Prove that $\frac{1}{90} < \sqrt{2024} - \sqrt{2023} <\frac{1}{88}$

Question

Question a) Prove that $$\frac{1}{90} < \sqrt{2024} - \sqrt{2023} <\frac{1}{88}$$

Question b) Is $$\sqrt{2024} - \sqrt{2023}$$ smaller or larger than $$\frac{1}{89}$$

No calculator is allowed, obviously and neither are methods of guessing the roots, we must move from the true inequality $0<1$ to that which we must prove.

Guessing that we must start by squaring both sides which is how i started.

$$\frac{1}{90^2} < 4047-2*\sqrt{2024}*\sqrt{2023} <\frac{1}{88^2}$$

i focus on the left side of the equation to first succesfuly prove that and get one side to one

$$1 < 90^2 (4047-2*\sqrt{2024}*\sqrt{2023})$$

This was were i got stuck maybe this is the intended transformation? $$1 < 90^2 (2*45^2 -3 -2*\sqrt{45^2-1)(45^2-2)})$$

Now thanks to help i have a basic proof. We start by multiplying by the conjugate

$$\frac{\sqrt{2024} + \sqrt{2023}}{90} < (\sqrt{2024} - \sqrt{2023})(\sqrt{2024} + \sqrt{2023})<\frac{\sqrt{2024} + \sqrt{2023}}{88}$$

Simplify $$\frac{\sqrt{45^2 -1} + \sqrt{45^2 -2}}{2*45} < 1 <\frac{\sqrt{45^2 -1} + \sqrt{45^2 -2}}{2*(45-1)}$$

Now simple proof says that in LL we have two numbers smaller than 45 added togheter in the numerator and larger than that in the denominator. While in RL we have that these must be larger than 44 and thus a denominator that is smaller than the numerator. QED LL is smaller than one and RL is larger and thus the original statement is true.

Any help is appreciated! This was a question on a test on the chapter inequalites My teacher should provide a solution in about one week.

Answer 1

Let $x = \frac{1}{\sqrt{2024} - \sqrt{2023}}$. Rationalizing the denominator gives:

$$x = \frac{\sqrt{2024} + \sqrt{2023}}{(\sqrt{2024} - \sqrt{2023})(\sqrt{2024} + \sqrt{2023})}$$ $$= \frac{\sqrt{2024} + \sqrt{2023}}{2024 - 2023}$$ $$= \sqrt{2024} + \sqrt{2023}$$

Note that $44^2 = 1936$ and $45^2 = 2025$. So each square root must be in this interval.

$$44 < \sqrt{2023} < 45$$ $$44 < \sqrt{2024} < 45$$

Adding them gives:

$$88 < \sqrt{2024} + \sqrt{2023} < 90$$ $$88 < x < 90$$

Since all values have the same sign, taking the reciprocals inverts the order:

$$\frac{1}{88} > \frac{1}{x} > \frac{1}{90}$$ $$\boxed{\frac{1}{90} < \sqrt{2024} - \sqrt{2023} < \frac{1}{88}}$$

If we bisect the interval $(44, 45)$, we get:

$$44.5^2 = 1980.25$$ $$44.5 < \sqrt{2023} < 45$$ $$44.5 < \sqrt{2024} < 45$$ $$89 < x < 90$$

$$\boxed{\frac{1}{90} < \sqrt{2024} - \sqrt{2023} < \frac{1}{89}}$$

Answer 2

You need the identity $(a+b)^2=a^2+b^2+2ab$ and some basic arithmetic to following along this post. You can simply evaluate $45^2=(40+5)^2=40^2+5^2+2(40)(5)=2025$.

Let $a=\sqrt{2024}$ and $b=\sqrt{2023}$. We expand $44^2=(45-1)^2=1936$ and $44.5^2=\left(45-\frac12\right)^2=2025+1/4-45=1980.25$. Since $\sqrt x$ is increasing we have $a,b>44.5$ and $a,b<45$. We also note that $(a+b)(a-b)=2024-2023=1$.

First question: Multiply each side of the inequality by $(a+b)$ so we have to show $\frac{a+b}{90}<1$ and $\frac{a+b}{88}>1$. We have that $a, b< 45$ so that $\frac{a+b}{90}<1$. We also have $a,b>44$ and it follows $\frac{a+b}{88}>1$.

Second question: since we have $44.5<a,b$ it follows $89<a+b$ and multiplying this by $(a-b)$ gives $a-b<\frac1{89}$. Note the direction of the inequality doesn’t reverse since $a-b>0$.

Answer 3

$\sqrt{2024}-\sqrt{2023}=\dfrac1{\sqrt{2024}+\sqrt{2023}}>\dfrac1{2\sqrt{2025}}=\dfrac1{90}\,.$

Moreover,

$\begin{align}\sqrt{2024}-\sqrt{2023}&=\dfrac1{\sqrt{2024}+\sqrt{2023}}<\dfrac1{2\sqrt{2023}}=\\[3pt]&=\dfrac1{2\!\cdot\!45}\sqrt{\frac{45^2}{2023}}\overset{\color{blue}{(*)}}{<}\dfrac1{2\!\cdot\!45}\!\cdot\!\dfrac{45^2}{2023}=\\[3pt]&=\dfrac{45}{4046}<\dfrac{45}{4005}=\dfrac1{89}\,.\end{align}$

$\color{blue}{(*)}$ because $\;\sqrt x<x\;$ for any $\,x\!>\!1\;$ and $\;\frac{45^2}{2023}\!=\!\frac{2025}{2023}\!>\!1\,.$