1/a+1/ab=1 this means b+1/ab=1 b+1=ab 1=b(a-1) as a and b are naturals,the only possible pair of is b=1 and a-1=1 shich gives us the only solution i.e a=2 and b=1
My reasoning is that 1/a must be greater than 1/ab to make this work. Also, 1/a must be 1/2 or greater. If a = 1, the equation fails. If a = 2 (the last possible option), b = 1.
1/a+1/ab=1
this means b+1/ab=1
b+1=ab
1=b(a-1)
as a and b are naturals,the only possible pair of is b=1 and a-1=1 shich gives us the only solution i.e a=2 and b=1
A=2, B=1
1/2 + 1/2*1 = 1/2+1/2 = 2
1/a+1/ab=1
b/ab+1/ab=1
b+1=ab
b=2 & a=1
(1/a)+(1/ab)=1
(b/ab)+(1/ab)=1
(1+b)/ab=1
1+b=ab
(1+b)/b=a
(1/b)+1=a
1+b=((1/b)+1)b
(1+b)/b=(1/b)+b
(1/b)+1=(1/b)+b
*1=b*
1+1=a
*a=2*
(1/a)(1 + 1/b) = 1
(1 + b) = ab or a = (1 + b)/b
b = 1 gives a = 2; b > 1 gives a = fraction; b < -1 gives a = fraction, so this is the only solution.
Yes.... you are right.
@@AplusB7No hes not, there are infinity solutions
a=2, b=1 или a=1, b=-2.
a = n
and a is not equal to zero
b = 1/(a-1)
So there infinitely many solutions
How about a = 2 and b = 1? I think thats the only solution.
My reasoning is that 1/a must be greater than 1/ab to make this work. Also, 1/a must be 1/2 or greater. If a = 1, the equation fails. If a = 2 (the last possible option), b = 1.