realjoejanuary
New member
- Joined
- Sep 22, 2022
- Messages
- 1
a(squared) / b ( squared) = 2
a( squared) = 2* b(squared)
a(squared) is an even integer
a is an even integer
Back to the original equation
a( squared) = 2*b(squared)
a(squared) / 2 = b(squared)
b(squared) = a(squared) / 2
If b(squared) = some number / 2 then
b( squared) is an even integet
so b is an even integer
Since a & b are both even integers a/b
cannot b the square root of 2. ( reducable to
some other number which will put us through same process)
YET
The proofs I've looked at go through extra steps after showing a is a multiple of some number to show b is also a multiple of some number. Isn't it enough to show that b(squared) can be derived by dividing a( squared) by 2 making b(squared) an even number?
a( squared) = 2* b(squared)
a(squared) is an even integer
a is an even integer
Back to the original equation
a( squared) = 2*b(squared)
a(squared) / 2 = b(squared)
b(squared) = a(squared) / 2
If b(squared) = some number / 2 then
b( squared) is an even integet
so b is an even integer
Since a & b are both even integers a/b
cannot b the square root of 2. ( reducable to
some other number which will put us through same process)
YET
The proofs I've looked at go through extra steps after showing a is a multiple of some number to show b is also a multiple of some number. Isn't it enough to show that b(squared) can be derived by dividing a( squared) by 2 making b(squared) an even number?