Question

disprove that a^2>b^2 for all a>b by finding a suitable counter example​

Answer 1

Answer:

Step-by-step explanation:

BRAINLIEST BRAINLIEST BRAINLIEST

Answer 2

Answer:

The given inequality does not hold for negative real numbers

Step-by-step explanation:

The inequality $a^2>b^2$ for all a>b holds only for positive real numbers but it does not hold for negative real numbers

For example,

5 > 4 implies$\:5^2>4^2$

-4 > -5 does not imply$\:(-4)^2>(-5)^2$