Question

Prove that:-
$1.{(a + b)}^{0} = 1$
$2. {(a - b)}^{0} = ?$
✅Explanation Required!✅
​

Answer 1

Answer:

Step-by-step explanation:

ANSWER:-

We know that:-

Anything to the power zero is one. It does not depend which or whatever symbol it has.

Derivation, to show anything equal to zero is one:-

For this, let us take an exponent $x$ with power $y$ $\implies x^y$._____(1)

Another exponent , let also be $x^y$._________(2)

Now, dividing (1) by (2) or (2) by (1) [Both are the same, so it doesn't matter],

$\frac{x^y}{x^y}$, in this step, if we cancel it off we get 1.

$\boxed{\sf{\frac{x^y}{x^y}=1}}$ _______________(3)

Now, dividing it manually, we get

$\boxed{\sf{x^{(y-y)} \implies x^0 }}$______________(4)

Now, comparing the results of (3) and (4), we get

$\sf{x^0 = 1}$

Hence Proved!

So, it does not matter, whether the base is negative or positive, whether it is an identity or not, Anything raise to the power zero is one.

So,

$\sf{(a-b)^0 = 1}$.

Another Thing to Note:-

• We know that exponent shows us how much time the number is multiplied.
• Another exceptional case is any number with the power 1.
• Here, if any number has power one, it remains the same

Remember:-

• Identities powers' lowest base is 2, that means squaring is the first phase which changes the number.
• These power also help to make Quadratic Equations.

Answer 2

Answer:

Given:

Prove that:-

→ ${\sf{\boxed{\sf{1.(a + b)^{0} = 1}}}}$

→ ${\sf{\boxed{\sf{2.(a - b)^{0} = ?}}}}$

Find:

→ Find the value for the second one.

Note:

→ Anything which as the power raised zero would be only 1.

Required-Answer:

→ ${\sf{\boxed{\sf{2.(a - b)^{0} = 1}}}}$

Proof:

→ If your divide any number with the same number itself you get the number 1.

Example (1):

→ ${\sf{\boxed{\sf{\dfrac{x^{1}}{y^{1}} }}}}$

→ ${\sf{\boxed{\sf{1 ÷ 1 = 1}}}}$

Example (2):

→ ${\sf{\boxed{\sf{\dfrac{m - n^{0}}{m - n^{0}} }}}}$

→ ${\sf{\boxed{\sf{1}}}}$

Similarly,

→ ${\sf{\boxed{\sf{(a - b)^{0}= 1}}}}$

HENCE PROVED!!!