Proving an inequality for real numbers

Question

Let $y$ be a non-negative real number and let $q$ be a positive rational number. I want to prove that there exists $\varepsilon>0$ (real) such that $(y+\varepsilon)^{n} < q + y^{n}$, where $n\geq 1$ is an integer. However, I am trying to prove this $\textbf{without}$ using the following:

1) The binomial theorem

2) The following identity $b^{n} - a^{n} = (b-a)(b^{n-1} + b^{n-1}a + ... + a^{n-1})$

3) Properties of nth-roots of real numbers (I am hoping to use the inequality in a proof about nth-roots, see here prove existence of nth roots for non-negative real numbers)

I have tried to show it by contradiction by assuming that $(y+\varepsilon)^{n} \geq q + y^{n}$ for all $\varepsilon>0$. One of the ways I have tried is inducting on $n$ on the hypothesis "there exists $\varepsilon>0$ such that $(y+\varepsilon)^{n} < q + y^{n}$, where the base case $n=1$ is obvious. Then I tried to use contradiction in the induction step. So far this has not succeeded.

Properties that I can use without risk of circularity is the order properties of the real numbers, as well as properties of exponentiation of real numbers with integer exponents. Properties of Cauchy sequences of rational numbers can also be used. However, limits cannot be used, as they are not developed until the next chapter in the book I am working on (so far in my textbook, the real numbers have been constructed as $\textbf{formal}$ limits of equivalent Cauchy sequences of rational numbers).

Thanks to everybody who read the post. All hints/feedback is appreciated.

Answer 1

This is just a comment on the OP's answer to their own question: In this case, the proof can be marginally simplified if you strengthen the assertion by dropping the assumption the $q$ is rational. After all, if it's true for all positive reals, then it's automatically true for all positive rationals. This obviates the need to invoke proposition 5.4.14: One can simply let

$$q_0={q\over2(y+1)}\quad\text{and}\quad\epsilon_1={q\over2y^k}$$

and argue inductively that there exists an $\epsilon_0$ for which

$$(y+\epsilon_0)^k-y^k\lt q_0$$

so that, with $\epsilon=\min(1,\epsilon_0,\epsilon_1)$, we have

$$\begin{align} (y+\epsilon)^{k+1}-y^{k+1}&=(y+\epsilon)((y+\epsilon)^k-y^k)+\epsilon y^k\\ &\le(y+1)((y+\epsilon_0)^k-y^k)+\epsilon_1y^k\\ &\lt(y+1)q_0+\epsilon_1y^k\\ &\le{q\over2}+{q\over2}\\ &=q \end{align}$$

(Note, the strict inequality appears in the middle line; the first inequality uses the definition of $\epsilon$ and the third one uses the definitions of $q_0$ and $\epsilon_1$.)

Answer 2

This is my attempt after helpful guidance in the comments above. For completeness, proposition 5.4.14 just states that between any two real numbers there exists a rational number.

We shall prove the following by induction: For any non-negative real number $y$ and for any positive rational number $q$ there exists $\varepsilon>0$ such that $(y+\varepsilon)^{n} - y^{n} < q$. The case $n=1$ is obvious. Now suppose the statement has been proven for $n=k$. We must show that it holds for $n=k+1$. Note that $$(y+\varepsilon)^{k+1} - y^{k+1} = (y+\varepsilon)((y+\varepsilon)^{k} - y^{k}) + y^{k}\varepsilon$$ Let $q_{0}$ be a positive rational number smaller than $q/(2(y+1))$. Such a number exists by proposition 5.4.14. By our induction hypothesis, there exists $\varepsilon_{0}$ such that $(y+\varepsilon_0)^{k} - y^{k} < q_{0}$. There also exists $\varepsilon_{1}$ such that $\varepsilon_{1} < q/(2y^{k}) $ (prop. 5.4.14 again). Hence, letting $\varepsilon = $min$(1, \varepsilon_{0}, \varepsilon_{1})$, we get that $$(y+\varepsilon)^{k+1} - y^{k+1} < (y+1)((y+\varepsilon)^{k} - y^{k}) + y^{k}\varepsilon < q/2 + q/2 < q$$ This completes the induction.