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# Find an example, by finding values for a and b, to show that √a^2+b^2 ≠ a+b equation?

### 2 Answers

- husoskiLv 71 month ago
It's easier to find out when they *are* equal. I assume you meant √(a² + b²) on the left. If so, and if the two sides of that inequality are actually equal, then their squares are also equal:

a² + b² = (a + b)²

= a² + 2ab + b²

Subtract a² + b² and turn around to get:

2ab = 0

That's only true if and only if at least one of a, b is 0.

So the original values √(a² + b²) and (a + b) are unequal if and only if both a and b are nonzero.

So, just try any two nonzero numbers for a and b and you'll have your counterexample, such as a=3, b=4. √(3² + 4²) = √(9 + 16) = √25 = 5, but 3+4 = 7.

- PuzzlingLv 71 month ago
People sometimes make the mistake of thinking that √(a² + b²) = √a² + √b² = a + b. But there are mistakes in that thinking. We can show that isn't a valid equality by finding a counterexample.

Let's pick a couple values. The following numbers work well:

a = 3

b = 4

Calculate the left hand side:

√(a² + b²) = √(3² + 4²) = √(9 + 16) = √25 = 5

Calculate the right hand side:

a + b = 3 + 4 = 7

So clearly that is not a valid equality.

UPDATE:

Here's one more pair I picked.

a = 1

b = 2

√(1² + 2²) = √5

1 + 2 = 3

Clearly √5 ≠ 3

You should pick a couple values of your own to verify you know what to do.