Verified answer

For any real number a and b, a > 0 and b > 0, √(a²+b²) = a + b

FALSE

Any numerical example will show this to be false. For example: a = 3, b = 4

√(a²+b²) = √(9+16) = √25 = 5

a + b = 3 + 4 = 7

The only time √(a²+b²) = a + b is when one of a or b (or both) = 0, and the other is non-negative. However, neither a nor b can be = 0 since we are told they are both > 0

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For real numbers a and b, if a³ = b³, then a = b is always TRUE.

a³ = b³

a³ − b³ = 0

(a − b) (a² + ab + b²) = 0

So either (a − b) = 0 or (a² + ab + b²) = 0

If a² + ab + b² = 0, then

a = (−b ± √(b²−4b²)) / 2 = (−b ± √(−3b²)) / 2

Discriminant −3b² is either negative or 0

Since a is real, discriminant −3b² must be 0 ---> b = 0 ----> a = (−0 ± √0) / 2 = 0

a = b = 0

If a² + ab + b² ≠ 0, then

a − b = 0

a = b

Either way, a = b

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Eduardo, you asked Captain Matticus:

In the first case, couldn't I just have √(a²+b²) = (a^(2 x 1/2) + b^(2 x 1/2)) = a + b

Answer is ABSOLUTELY NOT.

Exponentiation is distributive over multiplication, but NOT over addition:

For a, b > 0:

(a² + b²)^(1/2) ≠ (a²)^(1/2) + (b²)^(1/2)

but

(a² * b²)^(1/2) = (a²)^(1/2) * (b²)^(1/2) = a^(2*1/2) * b^(2*1/2) = a * b

Check out rules of exponentiation.

You WILL find: (x * y)^n = x^n * y^n

You will NOT find: (x + y)^n = x^n + y^n -----> FALSE

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Here's another way to show that √(a²+b²) ≠ a + b

For a > 0, b > 0

(a + b)² = (a + b) (a + b)

(a + b)² = a (a + b) + b (a + b) -----> using distributive property

(a + b)² = a*a + a*b + b*a + b*b -----> using distributive property again

(a + b)² = a² + 2ab + b²

Now take square root of both sides:

a + b = √(a² + 2ab + b²) ≠ √(a²+b²)

Therefore, a + b ≠ √(a²+b²)

√(a²+b²) = a + b

(a²+b²) = (a + b)^2=a²+b²+2a*b

so it can't be true,

if any only if a or b equals zero

(a^2) + (b^2) = (a^2) + 2ab + (b^2);

0 = ab;

For any positive real number a and b, sqrt(a^2 + b^2) = a + b. FALSE STATEMENT.

Let a = b = 1. Then sqrt(1^2 + 1^2) = 1 + 1, meaning sqrt(2) = 2. Doesn't work. Here's what must happen:

sqrt(a^2 + b^2) = a + b

Square both sides:

a^2 + b^2 = a^2 + 2ab + b^2

0 = 2ab. Stop here.

That means at least one of a or b MUST be 0 for the original equation to hold. So the original statement is ALWAYS false.

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If a^3 = b^3, a = b. Well...

a^3 = b^3

a^3 - b^3 = 0

(a - b)(a^2 + ab + b^2) = 0

a^2 + ab + b^2 = 0 if BOTH a and b are 0, so a = b here (there are imaginary solutions, but disregard those).

a - b = 0 ---> a = b.

So, if a^3 = b^3, a = b. ALWAYS TRUE.

true, logic

~Aizen

for any real number a and b, a > 0 and b > 0, √(a²+b²) = a + b

FALSE

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If a³ = b³ , a = b

TRUE

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sqrt(a^2 + b^2) = a + b

Square both sides

a^2 + b^2 = (a + b)^2

a^2 + b^2 = a^2 + 2ab + b^2

0 = 2ab

0 = a * b

Either a = 0 , b = 0 or both a and b equal 0. However, since a > 0 and b > 0, then those cannot be the case. It's false.

a^3 = b^3

a^3 - b^3 = 0

(a - b) * (a^2 + ab + b^2) = 0

a - b = 0

a = b

a^2 + ab + b^2 = 0

a = (-b +/- sqrt(b^2 - 4b^2)) / 2

a = (-b +/- b * sqrt(-3)) / 2

a = -b * (1 +/- i * sqrt(3)) / 2

a and b are real, so a = -b * (1 +/- i * sqrt(3)) / 2 is excluded

If a^3 = b^3, then a = b is always true.