π≠a+b√2 when a and b are both rationals

Question

I'm trying to prove that not all irrational numbers are of the form c=a+b√2 when a,b are rational. I'm thinking if I prove that π does not have this form, them there's an infinity of irrationals which can be written like c=a+bπ with a,b rationals cannot be written in the other form. I'm trying to prove it through contradiction. Therefore I'm supposing π=a+b√2. Then π-b√2=a. Now we have two possibilities. If a≠0 then a is irrational through the properties of irrational numbers which is therefore a contradiction. If a=0 then that means π=b√2. Now, I imagine the next step would lead to showing that b is an irrational number. However I do not know how to proceed to show this is the case. Thanks in advance.