Students first learning about fractions are often taught to "cross-multiply" when dealing with fraction with non-like denominators, however, in Mathematica, with the function SameQ: $$\frac{x}{y}+\frac{a}{b} = \frac{ay+xb}{yb}$$ returns "False," which seems to indicate that cross multiplying does not work universally. Am I doing something wrong here?
(The function SameQ prints "True" if the equations on the left and right side are identical and returns "false" if they are not.)
On
Wolfram Alpha reports the following:
"TrueQ[x/y+a/b=(ay+xb)/(yb)]": False
"Is x/y+a/b=(ay+xb)/(yb)?": True
Which is rather strange. And it doesn't help to make those denominators non-zero, by replacing them with $1+y^2$ and $1+b^2$:
"TrueQ[x/(1+y^2)+a/(1+b^2)=(a(1+y^2)+x(1+b^2))/((1+y^2)(1+b^2))]": False
"Is x/(1+y^2)+a/(1+b^2)=(a(1+y^2)+x(1+b^2))/((1+y^2)(1+b^2))?": True
Mathematica checks that your equation is true for all $x$, $y$, $a$ and $b$ that are in the real numbers (I suppose, you should verify that with your Mathematica documentation).
Indeed,
$$\frac{x}{y}+\frac{a}{b} \ne \frac{ay+xb}{yb}$$
when $y = 0$ or $b=0$.
Added: It might interest you to know precisely from where comes the equation.
We first define $\frac{1}{x}$, as being the inverse element of $x$, denoted $x^{-1}$. This is the element $x^{-1}$ such that $x\cdot x^{-1} = 1$ . This element only exists if $x\ne 0$, because if otherwise, then $x\cdot x^{-1} = 0 \cdot x^{-1} = 0 \ne 1$ which is a contradiction.
If $x, y, a, b \in \mathbb{R}$, $y \ne 0$ and $x \ne 0$, we have
\begin{align} \frac{x}{y}+\frac{a}{b} &= \left ( \frac{x}{y}+\frac{a}{b} \right ) \cdot 1\\ &=\left ( \frac{x}{y}+\frac{a}{b} \right ) \cdot \frac{yb}{yb}\\ &=\left ( \frac{x}{y}\cdot yb +\frac{a}{b}\cdot yb \right ) \cdot \frac{1}{yb}\\ &=\left ( xb +ay \right ) \cdot \frac{1}{yb}\\ &=\frac{xb +ay}{yb} \end{align}
Note that this follows from the assumption that $y \ne 0$ and $x \ne 0$.