Why sometimes a number is greater than its square?

What's 2×2? It's 4.  What's 3×3? It's 9  What's 4×4? It's 16  What's 2.5×2.5? It's 6.25  What's  2√×2√ ?It's 2  What's −4×−4? It's 16  And so on

You can clearly observe that for a certain number, the result obtained by multiplying that certain number with itself is always greater than that certain number. Is it really though?  No, it's not always. Here let me give you an example,   What's  0.5×0.5 ?  It's  0.25     Looking at it graphically,  Below is the graph for  f(x)=x2   In the interval  (0,1)we can see that the  f(x) is less than x .        So clearly the result obtained by multiplying that certain number with itself is 𝑛𝑜𝑡  always greater than that certain number.  Or Simply,  The square of a number is not always greater than the number. This happens for numbers greater than  0 and less than  1.   To understand why this is the case, we first look what kind of numbers are between 0 and 1 .  There are numbers whose denominator is greater than the numerator.   So when we square such numbers, the denominator will still be greater than the numerator and as a result, the squared number will be less than 1.    35×35=925,       9<25     One thing which we can also notice is that for the squared number, the difference between the numerator and denominator (denominator > numerator) is greater than in the original number  in their lowest possible fraction  This implies that the squared number's ratio is less than the original number, which also answers why the squared number is less than the original number.    On the other hand, for numbers greater than 1, the numerator is greater than the denominator.    So when we square such numbers, the numerator will still be greater than the denominator and as a result, the squared number will be always greater than 1. Also the difference between the numerator and denominator (denominator < numerator) is greater than in the original number  in their lowest possible fraction  This implies that the squared number ratio is more than the original number, which also answers why the squared number is greater than the original number.

 

Now answering the question why is the square of numbers between  0 and 1  less than the number itself in a more typical mathematical way.

 

Let's suppose  ab is a number where  a<b  and   ab   lies between 0 and 1 as   discussed above. Note that  a,b ∈ R(Real number)  And also, let  ab in their lowest form which means that they don't have common factors except 1.  For eg: 8 and 9 and excludes the possibility of -8 and -9.       As  0<   ab<1    ⇒a<b⇒a−b<0⇒a−b=k    ……(i)      where k is some negative number  Now let's square  ab i.e.   a2b2     As,0<a2b2<1     Wait a minute, we don’t know that. As this is the 'typical mathematical way' so we have to prove this also.  So let's proceed,  0<a2b2<1⇒a2<b2⇒a<b       which is true.    Now, why not do for,  a2b2>1     So let's proceed,   a2b2>1⇒a2>b2⇒a>b     which is a contradiction as we are doing this for  a<b.     Well, I hear some sad equality case noise too, so why not do it for equality case too,  a2b2=1   So let's proceed,    a2b2=1⇒a2=b2⇒a=b     which is a contradiction as we are doing this for  a<b.     So now that we have proved only   0<a2b2<1   is true    0<a2b2<1⇒a2<b2⇒a2−b2<0⇒(a−b)(a+b)<0⇒(a−b)(a+b)=p   .….….(ii)               where  p is some negative number.     Now from …..(i) and ……(ii)  (a−b)(a+b)=p   k(a+b)=p     Now as  a+b>0                          remember if we add two positive real numbers we will always get a positive real number.    So, we can conclude that  k>p   How?  As we needed to multiply k with some positive number to make it equal to  p  which implies that   k   is less than  p.  But as   k  and  p  are negative numbers so  k  will be instead greater than  p.   For eg:  −5×3=−15⇒−5≻−15.   This proves that,  a−b>a2−b2  Thus the ratio will always be smaller for the squared number as the difference is greater. So, the squared number will be less.   Also, taking the other route ab>a2b2.   ab−1>a2b2−1     ab−bb>a2b2−b2b2     a−bb>a2−b2b2     a−bb(a−b)<a2−b2b2(a−b)              we divided it by a negative so the sign will be changed.    1b<a2−b2b2(a−b)     b<a2−b2a−b       b<a+b        which is true  Thus,  ab>a2b2.     Now let's do it another way,    Let us suppose  0<a<1            ∀a∈R       Taking the  2nd and 3rd  term,  a<1   0<1−a …….(i)    We know,  if a<b   and c>0    then,  ac<bc            So, multiplying both sides of …….(i) by a   0<(1−a)a   0<a−a2   a2<a     Q.E.D

 -Aakash Gurung