-0.000 000 000 742 147 671 96 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 671 96(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
-0.000 000 000 742 147 671 96(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:

|-0.000 000 000 742 147 671 96| = 0.000 000 000 742 147 671 96


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: 0.000 000 000 742 147 671 96.

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) 0.000 000 000 742 147 671 96 × 2 = 0 + 0.000 000 001 484 295 343 92;
  • 2) 0.000 000 001 484 295 343 92 × 2 = 0 + 0.000 000 002 968 590 687 84;
  • 3) 0.000 000 002 968 590 687 84 × 2 = 0 + 0.000 000 005 937 181 375 68;
  • 4) 0.000 000 005 937 181 375 68 × 2 = 0 + 0.000 000 011 874 362 751 36;
  • 5) 0.000 000 011 874 362 751 36 × 2 = 0 + 0.000 000 023 748 725 502 72;
  • 6) 0.000 000 023 748 725 502 72 × 2 = 0 + 0.000 000 047 497 451 005 44;
  • 7) 0.000 000 047 497 451 005 44 × 2 = 0 + 0.000 000 094 994 902 010 88;
  • 8) 0.000 000 094 994 902 010 88 × 2 = 0 + 0.000 000 189 989 804 021 76;
  • 9) 0.000 000 189 989 804 021 76 × 2 = 0 + 0.000 000 379 979 608 043 52;
  • 10) 0.000 000 379 979 608 043 52 × 2 = 0 + 0.000 000 759 959 216 087 04;
  • 11) 0.000 000 759 959 216 087 04 × 2 = 0 + 0.000 001 519 918 432 174 08;
  • 12) 0.000 001 519 918 432 174 08 × 2 = 0 + 0.000 003 039 836 864 348 16;
  • 13) 0.000 003 039 836 864 348 16 × 2 = 0 + 0.000 006 079 673 728 696 32;
  • 14) 0.000 006 079 673 728 696 32 × 2 = 0 + 0.000 012 159 347 457 392 64;
  • 15) 0.000 012 159 347 457 392 64 × 2 = 0 + 0.000 024 318 694 914 785 28;
  • 16) 0.000 024 318 694 914 785 28 × 2 = 0 + 0.000 048 637 389 829 570 56;
  • 17) 0.000 048 637 389 829 570 56 × 2 = 0 + 0.000 097 274 779 659 141 12;
  • 18) 0.000 097 274 779 659 141 12 × 2 = 0 + 0.000 194 549 559 318 282 24;
  • 19) 0.000 194 549 559 318 282 24 × 2 = 0 + 0.000 389 099 118 636 564 48;
  • 20) 0.000 389 099 118 636 564 48 × 2 = 0 + 0.000 778 198 237 273 128 96;
  • 21) 0.000 778 198 237 273 128 96 × 2 = 0 + 0.001 556 396 474 546 257 92;
  • 22) 0.001 556 396 474 546 257 92 × 2 = 0 + 0.003 112 792 949 092 515 84;
  • 23) 0.003 112 792 949 092 515 84 × 2 = 0 + 0.006 225 585 898 185 031 68;
  • 24) 0.006 225 585 898 185 031 68 × 2 = 0 + 0.012 451 171 796 370 063 36;
  • 25) 0.012 451 171 796 370 063 36 × 2 = 0 + 0.024 902 343 592 740 126 72;
  • 26) 0.024 902 343 592 740 126 72 × 2 = 0 + 0.049 804 687 185 480 253 44;
  • 27) 0.049 804 687 185 480 253 44 × 2 = 0 + 0.099 609 374 370 960 506 88;
  • 28) 0.099 609 374 370 960 506 88 × 2 = 0 + 0.199 218 748 741 921 013 76;
  • 29) 0.199 218 748 741 921 013 76 × 2 = 0 + 0.398 437 497 483 842 027 52;
  • 30) 0.398 437 497 483 842 027 52 × 2 = 0 + 0.796 874 994 967 684 055 04;
  • 31) 0.796 874 994 967 684 055 04 × 2 = 1 + 0.593 749 989 935 368 110 08;
  • 32) 0.593 749 989 935 368 110 08 × 2 = 1 + 0.187 499 979 870 736 220 16;
  • 33) 0.187 499 979 870 736 220 16 × 2 = 0 + 0.374 999 959 741 472 440 32;
  • 34) 0.374 999 959 741 472 440 32 × 2 = 0 + 0.749 999 919 482 944 880 64;
  • 35) 0.749 999 919 482 944 880 64 × 2 = 1 + 0.499 999 838 965 889 761 28;
  • 36) 0.499 999 838 965 889 761 28 × 2 = 0 + 0.999 999 677 931 779 522 56;
  • 37) 0.999 999 677 931 779 522 56 × 2 = 1 + 0.999 999 355 863 559 045 12;
  • 38) 0.999 999 355 863 559 045 12 × 2 = 1 + 0.999 998 711 727 118 090 24;
  • 39) 0.999 998 711 727 118 090 24 × 2 = 1 + 0.999 997 423 454 236 180 48;
  • 40) 0.999 997 423 454 236 180 48 × 2 = 1 + 0.999 994 846 908 472 360 96;
  • 41) 0.999 994 846 908 472 360 96 × 2 = 1 + 0.999 989 693 816 944 721 92;
  • 42) 0.999 989 693 816 944 721 92 × 2 = 1 + 0.999 979 387 633 889 443 84;
  • 43) 0.999 979 387 633 889 443 84 × 2 = 1 + 0.999 958 775 267 778 887 68;
  • 44) 0.999 958 775 267 778 887 68 × 2 = 1 + 0.999 917 550 535 557 775 36;
  • 45) 0.999 917 550 535 557 775 36 × 2 = 1 + 0.999 835 101 071 115 550 72;
  • 46) 0.999 835 101 071 115 550 72 × 2 = 1 + 0.999 670 202 142 231 101 44;
  • 47) 0.999 670 202 142 231 101 44 × 2 = 1 + 0.999 340 404 284 462 202 88;
  • 48) 0.999 340 404 284 462 202 88 × 2 = 1 + 0.998 680 808 568 924 405 76;
  • 49) 0.998 680 808 568 924 405 76 × 2 = 1 + 0.997 361 617 137 848 811 52;
  • 50) 0.997 361 617 137 848 811 52 × 2 = 1 + 0.994 723 234 275 697 623 04;
  • 51) 0.994 723 234 275 697 623 04 × 2 = 1 + 0.989 446 468 551 395 246 08;
  • 52) 0.989 446 468 551 395 246 08 × 2 = 1 + 0.978 892 937 102 790 492 16;
  • 53) 0.978 892 937 102 790 492 16 × 2 = 1 + 0.957 785 874 205 580 984 32;
  • 54) 0.957 785 874 205 580 984 32 × 2 = 1 + 0.915 571 748 411 161 968 64;

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:


0.000 000 000 742 147 671 96(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 671 96(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 671 96(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 1111 1111 111


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 1111 1111 =

100 1011 1111 1111 1111 1111


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 1111 1111


Decimal number -0.000 000 000 742 147 671 96 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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