Question

√2 is irrational
Prove 5+3√2 as irrational
Answer with steps.

Answer 1

Let's suppose √2 is a rational number.
Then we can write it √2 = a/b where a, b
are whole numbers, b not zero.
We additionally assume that this a/b is
simplified to lowest terms, since that can
obviously be done with any fraction.
Notice that in order for a/b to be in
simplest terms, both of a and b cannot be
even. One or both must be odd. Otherwise,
we could simplify a/b further.
From the equality √2 = a/b it follows that
2 = a 2/ b 2, or a2 = 2 · b2 . So the square
of a is an even number since it is two
times something.
From this we know that a itself is also an
even number. Why? Because it can't be
odd; if a itself was odd, then a · a would
be odd too. Odd number times odd number
is always odd. Check it if you don't believe
me!
Okay, if a itself is an even number, then a
is 2 times some other whole number. In
symbols, a = 2k where k is this other
number. We don't need to know what k is;
it won't matter. Soon comes the
contradiction.
If we substitute a = 2k into the original
equation 2 = a 2 /b 2 , this is what we get:
2 = (2k) 2 /b 2
2 = 4k 2 /b 2
2*b 2 = 4k 2
b 2 = 2k 2
This means that b 2 is even, from which
follows again that b itself is even. And
that is a contradiction!!!
WHY is that a contradiction? Because we
started the whole process assuming that
a/b was simplified to lowest terms, and
now it turns out that a and b both would
be even. We ended at a contradiction; thus
our original assumption (that √2 is
rational) is not correct. Therefore √2
cannot be rational.

Answer 2

hope this help you...........