Question

Continuous fraction of ✓ 2453

Answer 1

Step-by-step explanation:

Let's just do an example. Let's find the continued fraction for 5–√. 5–√≈2.23 or something, and a0 is the integer part of this, which is 2.

Then we subtract a0 from 5–√ and take the reciprocal. That is, we calculate 15√−2. If you're using a calculator, this comes out to 4.23 or so. Then a1 is the integer part of this, which is 4. So:

5–√=2+14+1⋮

Where we haven't figured out the ⋮ part yet. To get that, we take our 4.23, subtract a1, and take the reciprocal; that is, we calculate 14.23−4≈4.23. This is just the same as we had before, so a2 is 4 again, and continuing in the same way, a3=a4=…=4:

5–√=2+14+14+14+1⋮

This procedure will work for any number whatever, but for 5–√ we can use a little algebraic cleverness to see that the fours really do repeat. When we get to the 15√−2 stage, we apply algebra to convert this to 15√−2⋅5√+25√+2=5–√+2. So we could say that:

5–√2+5–√=2+12+5–√=4+12+5–√.

If we substitute the right-hand side of the last equation expression into itself in place of 2+5–√, we get:

2+5–√=4+14+12+5–√=4+14+14+12+5–√=4+14+14+14+12+5–√=⋯

and it's evident that the fours will repeat forever