Proving rational surd inequalities

[chwala]

Homework Statement

show that $\frac {1}{2\sqrt {n+1} }$< $(n+1)^{0.5} -n^{0.5}$

Relevant Equations

maths induction

my attempt, i am not good in this kind of questions ...i need guidance.

 

[PeroK]

I think using induction may be quite awkward in this case. Instead, why not rearrange the original inequality?

 

[chwala]

I think using induction may be quite awkward in this case. Instead, why not rearrange the original inequality?

ok let me re arrange and upload...i have a problem with the latex for this problem...can you amend the original question for me?

 

[chwala]

sorry perok, this is how the original question looks like, a colleague suggested that i use maths induction...so what exactly am i supposed to do...thanks in advance

 

[PeroK]

View attachment 269239

sorry perok, this is how the original question looks like, a colleague suggested that i use maths induction...so what exactly am i supposed to do...thanks in advance

That doesn't ask you to use induction for part a). Only part c).

 

[chwala]

ok let me type what i had done initially in a moment...hope you're safe during this trying times of corona virus...

 

[chwala]

is this right approach?

 

[etotheipi]

Showing that the inequality holds for $n=0$ is not sufficient to deduce it holds $\forall n \geq 0$.

 

[chwala]

Showing that the inequality holds for $n=0$ is not sufficient to deduce it holds $\forall n \geq 0$.

ok what approach should i take on this...see my proof above...is it correct?

 

[etotheipi]

You can try raising both sides of the inequality to the power of $-1$

 

[PeroK]

View attachment 269240

is this right approach?

I'm afraid I can't follow what you are trying to do here.

One approach is to assume that
$$\frac 1 {2\sqrt{n +1}} \ge \sqrt{n+1} - \sqrt n$$
And reach a contradiction, or a simple equation for $n$. The first step must be to multiply through by $\sqrt{n+1}$.

 

[chwala]

ok let me try and follow what you have indicated...

 

[chwala]

 

[PeroK]

The last two steps are simply not correct.

There are basic techniques here that you don't seem to have picked up. If you have a surd in the denominator, you try multiplying though by it; if you have one surd, you isolate it on one side of the equation; then you square the equation.

This is basic algebra.

 

[chwala]

 

[chwala]

that is my another approach...

 

[PeroK]

that is my another approach...

You're nearly there. Just one step to go.

 

[chwala]

You're nearly there. Just one step to go.

bingo...

 

[PeroK]

bingo...

But, look, you can cut out a lot of that unnecessary algebra. I'll do it without using contradiction:
$$\frac 1 {2\sqrt{n +1}} < \sqrt{n+1} - \sqrt n$$ $$\Leftrightarrow \ \ \frac 1 2 < n + 1 - \sqrt{n(n+1)}$$ $$\Leftrightarrow \ \ \sqrt{n(n+1)} < n + \frac 1 2$$ $$\Leftrightarrow \ \ n^2 + n < n^2 + n + \frac 1 4$$ $$\Leftrightarrow \ 0 < \frac 1 4$$
Therefore, the stated inequality is true for all $n \ge 0$.

 

[chwala]

ok...how would it look like with contradiction, ...
can we say $n^2+n+0.25≤n^2+ n$ is a contradiction, therefore the converse is true?

 

[PeroK]

ok...how would it look like with contradiction, ...
can we say $n^2+n+0.25≤n^2+ n$ is a contradiction, therefore the converse is true?

Yes, you cancel the $n^2 + n$ and leave a contradiction. If you don't use contradiction, then you need to take care that you have a two-way implication in each line.

 

[chwala]

Yes, you cancel the $n^2 + n$ and leave a contradiction. If you don't use contradiction, then you need to take care that you have a two-way implication in each line.

perok thanks for your time on this, i will endeavour to be more visible on this forum, cheers bingo

 

[chwala]

this is how my colleague handled part a of the question

 

[chwala]

this is how he did part (c) of the question, i would appreciate alternative ways...