mr0no said:
How can I be sure if I can apply a function to both sides of an equation and preserve equality?
for example:
x = y ⇔ x2 = y2
or
x = y ⇔ \sqrt{}x = \sqrt{}y
The implication is only guaranteed to be one way. If x = y, then you can safely "apply a function to both sides" because you know x = y. That is, let z = f(x). But, since we know x = y, we can replace x with y in this equation, so z = f(y). The result z is the same in both cases, so we know f(x) = f(y).
However, given f(x) = f(y), you cannot necessarily conclude x = y, because the function f may not have a unique inverse. For examples, x
2 = y
2 does not imply x = y, because it is possible that x = -y. However, this doesn't contradict what I said about because taking the square root function of both sides
does not give you "x = y". If you apply the square root function to both sides, what you actually get is |x| = |y|, which is true, because ##\sqrt{a^2} = |a|## for any real number a.
or
x < y ⇔ log(x) < log(y)
What's up with that?
The case is of inequalities is trickier and a different situation. If x < y, then you can only conclude f(x) < f(y) if f is a monotonically increasing function only the region that contains x and y. That is, if f'(x) > 0 on the entire region containing x and y.
For example, consider 0<x<y and f(t) = exp(t). Since exp(t) is always increasing, we can conclude exp(x) < exp(y). However, consider a function g(t) = exp(-t). This function is always decreasing, so x < y actually implies g(x) > g(y).
For a trickier example, let h(t) = sin(t), and let ##0 < x < y < \pi/2##. The sine function is monotonically increasing on this interval, so we can conclude that sin(x) < sin(y). However, if it were the case that 0 < x < y, but ##y > \pi/2##, then we cannot conclude that sin(x) < sin(y) because it may not be true, as sin(t) is decreasing on ##\pi/2<t<3\pi/2##.
