Metamath Proof Explorer


Theorem sqrt2irr0

Description: The square root of 2 is an irrational number. (Contributed by AV, 23-Dec-2022)

Ref Expression
Assertion sqrt2irr0
|- ( sqrt ` 2 ) e. ( RR \ QQ )

Proof

Step Hyp Ref Expression
1 sqrt2irr
 |-  ( sqrt ` 2 ) e/ QQ
2 sqrt2re
 |-  ( sqrt ` 2 ) e. RR
3 2 a1i
 |-  ( ( sqrt ` 2 ) e/ QQ -> ( sqrt ` 2 ) e. RR )
4 df-nel
 |-  ( ( sqrt ` 2 ) e/ QQ <-> -. ( sqrt ` 2 ) e. QQ )
5 4 biimpi
 |-  ( ( sqrt ` 2 ) e/ QQ -> -. ( sqrt ` 2 ) e. QQ )
6 3 5 eldifd
 |-  ( ( sqrt ` 2 ) e/ QQ -> ( sqrt ` 2 ) e. ( RR \ QQ ) )
7 1 6 ax-mp
 |-  ( sqrt ` 2 ) e. ( RR \ QQ )