Question

The multiplicative inverse of \[-1+\sqrt{2}\] is
(a) \[-1-\sqrt{2}\]
(b) \[1-\sqrt{2}\]
(c) \[1+\sqrt{2}\]
(d) \[\sqrt{2}\]
(e) \[2-\sqrt{2}\]

Answer 1

Hint: Here, we have to first remove root form of \[-1+\sqrt{2}\] which we can eliminate by taking conjugate of \[-1+\sqrt{2}\] i.e. changing sign of \[\sqrt{2}\] from plus to minus and multiplying numerator and denominator. By solving that equation, we will get inverse form like \[\dfrac{1}{y}\] . Thus, y will be our required multiplicative inverse.

Complete step-by-step answer:
In the question, we are asked to find the multiplicative inverse of \[-1+\sqrt{2}\] . So, here we will understand the meaning of multiplicative inverse.
In Mathematics, a multiplicative inverse or reciprocal for a number y denoted as \[\dfrac{1}{y}\] or \[{{y}^{-1}}\] which when multiplied by y results the multiplicative identity. For a real number we have to divide 1 by the number. For example: Let say any real number 6, the reciprocal is \[\dfrac{1}{6}\] . So, when \[\dfrac{1}{6}\times 6=1\] the result is 1.
Here, first we will take conjugate of \[-1+\sqrt{2}\] i.e. changing sign of \[\sqrt{2}\] from plus to minus and multiplying numerator and denominator. So, we will get
\[\Rightarrow \dfrac{\left( -1+\sqrt{2} \right)\left( -1-\sqrt{2} \right)}{\left( -1-\sqrt{2} \right)}\]
Here, using the identity of \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\] . So, we will get
\[\Rightarrow \dfrac{\left( {{\left( -1 \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}} \right)}{\left( -1-\sqrt{2} \right)}\]
\[\Rightarrow \dfrac{\left( 1-2 \right)}{\left( -1-\sqrt{2} \right)}=\dfrac{-1}{-1-\sqrt{2}}\]
Taking minus sign common from numerator and denominator, we get
\[\Rightarrow \dfrac{1}{1+\sqrt{2}}\]
Thus, the multiplicative inverse of \[-1+\sqrt{2}\] is \[1+\sqrt{2}\] .
Hence, option (c) is correct.

Note: Another approach is to multiply \[-1+\sqrt{2}\] with \[y\] which results in 1. In mathematical form, it is written as \[\left( -1+\sqrt{2} \right)\times y=1\] . So, on solving this we get
\[\left( -1+\sqrt{2} \right)\times y=1\Rightarrow y=\dfrac{1}{-1+\sqrt{2}}\] So, here to remove denominator part, we will take conjugate and multiplying in both numerator and denominator, we get
\[\Rightarrow y=\dfrac{1}{-1+\sqrt{2}}\dfrac{\left( -1-\sqrt{2} \right)}{\left( -1-\sqrt{2} \right)}=\dfrac{-1-\sqrt{2}}{1-2}=1+\sqrt{2}\]
Thus, we get the same answer by this method also.
Also, by simply solving this \[-1+\sqrt{2}\] i.e. \[-1+1.41=0.41\] we can write this as \[0.41\times y=1\Rightarrow y=\dfrac{1}{0.41}=2.414\] . So, now we can solve every option and see which option matches with 2.414. So, by solving \[1+\sqrt{2}\] i.e. \[1+1.41=2.41\] we get the answer. So, by option method we get the same answer.