Why is $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$?

Question

Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$

I can't seem to wrap my head around that.

Answer 1

$$n^2 = \dfrac{2n^2}{2}$$ $$\dfrac{2 n^2}{2} - \dfrac{n^2}{2} = \dfrac{2n^2 - n^2}{2}= \dfrac{n^2}{2}$$

In short, for any $X$: $\;(X = x; X = n^2; \text{ or}\;X = n^{917})\;\text{etc.}:\;$ $${\bf two}\text{ halves of X}\;- \;{\bf one} \text{ half of X}\; = \;{\bf one} \text{ half of X}$$

Answer 2

$$x-\frac x2=\frac x2$$ Take from something its half and you're left with half of it

Answer 3

Did you know that $$x-\frac{x}{2}=\frac{x}{2}$$ whatever $x$ you take?

Answer 4

That equation is saying $$n - \text{ half of } n =\text{ half of } n$$

If you substitute n for integers it's quite easy to see why it works.

$$1 - 1/2 = 1/2$$ $$2 - 2/2 = 2/2$$

Answer 5

Since $\left(1-\dfrac{1}{2}\right)=\dfrac{1}{2}$, then $x\cdot\dfrac{1}{2}=\left(x-\dfrac{x}{2}\right)=\dfrac{x}{2}$.

Answer 6

We know that $n^2=n^2$.

Hence $n^2=\frac{2n^2}{2}$ $$\implies n^2=\frac{n^2}{2}+\frac{n^2}2$$ $$ \implies n^2- \frac{n^2}2=\frac{n^2}2$$ as we wanted.

Answer 7

Subtracting half of something yields half of that thing.

Answer 8

Let $n= \sin x$, so that $n^2 = \sin^{2} x = 1- \cos^{2} x$. Thus $$ n^{2}- \frac{n^{2}}{2} = 1-\cos^{2} x - \frac{1-\cos^{2}x}{2}. $$

Multiplying both sides by $2$ yields

$$ 2n^{2} - n^{2} = 2-2\cos^{2} x - (1-\cos^{2}x) = 2-2\cos^{2} x - 1 +\cos^{2}x = 1 - \cos^{2} x = \sin^{2} x = n^{2}. $$

Now dividing both sides by $2$ gives

$$ n^{2} - \frac{n^{2}}{2} = \frac{n^{2}}{2}. $$

Answer 9

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Answer 10

This reduces to proving that $1-\frac 12 = \frac 12$. In order to do this, expand $1$ into the geometric series $$1=\sum_{i=1}^{\infty}\frac 1{2^i},$$ and divide both sides by $2$ to get $$\frac 12=\sum_{i=1}^{\infty}\frac 1{2^{i+1}}=\sum_{i=2}^{\infty}\frac 1{2^i}.$$ It follows that $$1-\frac 12 = \sum_{i=1}^{\infty}\frac 1{2^i}-\frac 1{2^1}=\sum_{i=2}^{\infty}\frac 1{2^{i}}=\frac 12.$$

Answer 11

We want to prove that $n^2 - \frac{n^2}{2} = \frac{n^2}{2}$. In order to better understand what we want to do, let us simplify this equation a little bit by adding $\frac{n^2}{2}$ to both sides. We get $n^2 = \frac{n^2}{2} + \frac{n^2}{2}$. So in order to solve the problem, we just need to show that $\frac{n^2}{2} + \frac{n^2}{2} = n^2$. In order to show that these two things are equal, let's start with the left hand side and show that it equal to the right hand side. Since we know how to add fractions with the same denominator, we get $$\frac{n^2}{2} + \frac{n^2}{2} = \frac{2n^2}{2} = \frac{2}{2}n^2= n^2$$ Now we're done!

Answer 12

Simple arithmetic: \begin{equation} n^2 - \dfrac{n^2}{2} = n^2(1 - \dfrac{1}{2}) = n^2(\dfrac{1}{2}) = \dfrac{1}{2}n^2 = \dfrac{n^2}{2} \end{equation}

Answer 13

Maybe an ancient greek-style of answer can be fun to read.

Let $A$ be the square with lenght of side $n$. Thererefore the area of $A$ is $n \cdot n = n^2$. Now draw one of the two diagonals. Note that the sum of the areas of the two right triangles you create (call it $B,C$) is equal to the area of $A$, and they have the same area, because they have two equal legs and the right angle between. Consider, for example $B$: the area of it is $\frac{n \cdot n}{2}$. So if you take from $A$ of area $n^2$, $B$ of area $\frac{n^2}{2}$ there is still the other, $C$ of area $\frac{n^2}{2}$.

Answer 14

Actually, if you consider an integer division (so called Euclidian division) the proposition becomes: $$ n^2 - \frac{n^2}{2} = \frac{n^2}{2} + \text{mod}{(n^2,2)} $$

Indeed, by definition of the Euclidian division of the integer $n^2$ by the integer $2$, one has: $$ n^2 = 2\times\frac{n^2}{2} + \text{mod}{(n^2,2)} $$ where $\text{mod}{(a,b)}$ denotes the remainder of the euclidian division of an integer $a$ by an integer $b$.

Now, by (cleverly enough) remarking that the term $2\times \frac{n^2}{2}$ equals to $\frac{n^2}{2} + \frac{n^2}{2}$ one can write: $$ \begin{aligned} n^2 &= 2\times\frac{n^2}{2} + \text{mod}{(n^2,2)} \\ n^2 &= \frac{n^2}{2} + \frac{n^2}{2} + \text{mod}{(n^2,2)} \\ \Leftrightarrow n^2 - \frac{n^2}{2}&= \frac{n^2}{2} + \text{mod}{(n^2,2)} \\ \end{aligned} $$ that ends the proof ;)

Answer 15

We proceed by induction on $n$. The base case for $n = 0$ is $0^2 - \frac{0^2}{2} = 0 =\frac{0^2}{2}$, which follows from the fact that $0$ is the additive identity in $\mathbb{Q}$. Suppose the claim holds for some $n = k$. Now \begin{aligned} (k+1)^2 - \frac{(k+1)^2}{2} &= k^2 + 2k + 1 - \frac{k^2}{2} - \frac{2k}{2} - \frac{1}{2} \\ &= \color{blue}{\left ( k^2 - \frac{k^2}{2} \right )} + k + \frac{1}{2} \\ &= \color{blue}{\frac{k^2}{2}} + k + \frac{1}{2} \quad \color{blue}{\text{(hypothesis)}} \\ &= \frac{k^2 + 2k + 1}{2} \\ &= \frac{(k+1)^2}{2}. \end{aligned} This concludes the inductive step, so our claim holds for all $n \in \mathbb{N}$.