Sun`s secret number

Today i`ve found an interesting test:

1. class SCJPPrepration1 {  
2.   public static void main(String[] args) {  
3.      Integer i = Integer.valueOf(10);  
4.      Integer j = Integer.valueOf(10);  
5.   
6.        
7.      System.out.println(i==j);  
8.        
9.      int magicSunOfficialSecretNumber =143;  
10.        
11.      Integer k = Integer.valueOf(magicSunOfficialSecretNumber);  
12.      Integer l = Integer.valueOf(magicSunOfficialSecretNumber);  
13.   
14.        
15.      System.out.println(k==l);  
16.   
17.   }  
18. }     

Never thought before that answer will be : true and FALSE. 

Important thing  that need to remember: 

Whenever boxing is applied , only one wrapper object exists in the program for primitive values of( boolean,byte,char and int or short):
1).boolean values true or false,
2).a byte,
3).a char in range \u0000 to \u007f,
4),and an int or short value in the range -128 and 127.

so for eg if a and b refer to two wrapper objects that box same value which is among one of those mentioned above ,then a==b is always true,i.e object and reference equality will give same result..
Let us take an example,

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1.   Byte b1=(byte)10;   
2.   Byte b2=(byte)10;  
3.   System.out.println(b1==b2);                //true  
4.   System.out.println(b1.equals(b2));        //true  

but in case if we violate one of the above 4 points ,for eg here let us take point 4,
and do

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1.       Integer k=143;  
2.            Integer l=143;  
3.            System.out.println(k==l);                //false because after boxing the range of int is bet[-128,127]  
4.            System.out.println(k.equals(l));        //true as objects have same value  

Thus this proves the result.