why if x in 1/n power >(<) y in 1/m power then x in c/n power >(<) y in c/m power? As you might guess this is one more stupid question from non-matematician, and you are right. I found this exercise in "Algebra and trigonometry book":
$7^{1/2}$ or $4^{1/4}$. After some googling I found out that to solve this I should take both numbers to 4 power and then compare the result. The problem is that its no so obvious for me...
lets say I found that 49 is bigger number than 4, but why if
$(7^{1/2})^4 > (4^{1/4})^4$ the > sign is still the same in case of
$7^{1/2} > 4^{1/4}$ - we multiplying both sides of inequality by different numbers. 
It must sounds very silly, but math for me is more like interesting stuff I occasionally omitted in school/university solving excercises without understanding them and now I go for it.
thx
 A: Suppose you have an inequality $$a>b$$ Now you square both parts: $$a^2>b^2$$ This is equivalent to $$aa>bb$$ You multiply both parts by different numbers, but if $a>b$, then you increase $a$ more than $b$. But this means that the inequality remains true.
The same holds further for greater powers.
A: It is because the fourth power function (or any positive power) is increasing.  So $(7^{1/2})^4=7^{1/2}\cdot (7^{1/2})^3\gt 7^{1/2}\cdot (4^{1/4})^3\gt 4^{1/4} \cdot (4^{1/4})^3$.  If $a,b,c,d$ are all greater than zero, $a \gt b, c \gt d$, then $ab \gt cd$
A: I believe your question is basically: "Why is it true that if $x^4 > y^4$ then $x > y$?".
Is may, or it may not, be clear to you that the implication in the other direction is true: If $x > y$ then $x^4 > y^4$. I hope at least your intuition says it is true. If you want proof, just notice that $x^4 = xxxx > xxxy > xxyy > xyyy > yyyy = y^4$.
Now, once you are acquainted with this fact, try reasoning backwards. Suppose you know that $x^4 > y^4$. Then, either $x > y$, and we are happy, or $x < y$, or $x = y$. If it was true that $x = y$, then it would follow that $x^4 = y^4$, and we know it's not the case. Likewise, if $x < y$, then $x^4 < y^4$, which again is not the case. The only option left is $x > y$, which is what we were after.
