-0.000 000 000 742 128 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 128 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
-0.000 000 000 742 128 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. Start with the positive version of the number:

|-0.000 000 000 742 128 4| = 0.000 000 000 742 128 4


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 128 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) 0.000 000 000 742 128 4 × 2 = 0 + 0.000 000 001 484 256 8;
  • 2) 0.000 000 001 484 256 8 × 2 = 0 + 0.000 000 002 968 513 6;
  • 3) 0.000 000 002 968 513 6 × 2 = 0 + 0.000 000 005 937 027 2;
  • 4) 0.000 000 005 937 027 2 × 2 = 0 + 0.000 000 011 874 054 4;
  • 5) 0.000 000 011 874 054 4 × 2 = 0 + 0.000 000 023 748 108 8;
  • 6) 0.000 000 023 748 108 8 × 2 = 0 + 0.000 000 047 496 217 6;
  • 7) 0.000 000 047 496 217 6 × 2 = 0 + 0.000 000 094 992 435 2;
  • 8) 0.000 000 094 992 435 2 × 2 = 0 + 0.000 000 189 984 870 4;
  • 9) 0.000 000 189 984 870 4 × 2 = 0 + 0.000 000 379 969 740 8;
  • 10) 0.000 000 379 969 740 8 × 2 = 0 + 0.000 000 759 939 481 6;
  • 11) 0.000 000 759 939 481 6 × 2 = 0 + 0.000 001 519 878 963 2;
  • 12) 0.000 001 519 878 963 2 × 2 = 0 + 0.000 003 039 757 926 4;
  • 13) 0.000 003 039 757 926 4 × 2 = 0 + 0.000 006 079 515 852 8;
  • 14) 0.000 006 079 515 852 8 × 2 = 0 + 0.000 012 159 031 705 6;
  • 15) 0.000 012 159 031 705 6 × 2 = 0 + 0.000 024 318 063 411 2;
  • 16) 0.000 024 318 063 411 2 × 2 = 0 + 0.000 048 636 126 822 4;
  • 17) 0.000 048 636 126 822 4 × 2 = 0 + 0.000 097 272 253 644 8;
  • 18) 0.000 097 272 253 644 8 × 2 = 0 + 0.000 194 544 507 289 6;
  • 19) 0.000 194 544 507 289 6 × 2 = 0 + 0.000 389 089 014 579 2;
  • 20) 0.000 389 089 014 579 2 × 2 = 0 + 0.000 778 178 029 158 4;
  • 21) 0.000 778 178 029 158 4 × 2 = 0 + 0.001 556 356 058 316 8;
  • 22) 0.001 556 356 058 316 8 × 2 = 0 + 0.003 112 712 116 633 6;
  • 23) 0.003 112 712 116 633 6 × 2 = 0 + 0.006 225 424 233 267 2;
  • 24) 0.006 225 424 233 267 2 × 2 = 0 + 0.012 450 848 466 534 4;
  • 25) 0.012 450 848 466 534 4 × 2 = 0 + 0.024 901 696 933 068 8;
  • 26) 0.024 901 696 933 068 8 × 2 = 0 + 0.049 803 393 866 137 6;
  • 27) 0.049 803 393 866 137 6 × 2 = 0 + 0.099 606 787 732 275 2;
  • 28) 0.099 606 787 732 275 2 × 2 = 0 + 0.199 213 575 464 550 4;
  • 29) 0.199 213 575 464 550 4 × 2 = 0 + 0.398 427 150 929 100 8;
  • 30) 0.398 427 150 929 100 8 × 2 = 0 + 0.796 854 301 858 201 6;
  • 31) 0.796 854 301 858 201 6 × 2 = 1 + 0.593 708 603 716 403 2;
  • 32) 0.593 708 603 716 403 2 × 2 = 1 + 0.187 417 207 432 806 4;
  • 33) 0.187 417 207 432 806 4 × 2 = 0 + 0.374 834 414 865 612 8;
  • 34) 0.374 834 414 865 612 8 × 2 = 0 + 0.749 668 829 731 225 6;
  • 35) 0.749 668 829 731 225 6 × 2 = 1 + 0.499 337 659 462 451 2;
  • 36) 0.499 337 659 462 451 2 × 2 = 0 + 0.998 675 318 924 902 4;
  • 37) 0.998 675 318 924 902 4 × 2 = 1 + 0.997 350 637 849 804 8;
  • 38) 0.997 350 637 849 804 8 × 2 = 1 + 0.994 701 275 699 609 6;
  • 39) 0.994 701 275 699 609 6 × 2 = 1 + 0.989 402 551 399 219 2;
  • 40) 0.989 402 551 399 219 2 × 2 = 1 + 0.978 805 102 798 438 4;
  • 41) 0.978 805 102 798 438 4 × 2 = 1 + 0.957 610 205 596 876 8;
  • 42) 0.957 610 205 596 876 8 × 2 = 1 + 0.915 220 411 193 753 6;
  • 43) 0.915 220 411 193 753 6 × 2 = 1 + 0.830 440 822 387 507 2;
  • 44) 0.830 440 822 387 507 2 × 2 = 1 + 0.660 881 644 775 014 4;
  • 45) 0.660 881 644 775 014 4 × 2 = 1 + 0.321 763 289 550 028 8;
  • 46) 0.321 763 289 550 028 8 × 2 = 0 + 0.643 526 579 100 057 6;
  • 47) 0.643 526 579 100 057 6 × 2 = 1 + 0.287 053 158 200 115 2;
  • 48) 0.287 053 158 200 115 2 × 2 = 0 + 0.574 106 316 400 230 4;
  • 49) 0.574 106 316 400 230 4 × 2 = 1 + 0.148 212 632 800 460 8;
  • 50) 0.148 212 632 800 460 8 × 2 = 0 + 0.296 425 265 600 921 6;
  • 51) 0.296 425 265 600 921 6 × 2 = 0 + 0.592 850 531 201 843 2;
  • 52) 0.592 850 531 201 843 2 × 2 = 1 + 0.185 701 062 403 686 4;
  • 53) 0.185 701 062 403 686 4 × 2 = 0 + 0.371 402 124 807 372 8;
  • 54) 0.371 402 124 807 372 8 × 2 = 0 + 0.742 804 249 614 745 6;

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).


5. 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:


0.000 000 000 742 128 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1001 00(2)

6. Positive number before normalization:

0.000 000 000 742 128 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1001 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 128 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1001 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1010 1001 00(2) × 20 =

1.1001 0111 1111 1101 0100 100(2) × 2-31


8. 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 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1101 0100 100


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. 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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1110 1010 0100 =

100 1011 1111 1110 1010 0100


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

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

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1110 1010 0100


Decimal number -0.000 000 000 742 128 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 1010 0100

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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