Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$

Question

I would like to prove $$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$

• I'm interested in more ways of proving it

My thoughts:

\begin{align} \sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\ \frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\ \frac{1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{1}{\sqrt{x +1}+\sqrt{x}}\\ \sqrt{x +1}+\sqrt{x} &\neq \sqrt{x+2}+\sqrt{x+1}\\ \sqrt{x} &\neq \sqrt{x+2}\\ \end{align}

• Is my proof correct?
• I'm interested in more ways of proving it.

Answer 1

Hint #1:
Assume that $\sqrt{x + 2} - \sqrt{x + 1} = \sqrt{x + 1} - \sqrt{x}$ for some $x > 0$.

Hint #2:
Derive a contradiction.

Hint #3:
This proof (by contradiction) results to some changes in the notation you used in your proof.

Answer 2

Your proof is correct, but I feel that this could be proved by contradiction.

Assume for contradiction $\exists x>0$ such that the equation $\sqrt{x+2}-\sqrt{x+1}=\sqrt{x+1}-\sqrt{x}$ is true. Then, \begin{align} \sqrt{x+2}-\sqrt{x+1}&=\sqrt{x+1}-\sqrt{x}\\ \frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&=\frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\ \frac{1}{\sqrt{x+2}+\sqrt{x+1}}&=\frac{1}{\sqrt{x +1}+\sqrt{x}}\\ \sqrt{x +1}+\sqrt{x} &=\sqrt{x+2}+\sqrt{x+1}\\ \sqrt{x} &=\sqrt{x+2}\\ x&=x+2 \end{align} This is not true and we have reached a contradiction. Thus the equation does not hold.

Answer 3

Assume the statement is true \begin{align} \sqrt{x + 2} - \sqrt{x + 1} &= \sqrt{x + 1} - \sqrt{x}\\ \sqrt{x + 2} + \sqrt{x} &= 2\sqrt{x + 1}\\ 2x + 2 + (2\sqrt{x^2 + 2x}) &= 4x + 4\\ \sqrt{x^2 + 2x} &= x + 1\\ x^2 + 2x &= x^2 + 2x + 1\\ 0 &= 1\\ \end{align} Which is a contradiction. Now I am not sure if I'm allowed to manipulate the equation like this so the proof might be invalid. The reason I though I could was I had seen simillar logic in proving the irrationality of $\sqrt{2}$.

Answer 4

Your proof is correct, because each inequality you write is equivalent to the previous one (it should be noted, probably).

Changing all $\ne$ into $=$ would make it a proof by contradiction, that's however unnecessary.

In a different way, you could just swap terms and square, again changing inequalities into equivalent ones: \begin{gather} \sqrt{x+2}+\sqrt{x}\ne 2\sqrt{x+1}\\[10px] x+2+x+2\sqrt{x(x+2)}\ne 4x+4\\[10px] \sqrt{x(x+2)}\ne x+1\\[10px] x^2+2x\ne x^2+2x+1\\[10px] 0\ne1 \end{gather}

Answer 5

By the MVT:

$$\sqrt {x+2} - \sqrt {x+1} = \frac{1}{2\sqrt {c_x}}\cdot 1, \ \ \ \ \sqrt {x+1} - \sqrt {x} = \frac{1}{2\sqrt {d_x}}\cdot 1.$$

Here $c_x\in (x+1,x+2), d_x\in (x,x+1).$ Because $1/\sqrt x$ strictly decreases, the left term minus the right term is negative.

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Concavity: Slopes of successive chords on a strictly concave graph are strictly decreasing, and $\sqrt x$ is strictly concave. Therefore $\sqrt{x +2} - \sqrt{x +1} < \sqrt{x +1}-\sqrt{x}$  for all $x\ge 0.$

Answer 6

By the mean value theorem - MVT, for all $x >0$, it exists $\zeta_x \in (x, x+1)$ such that $$\frac{1}{2\sqrt{\zeta_x}}((x+1)-x)=\frac{1}{2\sqrt{\zeta_x}}=\sqrt{x+1}-\sqrt{x}$$ As $y \to \frac{1}{2\sqrt{y}}$ is stricly decreasing, you have $$\frac{1}{2\sqrt{\zeta_{x+1}}}=\sqrt{x+2} -\sqrt{x+1} < \sqrt {x+1}-\sqrt{x}=\frac{1}{2\sqrt{\zeta_x}}.$$

Answer 7

I will show that if $\sqrt{x+a}-\sqrt{x+b} =\sqrt{x+b}-\sqrt{x+c} $ for two different values of $x$, then $a=c$ and $b = a+c =2a $. If the equality holds for one value of $x$, then $8x(a+c-2b) =(a-4b+c)^2-4ac $.

Since, in the original problem, $a=2, b=1, c=0$, this can not hold for two values of $x$. If it holds for one $x$, since $a+c-2b = 0$ and $(a-4b+c)^2-4ac =2^2-0 =4 $, this does not hold for any $x$. then

Suppose $\sqrt{x+a}-\sqrt{x+b} =\sqrt{x+b}-\sqrt{x+c} $, or $\sqrt{x+a}+\sqrt{x+c} =2\sqrt{x+b} $.

Squaring, $x+a+x+c+2\sqrt{(x+a)(x+c)} =4(x+b) $ or $2x-a+4b-c =2\sqrt{(x+a)(x+c)} $. Squaring again, $4x^2-4x(a-4b+c)+(a-4b+c)^2 =4(x+a)(x+c) =4(x^2+x(a+c)+ac) =4x^2+4x(a+c)+4ac $ or $4x(a+c+a-4b+c) =(a-4b+c)^2-4ac $ or $4x(2a+2c-4b) =(a-4b+c)^2-4ac $ or $8x(a+c-2b) =(a-4b+c)^2-4ac $.

If this is true for more than one $x$, then $a+c-2b = 0$ and $(a-4b+c)^2-4ac = 0 $.

From the first equation, $2b = a+c$. Substituting this in the second equation, $4ac =(a-4b+c)^2 =(a+c-2(a+c))^2 =(a+c)^2 =a^2+2ac+c^2 $ or $0 =a^2-2ac+c^2 =(a-c)^2 $, so that $a=c$.

Therefore, if $\sqrt{x+a}-\sqrt{x+b} =\sqrt{x+b}-\sqrt{x+c} $ for two different values of $x$, then $a=c$ and $b = a+c =2a $.