-0.000 000 000 742 147 678 86 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 678 86(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
-0.000 000 000 742 147 678 86(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:

|-0.000 000 000 742 147 678 86| = 0.000 000 000 742 147 678 86


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: 0.000 000 000 742 147 678 86.

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) 0.000 000 000 742 147 678 86 × 2 = 0 + 0.000 000 001 484 295 357 72;
  • 2) 0.000 000 001 484 295 357 72 × 2 = 0 + 0.000 000 002 968 590 715 44;
  • 3) 0.000 000 002 968 590 715 44 × 2 = 0 + 0.000 000 005 937 181 430 88;
  • 4) 0.000 000 005 937 181 430 88 × 2 = 0 + 0.000 000 011 874 362 861 76;
  • 5) 0.000 000 011 874 362 861 76 × 2 = 0 + 0.000 000 023 748 725 723 52;
  • 6) 0.000 000 023 748 725 723 52 × 2 = 0 + 0.000 000 047 497 451 447 04;
  • 7) 0.000 000 047 497 451 447 04 × 2 = 0 + 0.000 000 094 994 902 894 08;
  • 8) 0.000 000 094 994 902 894 08 × 2 = 0 + 0.000 000 189 989 805 788 16;
  • 9) 0.000 000 189 989 805 788 16 × 2 = 0 + 0.000 000 379 979 611 576 32;
  • 10) 0.000 000 379 979 611 576 32 × 2 = 0 + 0.000 000 759 959 223 152 64;
  • 11) 0.000 000 759 959 223 152 64 × 2 = 0 + 0.000 001 519 918 446 305 28;
  • 12) 0.000 001 519 918 446 305 28 × 2 = 0 + 0.000 003 039 836 892 610 56;
  • 13) 0.000 003 039 836 892 610 56 × 2 = 0 + 0.000 006 079 673 785 221 12;
  • 14) 0.000 006 079 673 785 221 12 × 2 = 0 + 0.000 012 159 347 570 442 24;
  • 15) 0.000 012 159 347 570 442 24 × 2 = 0 + 0.000 024 318 695 140 884 48;
  • 16) 0.000 024 318 695 140 884 48 × 2 = 0 + 0.000 048 637 390 281 768 96;
  • 17) 0.000 048 637 390 281 768 96 × 2 = 0 + 0.000 097 274 780 563 537 92;
  • 18) 0.000 097 274 780 563 537 92 × 2 = 0 + 0.000 194 549 561 127 075 84;
  • 19) 0.000 194 549 561 127 075 84 × 2 = 0 + 0.000 389 099 122 254 151 68;
  • 20) 0.000 389 099 122 254 151 68 × 2 = 0 + 0.000 778 198 244 508 303 36;
  • 21) 0.000 778 198 244 508 303 36 × 2 = 0 + 0.001 556 396 489 016 606 72;
  • 22) 0.001 556 396 489 016 606 72 × 2 = 0 + 0.003 112 792 978 033 213 44;
  • 23) 0.003 112 792 978 033 213 44 × 2 = 0 + 0.006 225 585 956 066 426 88;
  • 24) 0.006 225 585 956 066 426 88 × 2 = 0 + 0.012 451 171 912 132 853 76;
  • 25) 0.012 451 171 912 132 853 76 × 2 = 0 + 0.024 902 343 824 265 707 52;
  • 26) 0.024 902 343 824 265 707 52 × 2 = 0 + 0.049 804 687 648 531 415 04;
  • 27) 0.049 804 687 648 531 415 04 × 2 = 0 + 0.099 609 375 297 062 830 08;
  • 28) 0.099 609 375 297 062 830 08 × 2 = 0 + 0.199 218 750 594 125 660 16;
  • 29) 0.199 218 750 594 125 660 16 × 2 = 0 + 0.398 437 501 188 251 320 32;
  • 30) 0.398 437 501 188 251 320 32 × 2 = 0 + 0.796 875 002 376 502 640 64;
  • 31) 0.796 875 002 376 502 640 64 × 2 = 1 + 0.593 750 004 753 005 281 28;
  • 32) 0.593 750 004 753 005 281 28 × 2 = 1 + 0.187 500 009 506 010 562 56;
  • 33) 0.187 500 009 506 010 562 56 × 2 = 0 + 0.375 000 019 012 021 125 12;
  • 34) 0.375 000 019 012 021 125 12 × 2 = 0 + 0.750 000 038 024 042 250 24;
  • 35) 0.750 000 038 024 042 250 24 × 2 = 1 + 0.500 000 076 048 084 500 48;
  • 36) 0.500 000 076 048 084 500 48 × 2 = 1 + 0.000 000 152 096 169 000 96;
  • 37) 0.000 000 152 096 169 000 96 × 2 = 0 + 0.000 000 304 192 338 001 92;
  • 38) 0.000 000 304 192 338 001 92 × 2 = 0 + 0.000 000 608 384 676 003 84;
  • 39) 0.000 000 608 384 676 003 84 × 2 = 0 + 0.000 001 216 769 352 007 68;
  • 40) 0.000 001 216 769 352 007 68 × 2 = 0 + 0.000 002 433 538 704 015 36;
  • 41) 0.000 002 433 538 704 015 36 × 2 = 0 + 0.000 004 867 077 408 030 72;
  • 42) 0.000 004 867 077 408 030 72 × 2 = 0 + 0.000 009 734 154 816 061 44;
  • 43) 0.000 009 734 154 816 061 44 × 2 = 0 + 0.000 019 468 309 632 122 88;
  • 44) 0.000 019 468 309 632 122 88 × 2 = 0 + 0.000 038 936 619 264 245 76;
  • 45) 0.000 038 936 619 264 245 76 × 2 = 0 + 0.000 077 873 238 528 491 52;
  • 46) 0.000 077 873 238 528 491 52 × 2 = 0 + 0.000 155 746 477 056 983 04;
  • 47) 0.000 155 746 477 056 983 04 × 2 = 0 + 0.000 311 492 954 113 966 08;
  • 48) 0.000 311 492 954 113 966 08 × 2 = 0 + 0.000 622 985 908 227 932 16;
  • 49) 0.000 622 985 908 227 932 16 × 2 = 0 + 0.001 245 971 816 455 864 32;
  • 50) 0.001 245 971 816 455 864 32 × 2 = 0 + 0.002 491 943 632 911 728 64;
  • 51) 0.002 491 943 632 911 728 64 × 2 = 0 + 0.004 983 887 265 823 457 28;
  • 52) 0.004 983 887 265 823 457 28 × 2 = 0 + 0.009 967 774 531 646 914 56;
  • 53) 0.009 967 774 531 646 914 56 × 2 = 0 + 0.019 935 549 063 293 829 12;
  • 54) 0.019 935 549 063 293 829 12 × 2 = 0 + 0.039 871 098 126 587 658 24;

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:


0.000 000 000 742 147 678 86(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 678 86(10) =

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


0.000 000 000 742 147 678 86(10) =

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

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

1.1001 1000 0000 0000 0000 000(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 000


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 0000 =

100 1100 0000 0000 0000 0000


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 0000


Decimal number -0.000 000 000 742 147 678 86 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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