It might be a silly question but if $$a+\sqrt{a^2+1}= b+\sqrt{b^2+1},$$ then can I conclude that $a=b$? I thought about squaring both sides but I think it is wrong! Because radicals will not be removed by doing that! Can you help me with proving that $a=b$ or not? Actually I'm going to prove that $x+\sqrt{x^2+1}$ is a $1$-$1$ function.
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9The function $x\mapsto x+\sqrt{x^2+1}$ is strictly increasing on $\Bbb R$. – Angina Seng Dec 09 '18 at 18:17
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Hint: Square the equation $$a-b=\sqrt{b^2+1}-\sqrt{a^2+1}$$ two times. The result must be $$(a-b)^2=0$$
Dr. Sonnhard Graubner
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Hey Dr. I was really looking for you here! I had a question about my mathematical daily problems! – user602338 Dec 09 '18 at 18:26
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Dr. Sonnhard , our teacher really makes us tired with givin more and more proving questions like this! Would you please tell me how to improve mathematical part of my mind like you! I want to prove my puzzles in first sight! How to do that? – user602338 Dec 09 '18 at 18:28
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This depends on the given problem, you can write it here. – Dr. Sonnhard Graubner Dec 09 '18 at 18:30
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Yeah but you always solve these kind of problems. I know you're dealing with them for decades. But I usually dont have enough time to ask. And i have to stand on my own feet! – user602338 Dec 09 '18 at 18:32
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Let $f:\mathbb R \to \mathbb R $ such that :
$$f(x) = x + \sqrt{x^2+1}$$
Then, it is :
$$f'(x) = 1 + \frac{x}{\sqrt{x^2+1}}= \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}} > 0, \; \forall x \in \mathbb R $$
Thus the function $f(x)$ is strictly increasing for $x \in \mathbb R$ and thus it is "$1-1$", which then means that :
$$a + \sqrt{a^2+1} = b + \sqrt{b^2=1} \implies a=b$$
Rebellos
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Just an added question here. The map is not surjective since $0$ does not have a pre-image? – Yadati Kiran Dec 09 '18 at 18:31
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Alternatively, for $a,b\in \Bbb R$, \begin{align}a+\sqrt{a^2+1}&=b+\sqrt{b^2+1}\\&\implies \frac{1}{a+\sqrt{a^2+1}}=\frac{1}{b+\sqrt{b^2+1}}\wedge a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\\ &\implies \sqrt{a^2+1}-a=\sqrt{b^2+1}-b \wedge a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\\ &\implies \sqrt{a^2+1}-a=\sqrt{b^2+1}-b \wedge a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\\&\implies (a+\sqrt{a^2+1})-(\sqrt{a^2+1}-a)=(b+\sqrt{b^2+1})-(\sqrt{b^2+1}-b) \\&\implies 2a=2b\implies a=b. \end{align} It is also easy to see that $a=b\implies a+\sqrt{a^2+1}=b+\sqrt{b^2+1}$. That is, $$a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\iff a=b.$$
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Hint:
WLOG $a=\cot2A,b=\cot2B,0<A,B<\dfrac\pi2$
we have $\cot A=\cot B\implies\cot2A=?$
lab bhattacharjee
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