Finding inverse to elements
I am trying to find the inverse of the following elements $2+sqrt7$, $sqrt3-sqrt2$ and $1+isqrt5$.
I would be very much thankful if someone could help me with this one.
complex-numbers inverse
edited Dec 9 at 13:58
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Dec 9 at 9:11
Torbjörn Olsson
134
add a comment |
I am trying to find the inverse of the following elements $2+sqrt7$, $sqrt3-sqrt2$ and $1+isqrt5$.
I would be very much thankful if someone could help me with this one.
complex-numbers inverse
edited Dec 9 at 13:58
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Dec 9 at 9:11
Torbjörn Olsson
134
Hint: $a^2 - b^2 = (a + b)(a - b)$
– Alex Vong
Dec 9 at 12:04
add a comment |
I am trying to find the inverse of the following elements $2+sqrt7$, $sqrt3-sqrt2$ and $1+isqrt5$.
I would be very much thankful if someone could help me with this one.
complex-numbers inverse
edited Dec 9 at 13:58
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Dec 9 at 9:11
Torbjörn Olsson
134
I am trying to find the inverse of the following elements $2+sqrt7$, $sqrt3-sqrt2$ and $1+isqrt5$.
I would be very much thankful if someone could help me with this one.
complex-numbers inverse
complex-numbers inverse
edited Dec 9 at 13:58
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Dec 9 at 9:11
Torbjörn Olsson
134
edited Dec 9 at 13:58
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Dec 9 at 9:11
Torbjörn Olsson
134
edited Dec 9 at 13:58
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Dec 9 at 13:58
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Dec 9 at 13:58
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
asked Dec 9 at 9:11
Torbjörn Olsson
134
asked Dec 9 at 9:11
Torbjörn Olsson
134
asked Dec 9 at 9:11
Torbjörn Olsson
134
134
Hint: $a^2 - b^2 = (a + b)(a - b)$
– Alex Vong
Dec 9 at 12:04
add a comment |
Hint: $a^2 - b^2 = (a + b)(a - b)$
– Alex Vong
Dec 9 at 12:04
Hint: $a^2 - b^2 = (a + b)(a - b)$
– Alex Vong
Dec 9 at 12:04
Hint: $a^2 - b^2 = (a + b)(a - b)$
– Alex Vong
Dec 9 at 12:04
add a comment |
1 Answer
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$$frac{1}{2+sqrt{7}}=frac{2-sqrt{7}}{(2+sqrt{7})(2-sqrt{7})}=frac{2-sqrt{7}}{4-7}=-frac{2}{3}+frac{sqrt{7}}{3}$$
$$frac{1}{1+isqrt{5}}=frac{1-isqrt{5}}{(1+isqrt{5})(1-isqrt{5})}=frac{1-isqrt{5}}{1-i^25}=frac{1-isqrt{5}}{1+5}=frac{1}{6}-ifrac{sqrt{5}}{6}$$
The second one is an exercise for you.
answered Dec 9 at 13:43
piotor
385
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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$$frac{1}{2+sqrt{7}}=frac{2-sqrt{7}}{(2+sqrt{7})(2-sqrt{7})}=frac{2-sqrt{7}}{4-7}=-frac{2}{3}+frac{sqrt{7}}{3}$$
$$frac{1}{1+isqrt{5}}=frac{1-isqrt{5}}{(1+isqrt{5})(1-isqrt{5})}=frac{1-isqrt{5}}{1-i^25}=frac{1-isqrt{5}}{1+5}=frac{1}{6}-ifrac{sqrt{5}}{6}$$
The second one is an exercise for you.
answered Dec 9 at 13:43
piotor
385
add a comment |
$$frac{1}{2+sqrt{7}}=frac{2-sqrt{7}}{(2+sqrt{7})(2-sqrt{7})}=frac{2-sqrt{7}}{4-7}=-frac{2}{3}+frac{sqrt{7}}{3}$$
$$frac{1}{1+isqrt{5}}=frac{1-isqrt{5}}{(1+isqrt{5})(1-isqrt{5})}=frac{1-isqrt{5}}{1-i^25}=frac{1-isqrt{5}}{1+5}=frac{1}{6}-ifrac{sqrt{5}}{6}$$
The second one is an exercise for you.
answered Dec 9 at 13:43
piotor
385
add a comment |
$$frac{1}{2+sqrt{7}}=frac{2-sqrt{7}}{(2+sqrt{7})(2-sqrt{7})}=frac{2-sqrt{7}}{4-7}=-frac{2}{3}+frac{sqrt{7}}{3}$$
$$frac{1}{1+isqrt{5}}=frac{1-isqrt{5}}{(1+isqrt{5})(1-isqrt{5})}=frac{1-isqrt{5}}{1-i^25}=frac{1-isqrt{5}}{1+5}=frac{1}{6}-ifrac{sqrt{5}}{6}$$
The second one is an exercise for you.
answered Dec 9 at 13:43
piotor
385
$$frac{1}{2+sqrt{7}}=frac{2-sqrt{7}}{(2+sqrt{7})(2-sqrt{7})}=frac{2-sqrt{7}}{4-7}=-frac{2}{3}+frac{sqrt{7}}{3}$$
$$frac{1}{1+isqrt{5}}=frac{1-isqrt{5}}{(1+isqrt{5})(1-isqrt{5})}=frac{1-isqrt{5}}{1-i^25}=frac{1-isqrt{5}}{1+5}=frac{1}{6}-ifrac{sqrt{5}}{6}$$
The second one is an exercise for you.
answered Dec 9 at 13:43
piotor
385
answered Dec 9 at 13:43
piotor
385
answered Dec 9 at 13:43
piotor
385
answered Dec 9 at 13:43
piotor
385
385
add a comment |
add a comment |
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Hint: $a^2 - b^2 = (a + b)(a - b)$
– Alex Vong
Dec 9 at 12:04