Can someone explain to why the algebra does not work for this inequality

$\sqrt{1-y}\gt-\frac{1}{2} $.
$1-y\gt\frac{1}{4} $.
$y\lt\frac{3}{4} $.
I know that it should be $y \lt 1$ since the RHS is $\ge0$

  • $\begingroup$ What do you want in the first place? Isn't it you want to solve for $y$ such that $\sqrt{1-y}>-\frac{1}{2}$? $\endgroup$ – Juniven Apr 17 '17 at 8:18
  • $\begingroup$ yes, so why can't we square both sides and solve $\endgroup$ – matt Apr 17 '17 at 8:18
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    $\begingroup$ All you need to do is to ensure that $\sqrt{1-y}$ exists as a real number and this happen if $1-y\geq 0$, that is, $y\leq 1$. So, if $y\leq 1$ then $\sqrt{1-y}\geq 0>-\frac{1}{2}$. So, you don't need to proceed further by squaring both sides. $\endgroup$ – Juniven Apr 17 '17 at 8:21
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    $\begingroup$ In a nutshell, the inequality $a<b$ does not imply the inequality $a^2<b^2$ without further conditions. $\endgroup$ – MPW Apr 17 '17 at 15:32

When solving inequalities, you need to either only apply reversible transformations to your equations, or when you apply irreversible transformations, to do so very carefully.

When solving equalities, you can sometimes get away with irreversible transformations. For example, when you go from $$\sqrt{1-y} = -\frac{1}{2}$$ to $$1-y = \frac{1}{4}$$ you are implicitly arguing as follows: "I know that the function $f(x)=x^2$ is well-defined, so that if $g(y) = h(y)$, also $f(g(y)) = f(h(y))$. Therefore every solution to the equation on my first line will also be a solution on my second line. But! I also know that $f$ is not one-to-one (injective) so it is not necessarily true that every solution to the second equation is also a solution to the original equation. So I'd better check at the end that all solutions are actually valid for the original problem."

Things are much more subtle with inequalities. There are two things wrong with your calculation. First, it is true that if $g(y) > h(y)$, and both $g(y)$ and $h(y)$ are positive, then $g(y)^2 > h(y)^2$. The second condition is critical: for instance $1 > -2$ but $1\not > 4.$

So already squaring both sides of your inequality is invalid. But there is a second problem: let's say you were trying to solve $$\sqrt{1-y} > \frac{1}{2}$$ so that squaring both sides does actually make some sense. If you write $$1-y > \frac{1}{4}$$ you are again arguing, "if the original inequality holds, then the second inequality must also hold. Therefore the set of $y$ that satisfy my original inequality must be a subset of those that satisfy my second one." Once you determine that the second inequality is satisfied whenever $y > \frac{3}{4}$, you still need to go back and check which of these $y$ actually satisfy the first inequality.

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    $\begingroup$ The important thing isn't actually reversibility. A function has to be strictly monotonically increasing for it to preserve $\lt$ and $\gt$, and monotonically non-decreasing for it to preserve $\le$ and $\ge$. Monotonically decreasing functions reverse those inequalities, and it's straightforward to construct invertible functions that don't consistently preserve or reverse inequalities (by stitching increasing and decreasing segments together piecewise with jump discontinuities). Invertibility is only enough for $\ne$. $\endgroup$ – user2357112 Apr 17 '17 at 16:58
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    $\begingroup$ @user2357112 that is all true. I mean "reversible" in the sense that applying the operation to the inequality preserves the set of solutions, $a>b \Leftrightarrow f(a) > f(b)$, not in the sense of $f$ being invertible. $\endgroup$ – user7530 Apr 17 '17 at 18:22

You can square both sides of an inequality only if both are positive. If they are both negative you can square, but the direction of inequality is flipped. In your case neither is true.


We can only square if both sides are non-negative.

Consider the interval $x > -\frac{1}{2}$, ie $(-\frac{1}{2},\infty)$. We must split this into $(-\frac{1}{2},0)$ and $[0,\infty)$

Then we separate the inequality into 2 cases:

a) Both sides are negative: $-\frac{1}{2} < \sqrt{1-x} < 0 $ ... ie the magnitude is less than $\frac{1}{2}$

b) Both sides are non-negative: $\sqrt{1-x} \geq 0$

a) hypothetically expands to:

$\frac{1}{2} > -\sqrt{1-x} > 0 $ ... hypothetically in this case $-\sqrt{1-x}$ is already "positive" (because of the case assumption)

$(\frac{1}{2})^2 > (-\sqrt{1-x})^2 > (0)^2 $ ... it is now "legal" to square both sides since they are "positive"

$\frac{1}{4} > (1-x) > (0) $ ... and so on.

(Technically this is a farcical case, as $\sqrt{1-x}$ cannot be negative, but what I wanted to demonstrate was the logic ... if the given were different and the inequality has a valid 'negative' case)

for b) $\sqrt{1-x} \geq 0$

Since both sides are already non-negative, it is legal to square both sides.

$(1-x) \geq (0)^2$ ... but only for $1-x \geq 0$

$(1-x) \geq (0)$

$(x-1) \leq 0$

$x \leq 1$

In this case it only restates the condition, but I hope you get where it should lead if the given were different.

Again, I hope I've made it clear enough, the 'negative' case in this given inequality gives no valid solution.

I hope this resource would help: https://www.math.purdue.edu/files/academic/courses/2007fall/MA301/MA301Ch2.pdf

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    $\begingroup$ Use \sqrt{1-x} to get "$\sqrt{1-x}$". $\endgroup$ – user21820 Apr 17 '17 at 16:15
  • $\begingroup$ oh. LOL. yeah that works! I'm a bit new to MathJax, that's a really helpful comment! $\endgroup$ – Karl Apr 17 '17 at 16:17
  • $\begingroup$ See here for Math SE's own MathJax guide. $\endgroup$ – user21820 Apr 17 '17 at 16:22

This is the difference between equivalence and implication.

You can do operations that preserve equivalence, which addition, subtraction, division multiplication and monotonic functions do

$$x+1 = x^2+2x\Leftrightarrow\\0 = x^2+x-1$$

Then all solutions are preserved and no new solutions are added.

Implication on the other hand is only one way. We are sure to not lose solutions, but not sure to not introduce new ones. Squaring is such an operation. In fact for complex numbers squaring automatically doubles the number of solutions for polynomials.


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