Pre-requisite: Number System, Addition and Subtraction.
Target Audience: Standard V to Standard VII.
In Syllabus – Standard V.
Revision – Standard VI & VII.
Symbol
I
V
X
L
C
D
M
Decimal Value
One (1)
Five (5)
Ten (10)
Fifty (50)
Hundred (100)
Five Hundred (500)
Thousand (1000)
Rules to decipher roman numerals:
1.
Repetition of alphabets means addition.
II = 1 + 1 = 2
III = 1+1+1 = 3
XX = 10 + 10 = 20
XXX = 10+10+10 = 30
2.
If a roman alphabet appears to the LHS of another roman alphabet and decimal value of LHS
Roman alphabet is lesser than decimal value of RHS Roman alphabet, the decimal value of LHS is subtracted from decimal value of RHS. If LHS has repetitions, the according to the rule 1, repetitions are added first and then subtracted from decimal value of RHS.
IV = 5 – 1 = 4
XL = 50 -10 = 40
X’s decimal value is 10 and L’s decimal value is 50. Since LHS’s (X) decimal value (10) is lesser than RHS’s (50) decimal value (50), so the resultant decimal value is 50-10=40
.
3.
If a roman alphabet appears to the LHS of another roman alphabet and the decimal value of
LHS Roman alphabet is greater than the decimal value of RHS Roman alphabet, the decimal values of LHS and RHS are added to get the resultant value. If alphabets are repeated then they are added first according to the rule 1.

VI = 5+1 = 6
VIII = 5+1+1+1 = 5+3 = 8
XXXVII = 10 + 10 + 10 + 5 + 1 + 1 = 30 + 5 + 2 = 30 + 7 = 37
XVIII = 10 + 5 + 1 + 1 + 1 = 10 + 5 +3 = 10 + 8 = 18
In the last example, X (10) > V (5) > I (1), the result is addition of respective decimal values. Since
‘I’ is repeated thrice, the value is calculated first.
4.
If more than 2 roman alphabets appear in an arithmetic and the decimal values are such that, in
ABC, A > B and B < C, first since B is less than C and appears to the left, going by rule 2 B’s decimal value is subtracted from C’s decimal value. Then this temporary result is added to A’s decimal value.
XXXIX = 10 + 10 + 10 + (10 – 1) = 10 + 10 +10 + 9 = 39
LXIV = 50 + 10 + (5 – 1) = 50 + 10 + 4 = 64
L (50) > X (10), I (1) < V (5), therefore first IV will be calculated (5 – 1 = 4) and then added to L +
X.
Rules to construct roman numerals:
1.
Pick the largest Roman alphabet lower than the value of the number to convert to Roman numeral. Repeat the Roman alphabet until the sum adds up to the number or the sum not exceeding the number. If the sum is not equal to the number and adding another currently selected alphabet will exceed the number, chose the next largest Roman alphabet smaller than the current Roman alphabet. Make the chosen alphabet as the current Roman alphabet and repeat this procedure until the alphabets add up to give you the desired no.
3 – 1 + 1 + 1 = III
The next largest Roman alphabet smaller than 3 is I. Therefore ‘I’ is chosen and repeated 3 times to represent 3.
8 – 5 + 1 + 1 + 1 = VIII
The next largest Roman alphabet lesser than the value of ‘8’ is V. So V is chosen and used once.
Repeating V again will make the sum as 10, so V is not repeated. Instead next smallest Roman alphabet from the table (‘I’) is chosen and added to V. ‘I’ is repeated thrice to give a sum of 8.

26 – 10 + 10 + 5 + 1 = XXVI
The next smallest Roman alphabet lesser than 26 is X (10). So X is chosen.
# Numeral
1. X
2. XX
3. XXX
4.
5.
6
XXV
XXVV
XXVI
Value
10
10+10 = 20
10+10+10 = 30
10 + 10 + 5 = 25
10 + 10 + 5 + 5 =
30
10 + 10 + 5 + 1 =
26 compare
10 < 26 (ok)
20 < 26 (ok)
30 > 26 (not ok, choose next largest alphabet lesser than X). Revert to step 2.
25 < 26 (ok)
30 > 26 (not ok, choose next largest alphabet lesser than X). Revert to step 4.
26 = 26 (End here)
2.
Any roman alphabet is never repeated more than 3 times consecutive. Instead, the next biggest
Roman alphabet is picked and a subtraction is performed.
4 = 1 + 1 + 1 + 1 = IIII by rule 1(wrong)
4 = 5 – 1 = IV (right)
4 is addition of four ones but 4 ones cannot be written consecutive. So next highest roman alphabet (V) is chosen and a subtraction is performed, hence the result is IV.
9 = 5 + 1 + 1 + 1+ 1 = VIIII by first rule, but quadruple repetition not allowed. Instead of ‘ V’ selecting next biggest alphabet (X) and subtracting 1 (I).
9 = 10 – 1 = IX. (right)
3.
Only powers of 10 (1, 10, 100…I, X and C respectively) can be subtracted. That means non power of 10 can be in subtraction, but cannot assume the negative sign. (VXX = 15 is wrongs as it translates to (10-5) + 10, non power of 10 has negative sign here. But 4 = 5 -1 = IV is valid here as the alphabet bearing negative sign is 1 = 10
0 and 5 has positive sign.)
45 cannot be written as 50 – 5, that is, VL (wrong)

By first rule, select X as it is largest alphabet smaller than the value of 45. Since X cannot be repeated 4 times to give 40 by second rules, next largest alphabet (L) greater than current (X) is selected and a subtraction is performed.
Now, 45 = 50 – 1 -1 -1 – 1 – 1 = IIIIIL, but by second rule ‘I’ cannot be repeated more than 3 times by second rule.
Again, 45 = 50 – 5, but 5 is not power of 10. Therefore 5 cannot be used in subtraction.
45 = (50 – 10) + 5 = XLV. (Right)
4.
Don’t subtract an alphabet from another, which is more than 10 times smaller than the other.
Subtracting ‘I’ from V and X is valid, but from L and C is invalid.
Sample questions for homework and assessment papers:
1.
What Roman alphabet is used to represent decimal 10? a) C b) D c) X d) V
2.
Convert the Roman numeral LXIX to decimal number.
3.
Convert decimal 73 to Roman numeral.
4.
Mark true or false: a) IX means, 1 + 10 = 11. (true/false) b) ‘14’ is written as XIIII in Roman numerals. (true/false) c) VII = 5 + 1 + 1. (true/false) d) VIII = 5 -1 -1 -1. (true/false)
5.
e) XX = 1010. (true/false)
Side A Side B
X
I
L
V
C
100
50
5
1
10