Proof by Contradiction
Assume: #1+sqrt{2}# is a rational number.
#=># There exist integers #m# and #n# such that #1+sqrt{2}=m/n#.
#=>sqrt{2}=m/n-1={m-n}/n#,
which means that #sqrt{2}# is a rational number.
#=># There exist integers #a# and #b# such that #sqrt{2}={a}/{b}#,
which is reduced to lowest terms.
#=> (sqrt{2})^2=(a/b)^2 => 2=a^2/b^2 => 2b^2=a^2#
#=> a^2# is even #=> a# is even
#=># There exists an integer #k# such that #a=2k#.
#=> 2b^2=(2k)^2 => 2b^2=4k^2 => b^2=2k^2#
#=> b^2# is even #=> b# is even
Since both #a# and #b# are even, #a/b# is NOT in lowest terms, which is a contradiction.
Hence, #1+sqrt{2}# is an irrational number.
I hope that this was helpful.
