Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm working on a computation which depends on the idea that given two natural numbers $x$ and $y$ where $y > x$, the product $x(y - x)$ will always be greater than $y$.

Is there a proof of this ? My elementary math is a bit rusty.

The simple evaluation gives : $xy - x^2$

I can't seem to formalize this relation with respect to $y$. Could somebody give me a refresher on the proof strategies for such a problem ?

*EDIT: * Apologies. I got lost in writing it here. It's actually $x (y - x + 1)$. So basically given two numbers this would result in the value above plus a summation series that is solvable using $x(x+1)/2$. Does that make more sense ? (P.S: thank you for the quick response.)

share|cite|improve this question
Let $y = 2$ and $x = 1$. Then $y > x$, but $x (y - x) = 1 \cdot (2-1) = 1$ and $y = 2$. Your statement is wrong. – k.stm Jan 20 '13 at 10:53
up vote 7 down vote accepted

$$x(y-x+1)>y\iff xy-y>x(x-1)$$ $$\iff y(x-1)>x(x-1)\iff (y-x)(x-1)>0$$

which will be true if $(y>x$ and $x>1)$ or if $(y<x$ and $x<1)$

i.e., if $y>x>1$ or $y<x<1$

share|cite|improve this answer
Yippee... I just saved a variable cause of this. Thanks. – Saad Farooq Jan 20 '13 at 11:27
@SaadFarooq, my pleasure. – lab bhattacharjee Jan 20 '13 at 11:28

Try $y=x+1$, then the product is $x$, which is not greater than $y$. Hence you try to prove something wrong.

share|cite|improve this answer

Try $x=2$ and $y=3$. Then $x(y-x)=2<y$.

share|cite|improve this answer

A different way to look at the problem: $$x(y-x+1)=-x^2+(y+1)x$$ with fixed $y$ is a quadratic function in $x$. The coefficient by $x^2$ is negative, so the maximal value of it will be at $x=\frac{y+1}{2}$, which is smaller than $y$ if $y>1$, and the value is equal to $\frac{(y+1)^2}{4}$, which is clearly larger than $y$ for $y\geq 3$, for example.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.