Vtgyjfy

It is clear that $sqrt 2$ is irrational and $sqrt 3$ is irrational too, and it is quite easy to prove it.

I concluded that the square root of any prime number $sqrt p$ is an irrational number.

I tried to prove it but stopped at one point $sqrt p$

$ a/b$ = $sqrt p $

$ a^2/b^2$ = $p $

$ a^2$ = $p b^2$

$ a^2$ = $p b^2$

Now I need to prove that if $ a^2$ is divisible by $p $ Then $ a$ should be divisible by $p $.

But I do not know how to do that.

Can anyone help me here?

I suppose you should have assumed that $a,b$ are coprims. After you have that $b^2|a^2$ ($b$ divides $a$), then by the Fundamental Theorem of Arithmetic $a,b$ can be expressed as:
$$displaystyle a =prod_{p_i text{ prime}\ alpha_i inmathbb{N}}p_i^{alpha_i}\
displaystyle b =prod_{q_i text{ prime}\ beta_i inmathbb{N}}q_i^{beta_i}$$
obviously $forall i,j : p_ineq q_j$ because $a,b$ are coprimes.

Then:
$$a^2 =prod_{p_i text{ prime}\ alpha_i inmathbb{N}}p_i^{2alpha_i} \
b^2 =prod_{q_i text{ prime}\ beta_i inmathbb{N}}q_i^{2beta_i}$$
but there is a contradiction in the affirmation $b^2|a^2$ because there is not exists a prime in common.

Use the result:- If a prime p divides ab, then p divides a or p divides b. In your question a and b are same.

First of all, the classic case $sqrt 2$ is easy, and somewhat peculiar in that it relies only on the even-/odd-ness of $a$ and $b$ to establish that

$2 mid ab Longrightarrow [2 mid a] vee [2 mid b], tag 1$

of which

$2 mid a^2 Longrightarrow 2 mid a tag 2$

is a special case. For writing

$a = 2m + 1, ; b = 2n + 1, tag 3$

we have

$ab = 4mn + 2m + 2n+ 1 = 2(2mn + m + n) + 1, tag 4$

which shows the product of two odd integers is odd; so $2 mid ab$ if and only if at least one of $a$, $b$ is even, which is what we need to ensure that (1) binds.

Even the case $sqrt 3$ appears to be somewhat more difficult; instead of looking at mere even-/odd-ness, we must consider all cases of possible remainders when $a$ and $b$ are divided by $3$:

$a = 3m + r_1, ; b = 3n + r_2, ; r_1, r_2 in {0, 1, 2 }; tag 5$

then

$ab = (3m + r_1)(3n + r_2)$
$= 9mn + 3(mr_2 + nr_1) + r_1r_2 = 3(3mn + mr_2 + nr_1) + r_1r_2, tag 6$

whence, since $r_1r_2 in {0, 1, 2, 4}$,

$3 mid ab Longleftrightarrow r_1r_2 =0; tag 7$

inspection of (5)-(7) reveals that

$3 mid ab Longleftrightarrow [3 mid a] vee [3 mid b], tag 8$

which is again sufficient to show that $sqrt 3 notin Bbb Q$; though similar to the case $sqrt 2$, here we have to inspect three possible products $r_1r_2 in {1, 2, 4 }$ (assuming $r_1 ne 0 ne r_2$) to ensure that (8) is the case.

The situation grows more complex as $p in Bbb P$ increases beyond $3$; the number of product remainders $r_1r_2$ continues to grow and, as of this point, we have presented no alternative but to examine the manually, one case at a time. There $p - 1$ possible non-zero remainders $r_1$, $r_2$ when we write

$a = pm + r_1, ; b = pn + r_2; tag 9$

thus the number of products to inspect is $p(p -1) / 2$; we needn't consider in detail the cases $r_i = 0$ since then $p$ divides at least one of $a$, $b$.

Fortunately we can invoke some elementary number theory/field theory to address the cases of ever-increasing $p$. Recall that, since $p in Bbb P$, the ring of integers mod $p$, $Bbb Z_p$, is a field, and that the non-zero elements of $Bbb Z_p$ are precisely represented by the set of integers ${1, 2, ldots, p - 1}$; furthermore, in the case of general $p$ we have via (9)

$ab = p^2mn + p(mr_2 + nr_1) + r_1 r_2 = p(pmn + mr_2 + nr_1) + r_1r_2; tag{10}$

thus

$p mid ab Longleftrightarrow p mid r_1r_2; tag{11}$

but

$p mid r_1r_2 Longleftrightarrow r_1r_2 equiv 0 mod p, tag{12}$

that is,

$r_1r_2 = 0 in Bbb Z_p, tag{13}$

which, since $Bbb Z_p$ is a field, we have

$r_1 = 0 in Bbb Z_p ; text{or} ; r_2 = 0 in Bbb Z_p; tag{14}$

that is,

$r_1 equiv 0 mod p ; text{or} ; r_2 equiv 0 mod p, tag{15}$

which, according to (9), is sufficient for

$p mid a ; text{or} ; p mid b. tag{16}$

We have thus proved, for every prime $p$, that

$p mid ab Longleftrightarrow [p mid a] vee [p mid b]; tag{17}$

of course, taking $a = b$ in this statement yields

$p mid a^2 Longleftrightarrow p mid a, tag{18}$

which, as pointed out by our OP asmgx, is sufficient to complete the proof that $sqrt p$ is irrational: from

$a^2 = pb^2 tag{19}$

we have

$p mid a^2 Longrightarrow p mid a Longrightarrow a = pc Longrightarrow p^2c^2 = pb^2 Longrightarrow pc^2 = b^2 Longrightarrow p mid b^2 Longrightarrow p mid b, tag{20}$

so we now have both $p mid a$ and $p mid b$; but this contradicts the usual assumption that

$sqrt p = dfrac{a}{b}, ; gcd(a, b) = 1, tag{21}$

thus completing the demonstration that $sqrt p notin Bbb Q$.

Well, we've more-or-less completed a proof that $p mid a^2$ implies $p mid a$, and used it to show that $sqrt p notin Q$. So in that sense we could end our discussion here. But there is yet another attack on these problems which I want to share before closing, one based on some simple notions from the theory of rings, to wit:

Everything we have done above is very arithmetic oriented; even our invocation of $Bbb Z_p$ is not much more than a way of describing arithmetical ideas and calculations. But follows another proof which is perhaps more in keeping with more modern ideas from abstract algebra.

We need a bit of ideal theory in $Bbb Z$; viz, $Bbb Z$ is a principal ideal domain; this means that every ideal

$I subset Bbb Z tag{22}$

satisfies

$I = (d) = Bbb Z d tag{23}$

for some $d in Bbb Z$.

Now let $p$ be irreducible in $Bbb Z$; then $(p)$ is maximal; for if $J$ is an ideal such that

$(p) subset J subsetneq Bbb Z, tag{24}$

we have for some $j in Bbb Z$,

$J = (j), tag{25}$

whence

$(p) subset (j) Longrightarrow p = cj; tag{26}$

then the irreducibility of $p$ implies that either $c$ or $j$ is a unit in $Bbb Z$; if $j$ is a unit we have $k in Bbb Z$ with

$kj = jk = 1; tag{27}$

but then

$1 = kj in (j), tag{28}$

so then

$z in Bbb Z Longrightarrow z = z cdot 1 in (j) Longrightarrow J = Bbb Z, tag{29}$

contradicting (24); so (27) is impossible; $j$ cannot be a unit. If $c$ is a unit, then

$p = cj Longrightarrow j = c^{-1}p Longrightarrow j in (p) Longrightarrow (j) = (p), tag{30}$

and therefore $(p)$ is maximal.

It follows that $Bbb Z / (p)$ is a field, hence it is an integral domain, hence
$(p)$ is a prime ideal in $Bbb Z$, $p$ is a prime in $Bbb Z$, and thus

$p mid ab Longrightarrow [p mid a] vee [p mid b]; tag{31}$

therefore

$p mid a^2 Longrightarrow p mid a; tag{32}$

now our proof that $sqrt p notin Bbb Q$ goes through, for if

$sqrt p = dfrac{r}{s},; gcd(r, s) = 1, tag{33}$

then

$p = dfrac{r^2}{s^2}, tag{34}$

$ps^2 = r^2 Longrightarrow p mid r^2 Longrightarrow p mid r Longrightarrow r = tp Longrightarrow r^2 = t^2 p^2$
$Longrightarrow ps^2 = p^2 t^2 Longrightarrow s^2 = pt^2 Longrightarrow p mid s^2 Longrightarrow p mid s, tag{35}$

contradicting (33); so once again we have that $sqrt p notin Bbb Q$.

Nota Bene: One of the real issues here is the different ways of defining "prime"; the most elementary is that $p in Bbb P$ if and only if its only divisors, up to signs, are itself $p$ and $1$; in more modern terms, however, we would call such $p$ irreducible; that is, $p$ is irreducible if $p = uv$ implies one of $u$, $v$ is a unit, a divisor of $1$. The second way of defining prime is that $p in Bbb P$ if and only if $p mid ab$ implies $p mid a$ or $p mid b$. In the above we have mostly begun with the assumption that $p$ is irreducible and derived the second property, $p mid ab Longleftrightarrow [p mid a] vee [p mid b]$ as a needed consequence. End of Note.