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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.

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  • $\begingroup$ Why don't you try induction on $n$ directly instead of contradiction? Also I don't see why you would want $y$ to be real and $q$ to be rational. $\endgroup$ Apr 27 '20 at 0:53
  • $\begingroup$ Btw Why do you prohibit the basic algebraic identities? They are nothing more than consequences of elementary properties of addition and multiplication. By prohibiting them you essentially prohibit to use any features of $+, \times $. $\endgroup$ Apr 27 '20 at 0:56
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    $\begingroup$ If you have any issues with idea let me know. Also write the proof as an answer here if you succeed. $\endgroup$ Apr 27 '20 at 1:08
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    $\begingroup$ For the second term we need $h<q/(2y^n)$ and for first term we already know that there is an $h_0>0$ such that $(y+h_0)^n-y^n<q/(2(y+1))$. Now let $h$ be such that $0<h<\min(1,h_0,q/(2y^n))$ and that value of $h$ works. $\endgroup$ Apr 27 '20 at 1:37
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    $\begingroup$ Looking at your questions I appreciate the effort you have put in understanding the existence of nth roots of a real number. Don't worry your efforts will bring fruit. $\endgroup$ Apr 27 '20 at 1:39
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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 $\epsilon_1$.)

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  • $\begingroup$ Thank you so much for the additional input :) $\endgroup$
    – tfjaervik
    Apr 27 '20 at 13:41
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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.

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    $\begingroup$ Fixed a typo and some formatting. +1 and this is how one enjoys writing proofs (even if a little help is needed). $\endgroup$ Apr 27 '20 at 11:13
  • $\begingroup$ You may accept your own answer so that the question gets removed from the unanswered queue or you can wait for better answers. $\endgroup$ Apr 27 '20 at 11:14
  • $\begingroup$ You must however wait 48 hours before you can accept your own answer. $\endgroup$ Apr 27 '20 at 11:26
  • $\begingroup$ To be clear, where you write $q/2(y+1)$ you actually mean $q/(2(y+1))$, right? $\endgroup$ Apr 27 '20 at 11:50
  • $\begingroup$ @ParamanandSingh Really appreciate your help :) I will accept the answer as soon as 48 hours have passed. $\endgroup$
    – tfjaervik
    Apr 27 '20 at 11:57

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