9.501 955 785 322 934 389 101 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 9.501 955 785 322 934 389 101 4(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
9.501 955 785 322 934 389 101 4(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 9.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

9(10) =

1001(2)


3. Convert to binary (base 2) the fractional part: 0.501 955 785 322 934 389 101 4.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.501 955 785 322 934 389 101 4 × 2 = 1 + 0.003 911 570 645 868 778 202 8;
  • 2) 0.003 911 570 645 868 778 202 8 × 2 = 0 + 0.007 823 141 291 737 556 405 6;
  • 3) 0.007 823 141 291 737 556 405 6 × 2 = 0 + 0.015 646 282 583 475 112 811 2;
  • 4) 0.015 646 282 583 475 112 811 2 × 2 = 0 + 0.031 292 565 166 950 225 622 4;
  • 5) 0.031 292 565 166 950 225 622 4 × 2 = 0 + 0.062 585 130 333 900 451 244 8;
  • 6) 0.062 585 130 333 900 451 244 8 × 2 = 0 + 0.125 170 260 667 800 902 489 6;
  • 7) 0.125 170 260 667 800 902 489 6 × 2 = 0 + 0.250 340 521 335 601 804 979 2;
  • 8) 0.250 340 521 335 601 804 979 2 × 2 = 0 + 0.500 681 042 671 203 609 958 4;
  • 9) 0.500 681 042 671 203 609 958 4 × 2 = 1 + 0.001 362 085 342 407 219 916 8;
  • 10) 0.001 362 085 342 407 219 916 8 × 2 = 0 + 0.002 724 170 684 814 439 833 6;
  • 11) 0.002 724 170 684 814 439 833 6 × 2 = 0 + 0.005 448 341 369 628 879 667 2;
  • 12) 0.005 448 341 369 628 879 667 2 × 2 = 0 + 0.010 896 682 739 257 759 334 4;
  • 13) 0.010 896 682 739 257 759 334 4 × 2 = 0 + 0.021 793 365 478 515 518 668 8;
  • 14) 0.021 793 365 478 515 518 668 8 × 2 = 0 + 0.043 586 730 957 031 037 337 6;
  • 15) 0.043 586 730 957 031 037 337 6 × 2 = 0 + 0.087 173 461 914 062 074 675 2;
  • 16) 0.087 173 461 914 062 074 675 2 × 2 = 0 + 0.174 346 923 828 124 149 350 4;
  • 17) 0.174 346 923 828 124 149 350 4 × 2 = 0 + 0.348 693 847 656 248 298 700 8;
  • 18) 0.348 693 847 656 248 298 700 8 × 2 = 0 + 0.697 387 695 312 496 597 401 6;
  • 19) 0.697 387 695 312 496 597 401 6 × 2 = 1 + 0.394 775 390 624 993 194 803 2;
  • 20) 0.394 775 390 624 993 194 803 2 × 2 = 0 + 0.789 550 781 249 986 389 606 4;
  • 21) 0.789 550 781 249 986 389 606 4 × 2 = 1 + 0.579 101 562 499 972 779 212 8;
  • 22) 0.579 101 562 499 972 779 212 8 × 2 = 1 + 0.158 203 124 999 945 558 425 6;
  • 23) 0.158 203 124 999 945 558 425 6 × 2 = 0 + 0.316 406 249 999 891 116 851 2;
  • 24) 0.316 406 249 999 891 116 851 2 × 2 = 0 + 0.632 812 499 999 782 233 702 4;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.501 955 785 322 934 389 101 4(10) =

0.1000 0000 1000 0000 0010 1100(2)

5. Positive number before normalization:

9.501 955 785 322 934 389 101 4(10) =

1001.1000 0000 1000 0000 0010 1100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


9.501 955 785 322 934 389 101 4(10) =

1001.1000 0000 1000 0000 0010 1100(2) =

1001.1000 0000 1000 0000 0010 1100(2) × 20 =

1.0011 0000 0001 0000 0000 0101 100(2) × 23


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): 3

Mantissa (not normalized):
1.0011 0000 0001 0000 0000 0101 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

3 + 2(8-1) - 1 =

(3 + 127)(10) =

130(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

130(10) =

1000 0010(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 001 1000 0000 1000 0000 0010 1100 =

001 1000 0000 1000 0000 0010


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
1000 0010

Mantissa (23 bits) =
001 1000 0000 1000 0000 0010


Decimal number 9.501 955 785 322 934 389 101 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0010 - 001 1000 0000 1000 0000 0010

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111