Is there any possibility?

Question

We can write
1 + a^(2) = b^(2)
like 1 + (0.75)^(2) = (1.25)^(2)
Can we write
1 + v^(2) = c^(2) + n
"v","c" & "n" are rational numbers.

Answer 1

1 + v² = c² + n
1 + v² - c² = n
1 + (v + c)(v - c) = n

So a rational n can be found for each v and c (provided v and c are both rational)

e.g.

v = 7
c = 2/3

1 + (7 + 2/3)(7 - 2/3) = n
1 + (23/3)(19/3) = n
1 + 437/9 = n

n = 446/9 which is rational

Note: where 1 + a² = b² works you can add n = 2 to the right hand side. This will leave the formula as 1 + a² = b² + 2. Subtract 1 from each side: a² = b² + 1 which is basically the same equation

Answer 2

No.....what you do to one side of an equation you must do to the other.

But you *can* do the following:

From: 1 + a² = b²

1 + v² = (k + n)²

Where a = v, and b = k + n

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Answer 3

Sure, there would be a bunch of possibilities, starting with your first equation and having 0 for n.

Or, just take two random integers, for v and c and you'll always get another integer for n.

1+v^2 = c^2+n
n = 1+v^2-c^2 = another integer

Answer 4

No.....what you do to one side of an equation you must do to the other.

But you *can* do the following:

From: 1 + a² = b²

1 + v² = (k + n)²

Where a = v, and b = k + n

Answer 5

Ys if it was: 1*(2)+a*(2)=b*(2) | /(2)
1+a=2
v*(2)-c*(2)=n-1
2*(v-c)=n-1
v-c=(n-1)/2
and if all this is 0 you can write like this.

Answer 6

no the no in ur example satisfy that eqn bt its nt true for all rational no.

Answer 7

we can as it was arithmetic operations

Answer 8

no