With sum of squared error (SSE), how are larger errors are penalized more than smaller errors?

I read this sentence about the sum of squared error SSE: "The square ensures the error is always positive and larger errors are penalized more than smaller errors". I want to understand what is meant by "larger errors are penalized more than smaller errors", and how to prove that?

• If you square small and bigger values, then the latter will change more then the first one. Plot $x$ vs $x^2$ to convince yourself.
– Tim
Aug 16 '17 at 5:38
• Consider accepting @Tim answer. If the error is $1$ its square is still $1$, but if the error is $100$ its square is $10000$ which is significantly larger. Aug 20 '17 at 19:08
• Let an error be $\epsilon$, so that its size is $|\epsilon|$. Then its square can be factored into two terms, $$\epsilon^2 = \omega \times |\epsilon|,$$ with $\omega = |\epsilon|$. This shows how the square of any number can be considered to be its size multiplied by a weight $\omega \ge 0$. Notice that these weights are directly proportional to the sizes of the errors. That's all there is to the quotation--it is a mathematical triviality, merely involving a reinterpretation of the square of a number.
– whuber
Aug 20 '17 at 21:05
• "The sum of squared error" is not a question. You have to actually ask a question.
– smci
Aug 21 '17 at 0:17

Plot $x$ against $x^2$. Smaller values squared change less, then larger values squared. The steepness of the slope of $x^2$ increases as $|x|$ grows. So if you square small errors, the penalty is smaller, then if you squared large errors. There is nothing to prove in here, it is just how the square function works. 