiro May 8 '15 at 8:15
• @Mehrdad Fundamentally, when calculating we have to use rational approximations. If you try to approximate $a$ and/or $b$ in $a^b$ with rationals you are still not done approximating because the power itself will need a high-degree root extraction and you have to approximate that too. With this method you only need to do two approximations, one for $x\approx \log a$ and the other for $y\approx e^{bx}$. – Mario Carneiro May 8 '15 at 8:21
• Shouldn't this read "for integer n", not x? – Patrick White May 8 '15 at 16:00

If the number is positive, then raising it to the power of an irrational number is well defined. This is because an irrational number can be defined as a converging sequence of rational numbers (like 3, 3.1, 3.14, 3.141, 3.1415 etc), and as these approach the irrational number then the power of these rational numbers also converges to a fixed value, which is the power of the irrational number.

If the number is negative, then the power is not defined. This is because whilst the irrational number can be defined a converging sequence of rational numbers, the power of these numbers does not converge.

• Ultimately, they are defined in the complex realm. However, most students do not see or need to understand this until upper level college math courses. Frequently, full courses in complex numbers are electives for math majors. See wolframalpha.com/input/?i=%28-3%29%5Epi – nickalh May 8 '15 at 8:47
• Nice example, thanks. @Peter Great! that made it so clear. – Bharath May 8 '15 at 9:50

Observe that $$a^r=e^{r\ln a}$$ and use the Taylor series for $e^x$.

Let $$x=a.a_1a_2a_3... and$$ $$y=b.b_1b_2d_3...>0$$. You can approximate x^y as a limit of the sequence $$a^b,a.a_1^{(b.b_1)},a,a_1a_2^{(b.b_1b_2)},....$$ As for physics application suppose you are dealing with an experiment which is governed by the differential equation $$yy'-(y')^2=(y^2)/x$$ One of the solutions is $$x^x$$ and if $$x=2^.5$$ then you get to evaluate an expression of the type you have mentioned.