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

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  • 1 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.

  • 1 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.

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