Justifying/falsifying $|x^2-16| < 1$

Question

Suppose $0 < |x-4| < 0.5$. Then is $|x^2-16|<1$ true?

What is wrong with the following reasoning?

• We have $|x^2-16|=|x-4||x+4| < 0.5 | x+4|$. So to find out whether $|x^2-16|<1$ is true, we can instead find whether $|x+4| < 2$ is true or not. (If yes, then by the above we will have $|x^2-16| < 0.5 |x+4| < 0.5 \cdot 2 = 1$. If no, then we will start looking for a counterexample to $|x^2-16|<1).$ To find whether $|x+4|<2$ is true, note that we have $-0.5 < x - 4 < 0.5$, and thus $8-0.5 < x + 4 < 8 + 0.5$. From this we can conclude that $|x+4| <8 + 0.5;$ this suggests that $|x+4| < 2$ doesn't need to be true. This also suggests that a counterexample to $|x^2-16|<1$ may be such that $x+4$ is between $8-0.5$ and $8 + 0.5$ but not between $-2$ and $2.$ So, let's try $x=4.01.$ This satisfies the last two conditions and also the given supposition, yet, it clearly isn't actually a counterexample to $|x^2-16|<1.$

What is wrong here? How come that $4.01$ justifies the fact that $|x+4|$ is not always less than $2$ for $x$s satisfying $0 < |x-4| < 0.5$ and yet it fails to justify the fact that $|x^2-16|$ is not always less than $1$ for $x$s satisfying $0 < |x-4| < 0.5$? Was it wrong to say at the beginning that the original problem reduces to finding out whether $|x+4| < 2$ is true or not under the assumption $0 < |x-4| < 0.5$? If so, why is it wrong, and what is the correct approach?

Answer 1

Addendum added to respond to the comment question of user125234.

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Your attempt to use $~x = 4.01~$ to generate a counter example failed, because $~|4.01 - 4|~$ is nowhere near $~0.5.~$ This explains why $~|4.01 - 4| \times |4.01 + 4|~$ fails to produce a product that is $~\geq 0.5.$

From your own analysis, you saw that $~8 - 0.5 < |x + 4| < 8 + 0.5.~$

Further, from your own analysis, you saw that in order to prove the conjecture true (with the conjecture actually being false), you would need to prove that $~|x+4| < 2.$

Here $~\dfrac{8}{2} = 4,~$ and the specific example that you chose $~x = 4.01,~$ results in $~|x - 4|~$ being significantly less than $~\dfrac{0.5}{4}.$

This explains why your attempt at using $~x = 4.01~$ to generate a counter example failed.

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$\underline{\text{Addendum}}$

I don't understand the first paragraph. Yes, $~| ~4.01−4 ~|~$ is nowhere near $~0.5~$, but I thought to get a counterexample one needs to provide $~x_0~$ such that $| ~x0−4 ~|<0.5~$ and $~| ~x+4 ~| \geq 2.$

Why should the counterexample $~x_0~$ have the property that $~| ~x_0 - 4~| ~$ have the have the property that $~| ~x_0−4 ~| ~$ is near $~0.5,~$ (whatever near means)?

To resolve your confusion, I need to go back to basics.

Consider the following two bullet points:

• I-1 (i.e. the first inequality)
  $~| ~x - 4 ~| < 0.5.~$

• I-2 (i.e. the second inequality)
  $| ~x^2 - 16 ~| < 1.$

You are not being asked whether it is possible, for some specific value of $~x,~$ to have I-1 and I-2 both be true.

If you were being asked this, which you are not, then $~x = 4.01~$ would answer the question.

Instead you are being asked whether every value of $~x~$ that satisfies I-1 must also satisfy I-2.

In other words, you are being asked whether I-1 implies I-2.

In fact, it does not. This means that it is possible to construct some value of $~x~$ that satisfies I-1, but does not satisfy I-2.

Before I discuss how to construct the necessary counter example, I should focus on the question that you posed:

...How come that 4.01 justifies the fact that $|x+4|$ is not always less than $2$ for $x$s satisfying $0 < |x-4| < 0.5$ and yet it fails to justify the fact that $|x^2-16|$ is not always less than $1$ for $x$s satisfying $0 < |x-4| < 0.5$?

In other words, the specific question that you posed, may be re-stated as:
Why doesn't $~x = 4.01~$ provide a counter example?
That is, $~|4.01 - 4| < 0.5.~$
Under the assumption that I-1 does not always imply I-2,
why is it that $~|4.01^2 - 16|~$ is not greater than $~1~?$

My answer, before the Addendum, answers the above question, but (apparently) not in clear enough detail. I will provide the detail here. For simplicity, assume that we are only examining values of $~x~$ that are greater than $~4.$

Then, the question boils down to:
why is it that $~4.01 - 4 < 0.5,~$
and yet $~4.01^2 - 16~$ is not greater than $~1~?$

To answer that question, continue to assume that you are focusing only on values of $~x~$ such that $~x > 4.~$ You have that

$$(x - 4) \times (x + 4) = x^2 - 16. \tag1$$

So, you are given that in (1) above, the first LHS factor is $~< 0.5,~$ and you are wondering why the product of the two factors in (1) above is not greater than $~1.~$

You know that $~(x - 4) < 0.5.$
If you knew that $~(x+4) < 2,~$ which you do not know,
then you could conclude that $~(x-4) \times (x+4) < (0.5) \times 2 = 1.$

However, all that you know is that $~(x + 4) < 8.5.$

This implies that if $~(x - 4) < \dfrac{1}{8.5},~$ that $~(x - 4) \times (x + 4) < 1.$

However, you don't know that $~(x - 4) < \dfrac{1}{8.5}.~$

What you do know is that $~(x - 4) < \dfrac{1}{2}.~$

This means that the upper bound on $~(x - 4)~$ is approximately four times bigger than is needed, because $~\dfrac{8.5}{2}~$ is approximately $~4.$

You happened to select a value of $~x > 4,~$ namely $~x = 4.01,~$ such that $~(x - 4) < \dfrac{1}{8.5},~$ which explains why $~x = 4.01~$ fails to provide a counter example. It is because you happened to pick a value of $~x~$ such that $~(x - 4)~$ is too small.

In order to construct a counter example, that satisfies I-1, and violates I-2, with $~x > 4,~$ you need to select some value of $~x~$ such that $~(x - 4),~$ while still being less than $~0.5,~$ is closer to $~0.5.$

For example, consider $~x = 4.4.~$

Then,

$$4.4^2 - 16 = (4.4 - 4) \times (4.4 + 4) = 0.4 \times 8.4 > 1.$$

Now, directly constrast this with $~x = 4.01.$

Then

$$4.01^2 - 16 = (4.01 - 4) \times (4.01 + 4) = (0.01) \times 8.01 < 1.$$

So, $~x = 4.01~$ failed to provide a counter example because $~(4.01 - 4)~$ is nowhere near $~0.5.$

That is, $~4.01 - 4 < \dfrac{0.5}{4}.~$

Consider $~x = 4.125 \implies x - 4 = \dfrac{0.5}{4}.$

Then, $~x + 4 = 8.125.$ So, when $~x = 4.125,~$ you have that

$$x^2 - 16 = (x - 4) \times (x + 4) = \dfrac{0.5}{4} \times 8.125 > \dfrac{0.5}{4} \times 8 = 1. \tag2 $$

So, to construct a counter example, where $~x > 4,~$ as (2) above shows, you should look for some value of $~x~$ such that $~(x - 4)~$ is closer to $~0.5.~$ That is, $~x - 4 = \dfrac{0.5}{4}~$ represents a sufficiently large value of $~x.$

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In fact, to find the smallest value of $~x~$ such that $~x > 4,~$ and $~x^2 - 16 \geq 1,~$ you want

$$x = \sqrt{17} \implies x - 4 = \sqrt{17} - 4 \approx 0.1231.$$

Answer 2

Suppose $0 < |x-4| < 0.5.$ Then is $|x^2-16|<1$ true?

• We have $|x^2-16|=|x-4||x+4| < 0.5 | x+4|.$ So to find out whether $|x^2-16|<1$ is true, we can instead find whether $|x+4| < 2$ is true or not. (If yes, then by the above we will have $|x^2-16| < 0.5 |x+4| < 0.5 \cdot 2 = 1.$ If no, then we will start looking for a counterexample).

Yes, you have validly argued that based on the given premise $0 < |x-4| < 0.5,$ $$\color\red{|x+4| < 2}\implies\text{the proposed conclusion } |x^2-16|<1.$$

Your first logical leap is thinking that the converse $$|x+4| \not< 2\implies |x^2-16|\not<1.$$ is also true.

To find whether $|x+4|<2$ is true, note that we have $-0.5 < x - 4 < 0.5,$ and thus $8-0.5 < x + 4 < 8 + 0.5.$ From this we can conclude that $|x+4| <8 + 0.5;$ this suggests that $|x+4| < 2$ doesn't need to be true. This also suggests that a counterexample to $|x^2-16|<1$ may be such that $x+4$ is between $8-0.5$ and $8 + 0.5$ but not between $-2$ and $2.$ So, let's try $x=4.01.$ This satisfies the last two conditions and also the given supposition, yet, it clearly isn't actually a counterexample to $|x^2-16|<1.$

Your second logical leap is taking

• "this suggests that $\color\red{|x+4| < 2}$ doesn't need to be true"

to mean

• "this means that $\color\red{|x+4| < 2}$ is false."

Your third instance of illogic is unfoundedly thinking that a counterexample to any statement derived from the given premise is automatically a counterexample to the proposed conclusion! (Or, perhaps there is actually no third mistake and, merely because of the aforementioned first logical leap, you are still under the misconception that your intermediate result is equivalent to the proposed conclusion.)

TLDR: You have shown

• neither that every value of $x$ falsifying $\color\red{|x+4| < 2}$ also falsifies the proposed conclusion,
• nor even that every value of $x$ satisfying the given premise falsifies $\color\red{|x+4| < 2}.$

Answer 3

With $z:=x-4$, we rewrite $$0<|z|<\tfrac12\implies|z|\,|z+8|<1.$$

It is easy to see that as soon as $z>\frac18$ (but still $<\frac12$), the second inequation is violated. There is no need for long speeches.