Proof square root 2 irrational

realjoejanuary

New member
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?

Cubist

Senior Member
I've tried to make the op more readable by replacing (squared) with ²...

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a² / b² = 2
a² = 2*b²

a² is an even integer
a is an even integer

Back to the original equation

a² = 2*b²
a² / 2 = b²
b² = a² / 2

If b² = some number / 2 then
b² is an even integer
so b is an even integer

Since a & b are both even integers a/b cannot be the square root of 2. (reducible 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 b² can be derived by dividing a² by 2 making b² an even number?

pka

Elite Member
Here is as much more general proof.
If [imath]n\in\mathbb{Z}^+[/imath] and is not a square then [imath]\bf\sqrt{n}[/imath] is irrational.
If [imath]\sqrt{n}[/imath] were rational then [imath]\exists\{a,b\}\subseteq\mathbb{Z}^+[/imath] such [imath]\sqrt{n}=\dfrac{a}{b}[/imath].
Then [imath]b\sqrt{n}=a[/imath] is a positive integer.
Thus consider the set [imath]\mathcal{T}=\left\{t\in\mathbb{Z}^+| t\sqrt{n}\in\mathbb{Z}\right\}[/imath]
Thus because [imath]b\in\mathcal{T}\ne\empty[/imath] has a least element [imath]K[/imath].
If [imath]H=\left\lfloor {\sqrt n } \right\rfloor [/imath] then [imath]H\in\mathbb{Z}^+~\&~0<\sqrt{n}-H<1[/imath] can you explain?
But then [imath]K\sqrt{n}-KH<K[/imath] is that a contradiction to thr minimality of [imath]K~?[/imath]

[imath][/imath][imath][/imath][imath][/imath][imath][/imath]

mmm4444bot

Super Moderator
OP said:
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 b² can be derived by dividing a² by 2 making b² an even number?
This is the question posed.

[imath]\;[/imath]

Cubist

Senior Member
The following logic is flawed...
If b² = some number / 2 then
b² is an even integer
so b is an even integer
...look at what happens if "some number" is 50...

b² = 50/2 = 25 then b² is not an even integer (but your proof states that it is)

Steven G

Elite Member
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?
Your proof is complete nonsense mainly because you never stated that a and b are integers. That fact is what you based your whole "proof" on!

You stated If b(squared) = some number / 2 then b( squared) is an even integer
Some number over 2? You meant to say an integer over 2!

Steven G

Elite Member
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?
You need stop saying number when you mean integer.
Is 7 a multiple of 5? Yes 7=(7/5)*5

Given any two numbers (not including 0), a and b, you can write a as a multiple of b.
a=(a/b)b