Prove that if $|x|<1$ then $x^{6}<1$

Question

I am trying to prove that if $|x|<1$ then $x^{6}<1$ and that if $x^{6}<1$ then $|x|<1$.

For the first part, I thought of first considering $0<x<1$. Multiplying by $x$ (which is positive) I then get $0<x^{2}<x<1$. And repeating the multiplication I would end up with $x^{6}<1$.

Then I would consider $-1<x<0$. But now multiplying by $x$ (which is negative) gives $-x<x^{2}$ and I dont see how to get $x^{2}<1$ from here and then $x^{6}<1$. There is probably a more elegant way of going about this?

For the second part I thought of using the contrapositive and prove $x\leq -1$ or $x\geq 1$ then $x^{6}\geq 1$, and then I would try to use the previous proof to break this into two cases.

$x\leq -1$ then $x^{6}\geq 1$ and $x\geq 1$ then $x^{6}\geq 1$.

Answer 1

The argument that dxiv made in a comment is, I think, the most elegant argument. As comments are ephemeral, I'll reiterate it here (with some additional details): note that $$ \frac{|x|^6 - 1}{|x|-1} = |x|^5 + |x|^4 + |x|^3 + |x|^2 + |x| + 1. \tag{1}$$ This is actually a specific case of a much more general result: for any natural number $n$, $$ \frac{t^n-1}{t-1} = \sum_{j=0}^{n-1} t^j. $$ This can be proved by an induction argument after noticing that $$t^{n+1} - 1 = t^{n+1} - t^n + t^n -1 = t^n(t-1) + t^n - 1.$$ In this particular case, we take $t = |x|$. In any event, since all of the terms on the right-hand side of (1) are nonnegative (and $1>0$, i.e. $1$ is strictly positive), it follows that $$ \frac{|x|^6 - 1}{|x|-1} > 0. \tag{2}$$ Note that $$ \frac{a}{b} > 0 \iff (a>0 \land b > 0) \lor (a< 0\land b < 0). $$ That is, a fraction is positive if and only if both the numerator and denominator have the same sign. Applying this to (2), either $$ |x|^6 - 1 < 0 \qquad\text{and}\qquad |x| - 1 < 0, \tag{3} $$ or $$ |x|^6 - 1 > 0 \qquad\text{and}\qquad |x| - 1 > 0. $$ Therefore if we assume that $|x|<1$ it immediately follows from (3) that $|x|^6<1$, and vice versa.

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Alternatively, we can make your approach work. The first part of your argument is basically an induction argument. We can actually strengthen that result a bit and get the following:

Proposition: If $|x| < 1$ then $|x|^n < 1$ for any natural number $n$.

Proof: The proof is by induction. Assume that $|x|< 1$, and note that $|x|^1 < 1$, providing a base for the induction. Now suppose that $|x|^k < 1$. Then $$ |x|^k < 1 \implies |x|^k \cdot |x| < 1 \cdot |x| \implies |x|^{k+1} < |x|. $$ But $|x|<1$, which proves the result.$\tag*{$\blacksquare$}$

Taking $n = 6$ gives gives the result in the original question. Then, as you note, the converse requires an additional argument. Following your instinct of attempting contraposition, you might consider the following proposition:

Proposition: Let $n$ be a natural number.

• If $|x| < 1$ then $|x|^n < 1$,
• if $|x| = 1$ then $|x|^n = 1$, and
• if $|x| > 1$ then $|x|^n > 1$.

There may be more elegant arguments, but the induction proof above can be used, mutatis mutandis, to prove the two new statements. This gives a kind of trichotomy, from which you can conclude that if $|x|^n < 1$, then $|x| < 1$.

Answer 2

suppose $0<x<1$ and $x^6\ge1$

$x^6=x\cdot x^5$ and since $0<x<1$ multiplying by $x$ is like scaling a value making it $100\cdot x$% of the original value, which is smaller than $<1\cdot100$%.

So $x^6=x\cdot x^5<x^5<...<x\cdot x<x<1$ and that's impossible

Answer 3

The case $x=0$ is clear. So we can assume that $x \ne 0$. Then we have that $ \frac{1}{|x|}>1$. Hence there ist $t>0$ such that $ \frac{1}{|x|}=1+t.$

Then we get, by Bernoulli:

$\frac{1}{x^6}=\frac{1}{|x|^6}=(1+t)^6 \ge 1+6t >1$ and the result follows.

Answer 4

First, note that $f(x)=x^6-1$ only has real roots at $-1$ and $1$. If it would be necessary to prove this, you could factor into $(x^3-1)(x^3+1)$, apply the sum/difference of cubes formulas, show that the quadratic factors yield non-real roots, and then appeal to the Fundamental Theorem of Algebra to say that $-1$ and $1$ are the unique real roots.

Also note that $x^6\ge 0$, so $x^6-1\ge -1$.

Now note that $f(-2)=64>0$, $f(2)=64>0$, and $f(0)=-1<0$. Then by the Intermediate Value Theorem, $f(x)>0$ on $(-\infty,-1)\cup (1,\infty)$ and $f(x)<0$ on $(-1,1)$.

Then $$\lvert x\rvert<1 \iff -1<x<1\\ \iff -1<x^6-1<0\\ \iff 0<x^6<1 $$

Answer 5

The first part for $0<x<1$ is OK.

Then I would consider $−1<x<0$. But now multiplying by $x$ (which is negative) gives $−x<x^2$ and I dont see how to get $x^2<1$ from here and then $x^6<1$. There is probably a more elegant way of going about this?

Note that $-1<x<0$ multiplied by $x$ (which is negative) results in $-x\color{red}>x^2$ (because an inequality multiplied by a negative number will have reversed inequality sign).

Now multiply $-1<x<0$ by $-1$ to get: $1>-x$ (again note the inequality sign reversal).

Can you merge the two points and finish?

Also note, the case $x=0$ is trivial, which could be included in the first part as $0\le x<1$.