Multiplying the inequalities gives a weird result which is obviously wrong

Here's a chain of conclusions which leads to the wrong outcome but I don't understand at what point I made a mistake.

\begin{align} 1 < a < 2 &\quad\Rightarrow\quad 1 < \,a^2\, < 4 \\[4pt] 1 < a < 2 &\quad\Rightarrow\quad \frac12 < \frac1a < 1 \end{align}

Now I want to multiply the two sets of inequations that I have to the right of the '$$\Rightarrow$$' so here we have:

$$\frac12 < \frac{a^2}a < 4 \quad\Leftrightarrow\quad \frac12 < a < 4$$

Obviously that's wrong, because $$1 < a < 2$$, but I don't understand where I went wrong, leading to a false conclusion.

I appreciate any help.

• if $1<a<2$ then it's true that $1/2<a<4$, but the inequalities $1<a^2<4$ and $1/2<1/a<1$ are not independent, so multiplying them without regarding that $a$ is the same in both inequalities will weaken the inequality Jan 1 '20 at 18:01
• I think the title should say "inequalities" rather than "equations" Jan 1 '20 at 18:04
• You didn't reach a false conclusion. You reached a weak conclusion. If $1 < a < 2$ then $\frac 12 < 1 < a < 2 < 4$ and $\frac 12 < a < 4$ is TRUE but it is a weaker statement. Imagine someone said "I must find out what type of job Jane Smith of 221 Hudson street has" and you conclude she works and Carbonics Inc. And find out that everyone who works at Carbonics is a human being. So you conclude "Jane Smith is a human being!". That isn't wrong. It's weak. Jan 1 '20 at 19:03
• A simpler version: If $x > 1$ and $x > 2$ then you can multiply to conclude $x^2 > 2$. That's true, but it's weaker than another true conclusion of $x^2 > 4$. Jan 2 '20 at 5:04

You haven't reached a false conclusion (yet). You started with the assumption that $$1 and you concluded with $$\frac12. That's a true consequence (since the open intervals $$(1,2)\subset(\frac12,4)$$) and a valid argument to reach it. You'd be in a mess if you tried to reverse those steps, though.

• Hm, I didn't even think about that, thanks. PS I don't know why someone downvoted your answer but I restored the karma :) Jan 1 '20 at 18:27
– user694818
Jan 1 '20 at 18:29

It's obviously TRUE.

If $$1 < a < 2$$ and $$\frac 12 < 1$$ then $$\frac 12 < 1$$ and $$1 < a$$ so $$\frac 12 < a$$ must be true.

And $$2 < 4$$ and $$a < 2$$ so $$a < 2 < 4$$ so $$a < 4$$ must be true. So $$\frac 12 < a$$ true and $$a < 4$$ is true so $$\frac 12 < a < 4$$ must be TRUE.

And $$\frac 12 < a < 4$$ does not contradict $$1 < a < 2$$.

It's just that $$1 is a stronger more precise statement.

Consider this.

$$a= 36$$ is an even number.

$$a=36$$ is a perfect square.

$$a=36$$ is a multiple of $$3$$.

Fred comes and says "$$a$$ is a multiple of $$3$$". Then Jane comes along and says "$$a$$ is a perfect square". Michael says "$$a$$ is an even number."

Olav hears all three of these people and says "Wait! Those are three completely different statements! So only one of them can be true! Which one of you is telling the truth and which one is lying."

....

And they are all telling the truth.

$$\frac 12 < a < 4$$ means that it is possible that $$a$$ might be less than $$1$$ but it does not mean that it is. So $$\frac 12 < a$$ does not contradict $$1 < a$$.

This is analogous to "$$a$$ is a multiple of $$3$$". That means $$a$$ might be odd but it doesn't mean that it is. "$$a$$ is a multiple of $$3$$" does not contradict "$$a$$ is an even number".

So on.

=====

Another thing to note:

(Assume only positive values)

$$a < b$$ and $$c < d \implies ac < bd$$. But that is one directional. It doesn't go the other way that $$ac < bd \not \implies a< b$$ and $$c < d$$

So $$1< a < 2\implies (1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)\implies \frac 12 < a<4$$

That is true. And indeed $$1< a < 2 \implies \frac 12 < 1 < a < 2 < 4\implies \frac 12 < a < 4$$.

But it doesn't go the other way!

$$\frac 12 < a < 4 \not \implies 1 < a < 2$$

Snd $$\frac 12 < a <4 \not \implies (1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)$$

[although $$(1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)$$ does actually imply $$1 < a < 2$$.)

• Thank you. So am I right saying that if we have different variables, $a < x < b, c < y < d => ac < xy < bd$ ? Jan 2 '20 at 7:49
• If the variables are non-negative then, yes, that is correct. $a < x;c>0$ means $ac < cx$. $c < y;x>0$ means that $cx < xy$. Transitivity means $ac < xy$. $x<b;y>0$ means $xy < by$. And $y<d;b>0$ means $by<bd$. Transitivity means $xy<bd$. Jan 2 '20 at 16:06

The first two inequalities are, in fact, biconditional ($$\iff$$), assuming $$a > 0$$. But when you multiply them, you weaken your final inequality. As you can see, your result $$1/2 < a < 4$$ is not wrong, but the $$\impliedby$$ part is. Why would it be? Try to argue and see the flaw in your argument.

"$$\iff$$ inequalities" have to be manipulated carefully. Losing strength is really easy: $$0 < x <1 \implies -0.5 < x < 1.5$$, but $$x = 1.2$$ solves the second one but not the first. You did a lot of implicit steps when multiplying those; if you had $$-2, you wouldn't take squares nor reciprocals as easy as you did, and multiplying the inequalities would need aditional steps since they have negative numbers. That's why step-by-step calculations are necessary for strong inequalities.

• Why did this get downvoted? Jan 1 '20 at 19:17
• The first inequation is not biconditional, since $1 < a^2 < 4$ does not imply that $a \in (1, 2)$, consider the interval $(-1, -2)$. Jun 27 '20 at 21:32

When you find yourself scratching your head over a result like this, it pays sometimes to redo the problem using only variables

\begin{align} A < x < B &\quad\Rightarrow\quad A^2 < \,x^2\, < B^2 \\[4pt] A < x < B &\quad\Rightarrow\quad \frac1B < \frac1x < \frac1A \end{align} multiply them together and you get \begin{align} \frac{A^2}B < \,x\, < \frac{B^2}A \\[4pt] \end{align} or \begin{align} A\frac{A}B < \,x\, < B\frac{B}A \\[4pt] \end{align}

Now, since you know that $$A (since $$A), that means that $$\frac{A}{B} < 1$$.

That means that $$A\frac{A}B < A$$ and $$B\frac{B}A > B$$, and your final inequality will always be weaker than your original inequality.

• Thank you, great idea! Jan 2 '20 at 7:45