Two friends, Majka and Martina, rationalized the denominator of the fraction
$$\frac{1}{\sqrt2+\sqrt3}.$$
Each of them rationalized the denominator in her own way.
Martina:
(1) She found the conjugate of the denominator (the same binomial with an opposite middle sign). The conjugate of $\sqrt2+\sqrt3$ is $\sqrt2-\sqrt3$.
(2) She multiplied with the conjugate both the numerator and denominator of a fraction. $$\frac{1}{\sqrt2+\sqrt3}=\frac{1}{\sqrt2+\sqrt3}\cdot\frac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}$$
(3) She simplified the expression: $$\frac{1}{\sqrt2+\sqrt3}=\frac{1}{\sqrt2+\sqrt3}\cdot\frac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}=\frac{\sqrt2-\sqrt3}{\left(\sqrt2\right)^2-\left(\sqrt3\right)^2}=\frac{\sqrt2-\sqrt3}{2-3}=\frac{\sqrt2-\sqrt3}{-1}=\sqrt3-\sqrt2$$
Majka:
(1) She used the multiplicative identity to rationalize the denominator. (The identity property of multiplication states that when a number is multiplied by $1$, the product will be the number itself.) $$\frac{1}{\sqrt2+\sqrt3}=\frac{1}{\sqrt2+\sqrt3}\cdot1=\frac{1}{\sqrt2+\sqrt3}\cdot\frac{\sqrt2+\sqrt3}{\sqrt2+\sqrt3}$$
(2) She simplified the expression: $$\frac{1}{\sqrt2+\sqrt3}=\frac{1}{\sqrt2+\sqrt3}\cdot \frac{\sqrt2+\sqrt3}{\sqrt2+\sqrt3}=\frac{\sqrt2+\sqrt3}{\left(\sqrt2\right)^2+\left(\sqrt3\right)^2}=\frac{\sqrt2+\sqrt3}{2+3}=\frac{\sqrt2+\sqrt3}{5}$$
Select the true statement.
The Majka's solution is correct. All of Majka's steps are correct.
The Majka's solution is incorrect. The error is in step (2). The product $\left(\sqrt2+\sqrt3\right)\left(\sqrt2+\sqrt3\right)$ is not equal to $\left(\sqrt2\right)^2+\left(\sqrt3\right)^2$.
The Martina's solution is incorrect. The error is in step (2). The product $\left(\sqrt2+\sqrt3\right)\left(\sqrt2-\sqrt3\right)$ is not equal to $\left(\sqrt2\right)^2-\left(\sqrt3\right)^2$.
The Martina's solution is incorrect. The error is in step (2), because of $$\frac{\sqrt2-\sqrt3}{-1}\neq\sqrt3-\sqrt2.$$
Martina's solution is correct. To rationalize the denominator, she used the multiplicative identity. (The identity property of multiplication states that when a number is multiplied by $1$, the product will be the number itself.) In our case, $1$ is written as a fraction whose numerator and denominator are defined as conjugate to the denominator of the original fraction: $$\frac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}=1$$ She also used the formula for difference of squares, applicable when there is a binomial in the denominator: $$(a+b)(a-b)=a^2-b^2$$ In our case: $$\left(\sqrt2+\sqrt3\right)\left(\sqrt2-\sqrt3\right)=\left(\sqrt2\right)^2-\left(\sqrt3\right)^2$$