-0.000 000 000 742 136 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 136 9(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
-0.000 000 000 742 136 9(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:

|-0.000 000 000 742 136 9| = 0.000 000 000 742 136 9


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: 0.000 000 000 742 136 9.

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) 0.000 000 000 742 136 9 × 2 = 0 + 0.000 000 001 484 273 8;
  • 2) 0.000 000 001 484 273 8 × 2 = 0 + 0.000 000 002 968 547 6;
  • 3) 0.000 000 002 968 547 6 × 2 = 0 + 0.000 000 005 937 095 2;
  • 4) 0.000 000 005 937 095 2 × 2 = 0 + 0.000 000 011 874 190 4;
  • 5) 0.000 000 011 874 190 4 × 2 = 0 + 0.000 000 023 748 380 8;
  • 6) 0.000 000 023 748 380 8 × 2 = 0 + 0.000 000 047 496 761 6;
  • 7) 0.000 000 047 496 761 6 × 2 = 0 + 0.000 000 094 993 523 2;
  • 8) 0.000 000 094 993 523 2 × 2 = 0 + 0.000 000 189 987 046 4;
  • 9) 0.000 000 189 987 046 4 × 2 = 0 + 0.000 000 379 974 092 8;
  • 10) 0.000 000 379 974 092 8 × 2 = 0 + 0.000 000 759 948 185 6;
  • 11) 0.000 000 759 948 185 6 × 2 = 0 + 0.000 001 519 896 371 2;
  • 12) 0.000 001 519 896 371 2 × 2 = 0 + 0.000 003 039 792 742 4;
  • 13) 0.000 003 039 792 742 4 × 2 = 0 + 0.000 006 079 585 484 8;
  • 14) 0.000 006 079 585 484 8 × 2 = 0 + 0.000 012 159 170 969 6;
  • 15) 0.000 012 159 170 969 6 × 2 = 0 + 0.000 024 318 341 939 2;
  • 16) 0.000 024 318 341 939 2 × 2 = 0 + 0.000 048 636 683 878 4;
  • 17) 0.000 048 636 683 878 4 × 2 = 0 + 0.000 097 273 367 756 8;
  • 18) 0.000 097 273 367 756 8 × 2 = 0 + 0.000 194 546 735 513 6;
  • 19) 0.000 194 546 735 513 6 × 2 = 0 + 0.000 389 093 471 027 2;
  • 20) 0.000 389 093 471 027 2 × 2 = 0 + 0.000 778 186 942 054 4;
  • 21) 0.000 778 186 942 054 4 × 2 = 0 + 0.001 556 373 884 108 8;
  • 22) 0.001 556 373 884 108 8 × 2 = 0 + 0.003 112 747 768 217 6;
  • 23) 0.003 112 747 768 217 6 × 2 = 0 + 0.006 225 495 536 435 2;
  • 24) 0.006 225 495 536 435 2 × 2 = 0 + 0.012 450 991 072 870 4;
  • 25) 0.012 450 991 072 870 4 × 2 = 0 + 0.024 901 982 145 740 8;
  • 26) 0.024 901 982 145 740 8 × 2 = 0 + 0.049 803 964 291 481 6;
  • 27) 0.049 803 964 291 481 6 × 2 = 0 + 0.099 607 928 582 963 2;
  • 28) 0.099 607 928 582 963 2 × 2 = 0 + 0.199 215 857 165 926 4;
  • 29) 0.199 215 857 165 926 4 × 2 = 0 + 0.398 431 714 331 852 8;
  • 30) 0.398 431 714 331 852 8 × 2 = 0 + 0.796 863 428 663 705 6;
  • 31) 0.796 863 428 663 705 6 × 2 = 1 + 0.593 726 857 327 411 2;
  • 32) 0.593 726 857 327 411 2 × 2 = 1 + 0.187 453 714 654 822 4;
  • 33) 0.187 453 714 654 822 4 × 2 = 0 + 0.374 907 429 309 644 8;
  • 34) 0.374 907 429 309 644 8 × 2 = 0 + 0.749 814 858 619 289 6;
  • 35) 0.749 814 858 619 289 6 × 2 = 1 + 0.499 629 717 238 579 2;
  • 36) 0.499 629 717 238 579 2 × 2 = 0 + 0.999 259 434 477 158 4;
  • 37) 0.999 259 434 477 158 4 × 2 = 1 + 0.998 518 868 954 316 8;
  • 38) 0.998 518 868 954 316 8 × 2 = 1 + 0.997 037 737 908 633 6;
  • 39) 0.997 037 737 908 633 6 × 2 = 1 + 0.994 075 475 817 267 2;
  • 40) 0.994 075 475 817 267 2 × 2 = 1 + 0.988 150 951 634 534 4;
  • 41) 0.988 150 951 634 534 4 × 2 = 1 + 0.976 301 903 269 068 8;
  • 42) 0.976 301 903 269 068 8 × 2 = 1 + 0.952 603 806 538 137 6;
  • 43) 0.952 603 806 538 137 6 × 2 = 1 + 0.905 207 613 076 275 2;
  • 44) 0.905 207 613 076 275 2 × 2 = 1 + 0.810 415 226 152 550 4;
  • 45) 0.810 415 226 152 550 4 × 2 = 1 + 0.620 830 452 305 100 8;
  • 46) 0.620 830 452 305 100 8 × 2 = 1 + 0.241 660 904 610 201 6;
  • 47) 0.241 660 904 610 201 6 × 2 = 0 + 0.483 321 809 220 403 2;
  • 48) 0.483 321 809 220 403 2 × 2 = 0 + 0.966 643 618 440 806 4;
  • 49) 0.966 643 618 440 806 4 × 2 = 1 + 0.933 287 236 881 612 8;
  • 50) 0.933 287 236 881 612 8 × 2 = 1 + 0.866 574 473 763 225 6;
  • 51) 0.866 574 473 763 225 6 × 2 = 1 + 0.733 148 947 526 451 2;
  • 52) 0.733 148 947 526 451 2 × 2 = 1 + 0.466 297 895 052 902 4;
  • 53) 0.466 297 895 052 902 4 × 2 = 0 + 0.932 595 790 105 804 8;
  • 54) 0.932 595 790 105 804 8 × 2 = 1 + 0.865 191 580 211 609 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:


0.000 000 000 742 136 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1100 1111 01(2)

6. Positive number before normalization:

0.000 000 000 742 136 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1100 1111 01(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:


0.000 000 000 742 136 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1100 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1100 1111 01(2) × 20 =

1.1001 0111 1111 1110 0111 101(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 1110 0111 101


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 1111 0011 1101 =

100 1011 1111 1111 0011 1101


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 1111 0011 1101


Decimal number -0.000 000 000 742 136 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 0011 1101

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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