-0.000 000 000 742 147 918 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 918(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
-0.000 000 000 742 147 918(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:

|-0.000 000 000 742 147 918| = 0.000 000 000 742 147 918


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: 0.000 000 000 742 147 918.

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) 0.000 000 000 742 147 918 × 2 = 0 + 0.000 000 001 484 295 836;
  • 2) 0.000 000 001 484 295 836 × 2 = 0 + 0.000 000 002 968 591 672;
  • 3) 0.000 000 002 968 591 672 × 2 = 0 + 0.000 000 005 937 183 344;
  • 4) 0.000 000 005 937 183 344 × 2 = 0 + 0.000 000 011 874 366 688;
  • 5) 0.000 000 011 874 366 688 × 2 = 0 + 0.000 000 023 748 733 376;
  • 6) 0.000 000 023 748 733 376 × 2 = 0 + 0.000 000 047 497 466 752;
  • 7) 0.000 000 047 497 466 752 × 2 = 0 + 0.000 000 094 994 933 504;
  • 8) 0.000 000 094 994 933 504 × 2 = 0 + 0.000 000 189 989 867 008;
  • 9) 0.000 000 189 989 867 008 × 2 = 0 + 0.000 000 379 979 734 016;
  • 10) 0.000 000 379 979 734 016 × 2 = 0 + 0.000 000 759 959 468 032;
  • 11) 0.000 000 759 959 468 032 × 2 = 0 + 0.000 001 519 918 936 064;
  • 12) 0.000 001 519 918 936 064 × 2 = 0 + 0.000 003 039 837 872 128;
  • 13) 0.000 003 039 837 872 128 × 2 = 0 + 0.000 006 079 675 744 256;
  • 14) 0.000 006 079 675 744 256 × 2 = 0 + 0.000 012 159 351 488 512;
  • 15) 0.000 012 159 351 488 512 × 2 = 0 + 0.000 024 318 702 977 024;
  • 16) 0.000 024 318 702 977 024 × 2 = 0 + 0.000 048 637 405 954 048;
  • 17) 0.000 048 637 405 954 048 × 2 = 0 + 0.000 097 274 811 908 096;
  • 18) 0.000 097 274 811 908 096 × 2 = 0 + 0.000 194 549 623 816 192;
  • 19) 0.000 194 549 623 816 192 × 2 = 0 + 0.000 389 099 247 632 384;
  • 20) 0.000 389 099 247 632 384 × 2 = 0 + 0.000 778 198 495 264 768;
  • 21) 0.000 778 198 495 264 768 × 2 = 0 + 0.001 556 396 990 529 536;
  • 22) 0.001 556 396 990 529 536 × 2 = 0 + 0.003 112 793 981 059 072;
  • 23) 0.003 112 793 981 059 072 × 2 = 0 + 0.006 225 587 962 118 144;
  • 24) 0.006 225 587 962 118 144 × 2 = 0 + 0.012 451 175 924 236 288;
  • 25) 0.012 451 175 924 236 288 × 2 = 0 + 0.024 902 351 848 472 576;
  • 26) 0.024 902 351 848 472 576 × 2 = 0 + 0.049 804 703 696 945 152;
  • 27) 0.049 804 703 696 945 152 × 2 = 0 + 0.099 609 407 393 890 304;
  • 28) 0.099 609 407 393 890 304 × 2 = 0 + 0.199 218 814 787 780 608;
  • 29) 0.199 218 814 787 780 608 × 2 = 0 + 0.398 437 629 575 561 216;
  • 30) 0.398 437 629 575 561 216 × 2 = 0 + 0.796 875 259 151 122 432;
  • 31) 0.796 875 259 151 122 432 × 2 = 1 + 0.593 750 518 302 244 864;
  • 32) 0.593 750 518 302 244 864 × 2 = 1 + 0.187 501 036 604 489 728;
  • 33) 0.187 501 036 604 489 728 × 2 = 0 + 0.375 002 073 208 979 456;
  • 34) 0.375 002 073 208 979 456 × 2 = 0 + 0.750 004 146 417 958 912;
  • 35) 0.750 004 146 417 958 912 × 2 = 1 + 0.500 008 292 835 917 824;
  • 36) 0.500 008 292 835 917 824 × 2 = 1 + 0.000 016 585 671 835 648;
  • 37) 0.000 016 585 671 835 648 × 2 = 0 + 0.000 033 171 343 671 296;
  • 38) 0.000 033 171 343 671 296 × 2 = 0 + 0.000 066 342 687 342 592;
  • 39) 0.000 066 342 687 342 592 × 2 = 0 + 0.000 132 685 374 685 184;
  • 40) 0.000 132 685 374 685 184 × 2 = 0 + 0.000 265 370 749 370 368;
  • 41) 0.000 265 370 749 370 368 × 2 = 0 + 0.000 530 741 498 740 736;
  • 42) 0.000 530 741 498 740 736 × 2 = 0 + 0.001 061 482 997 481 472;
  • 43) 0.001 061 482 997 481 472 × 2 = 0 + 0.002 122 965 994 962 944;
  • 44) 0.002 122 965 994 962 944 × 2 = 0 + 0.004 245 931 989 925 888;
  • 45) 0.004 245 931 989 925 888 × 2 = 0 + 0.008 491 863 979 851 776;
  • 46) 0.008 491 863 979 851 776 × 2 = 0 + 0.016 983 727 959 703 552;
  • 47) 0.016 983 727 959 703 552 × 2 = 0 + 0.033 967 455 919 407 104;
  • 48) 0.033 967 455 919 407 104 × 2 = 0 + 0.067 934 911 838 814 208;
  • 49) 0.067 934 911 838 814 208 × 2 = 0 + 0.135 869 823 677 628 416;
  • 50) 0.135 869 823 677 628 416 × 2 = 0 + 0.271 739 647 355 256 832;
  • 51) 0.271 739 647 355 256 832 × 2 = 0 + 0.543 479 294 710 513 664;
  • 52) 0.543 479 294 710 513 664 × 2 = 1 + 0.086 958 589 421 027 328;
  • 53) 0.086 958 589 421 027 328 × 2 = 0 + 0.173 917 178 842 054 656;
  • 54) 0.173 917 178 842 054 656 × 2 = 0 + 0.347 834 357 684 109 312;

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:


0.000 000 000 742 147 918(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 918(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 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:


0.000 000 000 742 147 918(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 00(2) × 20 =

1.1001 1000 0000 0000 0000 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 1000 0000 0000 0000 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 1100 0000 0000 0000 0100 =

100 1100 0000 0000 0000 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 1100 0000 0000 0000 0100


Decimal number -0.000 000 000 742 147 918 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 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