-0.000 000 000 742 147 676 018 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 018(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
-0.000 000 000 742 147 676 018(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:

|-0.000 000 000 742 147 676 018| = 0.000 000 000 742 147 676 018


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: 0.000 000 000 742 147 676 018.

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) 0.000 000 000 742 147 676 018 × 2 = 0 + 0.000 000 001 484 295 352 036;
  • 2) 0.000 000 001 484 295 352 036 × 2 = 0 + 0.000 000 002 968 590 704 072;
  • 3) 0.000 000 002 968 590 704 072 × 2 = 0 + 0.000 000 005 937 181 408 144;
  • 4) 0.000 000 005 937 181 408 144 × 2 = 0 + 0.000 000 011 874 362 816 288;
  • 5) 0.000 000 011 874 362 816 288 × 2 = 0 + 0.000 000 023 748 725 632 576;
  • 6) 0.000 000 023 748 725 632 576 × 2 = 0 + 0.000 000 047 497 451 265 152;
  • 7) 0.000 000 047 497 451 265 152 × 2 = 0 + 0.000 000 094 994 902 530 304;
  • 8) 0.000 000 094 994 902 530 304 × 2 = 0 + 0.000 000 189 989 805 060 608;
  • 9) 0.000 000 189 989 805 060 608 × 2 = 0 + 0.000 000 379 979 610 121 216;
  • 10) 0.000 000 379 979 610 121 216 × 2 = 0 + 0.000 000 759 959 220 242 432;
  • 11) 0.000 000 759 959 220 242 432 × 2 = 0 + 0.000 001 519 918 440 484 864;
  • 12) 0.000 001 519 918 440 484 864 × 2 = 0 + 0.000 003 039 836 880 969 728;
  • 13) 0.000 003 039 836 880 969 728 × 2 = 0 + 0.000 006 079 673 761 939 456;
  • 14) 0.000 006 079 673 761 939 456 × 2 = 0 + 0.000 012 159 347 523 878 912;
  • 15) 0.000 012 159 347 523 878 912 × 2 = 0 + 0.000 024 318 695 047 757 824;
  • 16) 0.000 024 318 695 047 757 824 × 2 = 0 + 0.000 048 637 390 095 515 648;
  • 17) 0.000 048 637 390 095 515 648 × 2 = 0 + 0.000 097 274 780 191 031 296;
  • 18) 0.000 097 274 780 191 031 296 × 2 = 0 + 0.000 194 549 560 382 062 592;
  • 19) 0.000 194 549 560 382 062 592 × 2 = 0 + 0.000 389 099 120 764 125 184;
  • 20) 0.000 389 099 120 764 125 184 × 2 = 0 + 0.000 778 198 241 528 250 368;
  • 21) 0.000 778 198 241 528 250 368 × 2 = 0 + 0.001 556 396 483 056 500 736;
  • 22) 0.001 556 396 483 056 500 736 × 2 = 0 + 0.003 112 792 966 113 001 472;
  • 23) 0.003 112 792 966 113 001 472 × 2 = 0 + 0.006 225 585 932 226 002 944;
  • 24) 0.006 225 585 932 226 002 944 × 2 = 0 + 0.012 451 171 864 452 005 888;
  • 25) 0.012 451 171 864 452 005 888 × 2 = 0 + 0.024 902 343 728 904 011 776;
  • 26) 0.024 902 343 728 904 011 776 × 2 = 0 + 0.049 804 687 457 808 023 552;
  • 27) 0.049 804 687 457 808 023 552 × 2 = 0 + 0.099 609 374 915 616 047 104;
  • 28) 0.099 609 374 915 616 047 104 × 2 = 0 + 0.199 218 749 831 232 094 208;
  • 29) 0.199 218 749 831 232 094 208 × 2 = 0 + 0.398 437 499 662 464 188 416;
  • 30) 0.398 437 499 662 464 188 416 × 2 = 0 + 0.796 874 999 324 928 376 832;
  • 31) 0.796 874 999 324 928 376 832 × 2 = 1 + 0.593 749 998 649 856 753 664;
  • 32) 0.593 749 998 649 856 753 664 × 2 = 1 + 0.187 499 997 299 713 507 328;
  • 33) 0.187 499 997 299 713 507 328 × 2 = 0 + 0.374 999 994 599 427 014 656;
  • 34) 0.374 999 994 599 427 014 656 × 2 = 0 + 0.749 999 989 198 854 029 312;
  • 35) 0.749 999 989 198 854 029 312 × 2 = 1 + 0.499 999 978 397 708 058 624;
  • 36) 0.499 999 978 397 708 058 624 × 2 = 0 + 0.999 999 956 795 416 117 248;
  • 37) 0.999 999 956 795 416 117 248 × 2 = 1 + 0.999 999 913 590 832 234 496;
  • 38) 0.999 999 913 590 832 234 496 × 2 = 1 + 0.999 999 827 181 664 468 992;
  • 39) 0.999 999 827 181 664 468 992 × 2 = 1 + 0.999 999 654 363 328 937 984;
  • 40) 0.999 999 654 363 328 937 984 × 2 = 1 + 0.999 999 308 726 657 875 968;
  • 41) 0.999 999 308 726 657 875 968 × 2 = 1 + 0.999 998 617 453 315 751 936;
  • 42) 0.999 998 617 453 315 751 936 × 2 = 1 + 0.999 997 234 906 631 503 872;
  • 43) 0.999 997 234 906 631 503 872 × 2 = 1 + 0.999 994 469 813 263 007 744;
  • 44) 0.999 994 469 813 263 007 744 × 2 = 1 + 0.999 988 939 626 526 015 488;
  • 45) 0.999 988 939 626 526 015 488 × 2 = 1 + 0.999 977 879 253 052 030 976;
  • 46) 0.999 977 879 253 052 030 976 × 2 = 1 + 0.999 955 758 506 104 061 952;
  • 47) 0.999 955 758 506 104 061 952 × 2 = 1 + 0.999 911 517 012 208 123 904;
  • 48) 0.999 911 517 012 208 123 904 × 2 = 1 + 0.999 823 034 024 416 247 808;
  • 49) 0.999 823 034 024 416 247 808 × 2 = 1 + 0.999 646 068 048 832 495 616;
  • 50) 0.999 646 068 048 832 495 616 × 2 = 1 + 0.999 292 136 097 664 991 232;
  • 51) 0.999 292 136 097 664 991 232 × 2 = 1 + 0.998 584 272 195 329 982 464;
  • 52) 0.998 584 272 195 329 982 464 × 2 = 1 + 0.997 168 544 390 659 964 928;
  • 53) 0.997 168 544 390 659 964 928 × 2 = 1 + 0.994 337 088 781 319 929 856;
  • 54) 0.994 337 088 781 319 929 856 × 2 = 1 + 0.988 674 177 562 639 859 712;

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:


0.000 000 000 742 147 676 018(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 018(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:


0.000 000 000 742 147 676 018(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 -0.000 000 000 742 147 676 018 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