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Suppose I have a fraction: $$\frac{2^n}{2^{2n}+1}$$

I can simplify it to become: $$\frac{1}{2^{n}+\frac{1}{2^n}}$$

Now obviously, this is just dividing both the numerator and the denominator of the fraction by $2^n.$ My question is why I can do this. Can anyone explain the algebra behind this division to me?

EDIT: I tried thinking about the initial fraction as $2^n \cdot \frac{1}{2^{2n}+1}$ and the division operation as $\frac{2^n \cdot \frac{1}{2^{2n}+1}}{2^n}$, but I couldn't get anywhere with attempting to compute $\frac{\frac{1}{2^{2n}+1}}{2^n}$.

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  • $\begingroup$ you simply multiply numerator and numerator with $1/2^n$, so nothing change $\endgroup$ – newzad May 8 '13 at 22:12
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You’re just multiplying the original fraction by $1$ in a cleverly chosen disguise:

$$1=\frac{1/2^n}{1/2^n}\;,$$

so

$$\frac{2^n}{2^{2n}+1}=\frac{2^n}{2^{2n}+1}\cdot\frac{1/2^n}{1/2^n}=\frac1{2^n+\frac1{2^n}}\;.$$

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  • $\begingroup$ Clever indeed! Thanks! $\endgroup$ – xisk May 8 '13 at 22:13
  • $\begingroup$ @xisk: You’re welcome! $\endgroup$ – Brian M. Scott May 8 '13 at 22:14
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One way of looking at it, is that you are multiplying the numerator and the denominator by $\frac{1}{2^n}$. In the numerator you get $\frac{2^n}{2^n}$ which is $1$. In the denominator you get:

$$\frac{1}{2^n}\cdot (2^{2n} + 1) = \frac{2^{2n}}{2^n} + \frac{1}{2^n} = 2^{2n-n} + \frac{1}{2^n} = 2^n + \frac{1}{2^n}$$

and there you go:

$$\dfrac{1}{2^n + \dfrac{1}{2^n}}$$

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