<PRE><B>Question 706338</B>
Rational numbers can be expressed as p/q where both p and q are integers and q is not 0.&lt;P&gt;
1 can be expressed as 1/1.  It is rational.&lt;P&gt;
{{{sqrt(7)}}} is not a rational number.  There is no way to express it as p/q - see proof at the end.&lt;P&gt;
9 is rational = 9/1 (any integer is rational).  0 is rational (0/1).&lt;P&gt;
By definition, 6/7 is rational (p=6 and q=7)&lt;P&gt;
{{{sqrt(16)}}} = 4 or -4, so it is rational (4/1 or -4/1).&lt;P&gt;
That line over the 4 means the 4 repeats forever...it is a non terminating decimal that repeats.  Non terminating, repeating decimals are rational.&lt;P&gt;
It&#39;s easy to find the fraction:&lt;P&gt;
A:  x = .4(repeating)  &lt;P&gt;
B: 10x = 4.4(repeating) (multiplied A by 10.)&lt;P&gt;
Subtract:  B-A = 9x = 4  &lt;P&gt;
Solve for x...x=4/9.&lt;P&gt;
.24 is a terminating decimal.  Terminating decimals are rational.  .24 is 24/100 or simplified 6/25.&lt;P&gt;
The only number from B that isn&#39;t a rational number is {{{sqrt(7)}}}.
&lt;P&gt;Hope the solution helped.  Sometimes you need more than a solution.  Contact fcabanski@hotmail.com for online, private tutoring, or personalized problem solving (quick for groups of problems.)&lt;P&gt;


Proof by contradiction: suppose that root 7 (I&#39;ll write sqrt(7)) is a rational number, then we can write sqrt(7)=a/b where a and b are integers in their lowest form (ie they are fully cancelled). Then square both sides, you get 7=(a^2)/(b^2) rearranging gives (a^2)=7(b^2). Now consider the prime factors of a and b. Their squares have an even number of prime factors (eg. every prime factor of a is there twice in a squared). So a^2 and b^2 have an even number of prime factors. But 7(b^2) then has an odd number of prime factors. But a^2 can&#39;t have an odd and an even number of prime factors by unique factorisation. Contradiction X So root 7 is irrational.</PRE>