If the operation $\&$ has the following properties: $x\&0=0\forall x\in \mathbb{R}$ and $x\&(y+1)=x\&y+(x-y)\forall x,y\in \mathbb{R}$

Question

If the operation $\&$ has the following properties: $x\&0=0\forall x\in \mathbb{R}$ and $x\&(y+1)=x\&y+(x-y)\forall x,y\in \mathbb{R}$ Find the value of $2014\& 10$

I tried to solve this question as follows:

$x\&-1+x+1=0$

$x\&-1=-(x+1)$

Using $x=0$, gives you nothing. I know that the answer is $20095$, but I don't know how to work it out. Could you please explain to me how to solve this question?

Answer 1

$x\&(y+1)=x\&y+(x-y)$

$... = x\&(y-1) + (x-y) + (x-(y-1))= x\&(y-1) + 2x -2y + 1$

$=x\&(y-2) + 2x-2y + 1 -(x-(y-2))= x\&(y-2) + 3x-3y + 3$.

Claim: $x\&(y+1) = x\&(y-k) + (k+1)(x-y) + \sum_{j=0}^k j$.

Proof: by induction: If $x\&(y-k) + (k+1)(x-y) + \sum_{j=0}^k j$ the $x\&(y-k-1)+(x-(y-k-1) + (k+1)(x-y) + \sum_{j=0}^k j=x\&(y-k-1)+(k+2)(x-y) +\sum_{j=0}^{k+1}j$

so

So let $x=2014$ and $y = 9$ then

$x\&(y+1) = x\&(y-y) + (y+1)(x-y) + \sum_{j=0}^y j = $

$2014\&10 = 2014\&0 + 10(2014-9) +\sum_{j=0}^9 j = $

$0 + 10(2005) + \frac {9\cdot 10}2 =$

$20050 + 45 = 20095$.

.....

If we want we can express $x\&w$ directly as:

Claim 2: Set $k = y$ then

$x\&(y+1) = x\&(y-y) + (y+1)(x-y) + \sum_{j=0}^y j=$

$x\&0 + (y+1)(x-y) +\frac {y(y+1)}2=$

$(y+1)[(x-y)+\frac y2]=$

$(y+1)[(x-\frac y2)]$

If we let $w = y+1$ we have

$x\&w = w(x-\frac {w-1}2)$

So $2014\&10 = 10(2014 - \frac 92)=20140 - 45= 20095$.

Answer 2

Hint: Show by induction on $k$ that $$x\& (y+k)=x\& y +kx-\sum_{j=0}^{k-1}(y+j)$$

and then use that with $y=0$.