Math, asked by TheQuantumMan, 6 months ago

a+b=?
* Random letter answer not allowed.
* Wrong answer not allowed​

Answers

Answered by Akansha6414
0

Step-by-step explanation:

Step-by-step explanation: a,b are positive integers. The answer to the problem turns out to be a−b=0 as the only two obvious solutions are a=b=0,2 (and we select two as it can't be zero).

Step-by-step explanation: a,b are positive integers. The answer to the problem turns out to be a−b=0 as the only two obvious solutions are a=b=0,2 (and we select two as it can't be zero).My question is how do we prove this? I was trying

Step-by-step explanation: a,b are positive integers. The answer to the problem turns out to be a−b=0 as the only two obvious solutions are a=b=0,2 (and we select two as it can't be zero).My question is how do we prove this? I was tryinga+b=ab

Step-by-step explanation: a,b are positive integers. The answer to the problem turns out to be a−b=0 as the only two obvious solutions are a=b=0,2 (and we select two as it can't be zero).My question is how do we prove this? I was tryinga+b=abaa−1=b

Step-by-step explanation: a,b are positive integers. The answer to the problem turns out to be a−b=0 as the only two obvious solutions are a=b=0,2 (and we select two as it can't be zero).My question is how do we prove this? I was tryinga+b=abaa−1=bBut I can't proceed by substituting this in a−b and solving the polynomial as I would have to first assume that a−b=0.

Step-by-step explanation: a,b are positive integers. The answer to the problem turns out to be a−b=0 as the only two obvious solutions are a=b=0,2 (and we select two as it can't be zero).My question is how do we prove this? I was tryinga+b=abaa−1=bBut I can't proceed by substituting this in a−b and solving the polynomial as I would have to first assume that a−b=0.How do we prove this? How can we prove that the only value a and b can take is 2 (and hence a−b=0)? Alternatively, how can we prove that if a−b is non-zero, then a and b aren't integers? How can we prove that there are infinitely many solutions if the constraint is shifted to real numbers?

Answered by babitha638
0

a+b= Z

Because, it can be any number, i.e Integer, which includes Rational numbers , whole numbers, and natural numbers.

so, the answer may be a number which is an integer .

You didnt answered my question that I have asked in your answers.

Please tell me

Similar questions