Sam's Mistake

Algebra Level 4

Sam is excited to learn about the distributive law, and thinks that it applies to every possible operation. As such, he claims that $\sqrt{a^2 + b^2} = \sqrt{a^2} + \sqrt{b^2}$

How many ordered pairs of integers $(a, b$ ) are there, such that $-10≤a≤10, -10≤b≤10$ and $\sqrt{a^2 + b^2} = \sqrt{a^2} + \sqrt{b^2}?$

The answer is 41.

1 solution

Chaithanya Hr
Nov 18, 2014

Squaring the equation on both sides, we get $a^2$ + $b^2$ = $a^2$ + $b^2$ +2|ab|.

After cancelling $a^2$ + $b^2$ on both sides, we get 2|ab|=0, which is true only when one of the numbers i.e either 'a' or 'b' is 0. Therefore the ordered pairs would be

(0,0), (0,1), (0,2), (0,3),..........(0,10), (0,-1), (0,-2), (0,-3),......(0,-10), (1,0), (2,0), (3,0),...........(10,0), ( -1,0),(-2,0), (-3,0),......(-10,0). That gives us 41 ordered pairs.

Sam's Mistake

The answer is 41.

1 solution

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