-0.000 000 000 742 147 676 256 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 256(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
-0.000 000 000 742 147 676 256(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:

|-0.000 000 000 742 147 676 256| = 0.000 000 000 742 147 676 256


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: 0.000 000 000 742 147 676 256.

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) 0.000 000 000 742 147 676 256 × 2 = 0 + 0.000 000 001 484 295 352 512;
  • 2) 0.000 000 001 484 295 352 512 × 2 = 0 + 0.000 000 002 968 590 705 024;
  • 3) 0.000 000 002 968 590 705 024 × 2 = 0 + 0.000 000 005 937 181 410 048;
  • 4) 0.000 000 005 937 181 410 048 × 2 = 0 + 0.000 000 011 874 362 820 096;
  • 5) 0.000 000 011 874 362 820 096 × 2 = 0 + 0.000 000 023 748 725 640 192;
  • 6) 0.000 000 023 748 725 640 192 × 2 = 0 + 0.000 000 047 497 451 280 384;
  • 7) 0.000 000 047 497 451 280 384 × 2 = 0 + 0.000 000 094 994 902 560 768;
  • 8) 0.000 000 094 994 902 560 768 × 2 = 0 + 0.000 000 189 989 805 121 536;
  • 9) 0.000 000 189 989 805 121 536 × 2 = 0 + 0.000 000 379 979 610 243 072;
  • 10) 0.000 000 379 979 610 243 072 × 2 = 0 + 0.000 000 759 959 220 486 144;
  • 11) 0.000 000 759 959 220 486 144 × 2 = 0 + 0.000 001 519 918 440 972 288;
  • 12) 0.000 001 519 918 440 972 288 × 2 = 0 + 0.000 003 039 836 881 944 576;
  • 13) 0.000 003 039 836 881 944 576 × 2 = 0 + 0.000 006 079 673 763 889 152;
  • 14) 0.000 006 079 673 763 889 152 × 2 = 0 + 0.000 012 159 347 527 778 304;
  • 15) 0.000 012 159 347 527 778 304 × 2 = 0 + 0.000 024 318 695 055 556 608;
  • 16) 0.000 024 318 695 055 556 608 × 2 = 0 + 0.000 048 637 390 111 113 216;
  • 17) 0.000 048 637 390 111 113 216 × 2 = 0 + 0.000 097 274 780 222 226 432;
  • 18) 0.000 097 274 780 222 226 432 × 2 = 0 + 0.000 194 549 560 444 452 864;
  • 19) 0.000 194 549 560 444 452 864 × 2 = 0 + 0.000 389 099 120 888 905 728;
  • 20) 0.000 389 099 120 888 905 728 × 2 = 0 + 0.000 778 198 241 777 811 456;
  • 21) 0.000 778 198 241 777 811 456 × 2 = 0 + 0.001 556 396 483 555 622 912;
  • 22) 0.001 556 396 483 555 622 912 × 2 = 0 + 0.003 112 792 967 111 245 824;
  • 23) 0.003 112 792 967 111 245 824 × 2 = 0 + 0.006 225 585 934 222 491 648;
  • 24) 0.006 225 585 934 222 491 648 × 2 = 0 + 0.012 451 171 868 444 983 296;
  • 25) 0.012 451 171 868 444 983 296 × 2 = 0 + 0.024 902 343 736 889 966 592;
  • 26) 0.024 902 343 736 889 966 592 × 2 = 0 + 0.049 804 687 473 779 933 184;
  • 27) 0.049 804 687 473 779 933 184 × 2 = 0 + 0.099 609 374 947 559 866 368;
  • 28) 0.099 609 374 947 559 866 368 × 2 = 0 + 0.199 218 749 895 119 732 736;
  • 29) 0.199 218 749 895 119 732 736 × 2 = 0 + 0.398 437 499 790 239 465 472;
  • 30) 0.398 437 499 790 239 465 472 × 2 = 0 + 0.796 874 999 580 478 930 944;
  • 31) 0.796 874 999 580 478 930 944 × 2 = 1 + 0.593 749 999 160 957 861 888;
  • 32) 0.593 749 999 160 957 861 888 × 2 = 1 + 0.187 499 998 321 915 723 776;
  • 33) 0.187 499 998 321 915 723 776 × 2 = 0 + 0.374 999 996 643 831 447 552;
  • 34) 0.374 999 996 643 831 447 552 × 2 = 0 + 0.749 999 993 287 662 895 104;
  • 35) 0.749 999 993 287 662 895 104 × 2 = 1 + 0.499 999 986 575 325 790 208;
  • 36) 0.499 999 986 575 325 790 208 × 2 = 0 + 0.999 999 973 150 651 580 416;
  • 37) 0.999 999 973 150 651 580 416 × 2 = 1 + 0.999 999 946 301 303 160 832;
  • 38) 0.999 999 946 301 303 160 832 × 2 = 1 + 0.999 999 892 602 606 321 664;
  • 39) 0.999 999 892 602 606 321 664 × 2 = 1 + 0.999 999 785 205 212 643 328;
  • 40) 0.999 999 785 205 212 643 328 × 2 = 1 + 0.999 999 570 410 425 286 656;
  • 41) 0.999 999 570 410 425 286 656 × 2 = 1 + 0.999 999 140 820 850 573 312;
  • 42) 0.999 999 140 820 850 573 312 × 2 = 1 + 0.999 998 281 641 701 146 624;
  • 43) 0.999 998 281 641 701 146 624 × 2 = 1 + 0.999 996 563 283 402 293 248;
  • 44) 0.999 996 563 283 402 293 248 × 2 = 1 + 0.999 993 126 566 804 586 496;
  • 45) 0.999 993 126 566 804 586 496 × 2 = 1 + 0.999 986 253 133 609 172 992;
  • 46) 0.999 986 253 133 609 172 992 × 2 = 1 + 0.999 972 506 267 218 345 984;
  • 47) 0.999 972 506 267 218 345 984 × 2 = 1 + 0.999 945 012 534 436 691 968;
  • 48) 0.999 945 012 534 436 691 968 × 2 = 1 + 0.999 890 025 068 873 383 936;
  • 49) 0.999 890 025 068 873 383 936 × 2 = 1 + 0.999 780 050 137 746 767 872;
  • 50) 0.999 780 050 137 746 767 872 × 2 = 1 + 0.999 560 100 275 493 535 744;
  • 51) 0.999 560 100 275 493 535 744 × 2 = 1 + 0.999 120 200 550 987 071 488;
  • 52) 0.999 120 200 550 987 071 488 × 2 = 1 + 0.998 240 401 101 974 142 976;
  • 53) 0.998 240 401 101 974 142 976 × 2 = 1 + 0.996 480 802 203 948 285 952;
  • 54) 0.996 480 802 203 948 285 952 × 2 = 1 + 0.992 961 604 407 896 571 904;

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:


0.000 000 000 742 147 676 256(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 256(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:


0.000 000 000 742 147 676 256(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 -0.000 000 000 742 147 676 256 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