-0.000 000 000 742 147 677 66 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 66(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
-0.000 000 000 742 147 677 66(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:

|-0.000 000 000 742 147 677 66| = 0.000 000 000 742 147 677 66


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: 0.000 000 000 742 147 677 66.

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) 0.000 000 000 742 147 677 66 × 2 = 0 + 0.000 000 001 484 295 355 32;
  • 2) 0.000 000 001 484 295 355 32 × 2 = 0 + 0.000 000 002 968 590 710 64;
  • 3) 0.000 000 002 968 590 710 64 × 2 = 0 + 0.000 000 005 937 181 421 28;
  • 4) 0.000 000 005 937 181 421 28 × 2 = 0 + 0.000 000 011 874 362 842 56;
  • 5) 0.000 000 011 874 362 842 56 × 2 = 0 + 0.000 000 023 748 725 685 12;
  • 6) 0.000 000 023 748 725 685 12 × 2 = 0 + 0.000 000 047 497 451 370 24;
  • 7) 0.000 000 047 497 451 370 24 × 2 = 0 + 0.000 000 094 994 902 740 48;
  • 8) 0.000 000 094 994 902 740 48 × 2 = 0 + 0.000 000 189 989 805 480 96;
  • 9) 0.000 000 189 989 805 480 96 × 2 = 0 + 0.000 000 379 979 610 961 92;
  • 10) 0.000 000 379 979 610 961 92 × 2 = 0 + 0.000 000 759 959 221 923 84;
  • 11) 0.000 000 759 959 221 923 84 × 2 = 0 + 0.000 001 519 918 443 847 68;
  • 12) 0.000 001 519 918 443 847 68 × 2 = 0 + 0.000 003 039 836 887 695 36;
  • 13) 0.000 003 039 836 887 695 36 × 2 = 0 + 0.000 006 079 673 775 390 72;
  • 14) 0.000 006 079 673 775 390 72 × 2 = 0 + 0.000 012 159 347 550 781 44;
  • 15) 0.000 012 159 347 550 781 44 × 2 = 0 + 0.000 024 318 695 101 562 88;
  • 16) 0.000 024 318 695 101 562 88 × 2 = 0 + 0.000 048 637 390 203 125 76;
  • 17) 0.000 048 637 390 203 125 76 × 2 = 0 + 0.000 097 274 780 406 251 52;
  • 18) 0.000 097 274 780 406 251 52 × 2 = 0 + 0.000 194 549 560 812 503 04;
  • 19) 0.000 194 549 560 812 503 04 × 2 = 0 + 0.000 389 099 121 625 006 08;
  • 20) 0.000 389 099 121 625 006 08 × 2 = 0 + 0.000 778 198 243 250 012 16;
  • 21) 0.000 778 198 243 250 012 16 × 2 = 0 + 0.001 556 396 486 500 024 32;
  • 22) 0.001 556 396 486 500 024 32 × 2 = 0 + 0.003 112 792 973 000 048 64;
  • 23) 0.003 112 792 973 000 048 64 × 2 = 0 + 0.006 225 585 946 000 097 28;
  • 24) 0.006 225 585 946 000 097 28 × 2 = 0 + 0.012 451 171 892 000 194 56;
  • 25) 0.012 451 171 892 000 194 56 × 2 = 0 + 0.024 902 343 784 000 389 12;
  • 26) 0.024 902 343 784 000 389 12 × 2 = 0 + 0.049 804 687 568 000 778 24;
  • 27) 0.049 804 687 568 000 778 24 × 2 = 0 + 0.099 609 375 136 001 556 48;
  • 28) 0.099 609 375 136 001 556 48 × 2 = 0 + 0.199 218 750 272 003 112 96;
  • 29) 0.199 218 750 272 003 112 96 × 2 = 0 + 0.398 437 500 544 006 225 92;
  • 30) 0.398 437 500 544 006 225 92 × 2 = 0 + 0.796 875 001 088 012 451 84;
  • 31) 0.796 875 001 088 012 451 84 × 2 = 1 + 0.593 750 002 176 024 903 68;
  • 32) 0.593 750 002 176 024 903 68 × 2 = 1 + 0.187 500 004 352 049 807 36;
  • 33) 0.187 500 004 352 049 807 36 × 2 = 0 + 0.375 000 008 704 099 614 72;
  • 34) 0.375 000 008 704 099 614 72 × 2 = 0 + 0.750 000 017 408 199 229 44;
  • 35) 0.750 000 017 408 199 229 44 × 2 = 1 + 0.500 000 034 816 398 458 88;
  • 36) 0.500 000 034 816 398 458 88 × 2 = 1 + 0.000 000 069 632 796 917 76;
  • 37) 0.000 000 069 632 796 917 76 × 2 = 0 + 0.000 000 139 265 593 835 52;
  • 38) 0.000 000 139 265 593 835 52 × 2 = 0 + 0.000 000 278 531 187 671 04;
  • 39) 0.000 000 278 531 187 671 04 × 2 = 0 + 0.000 000 557 062 375 342 08;
  • 40) 0.000 000 557 062 375 342 08 × 2 = 0 + 0.000 001 114 124 750 684 16;
  • 41) 0.000 001 114 124 750 684 16 × 2 = 0 + 0.000 002 228 249 501 368 32;
  • 42) 0.000 002 228 249 501 368 32 × 2 = 0 + 0.000 004 456 499 002 736 64;
  • 43) 0.000 004 456 499 002 736 64 × 2 = 0 + 0.000 008 912 998 005 473 28;
  • 44) 0.000 008 912 998 005 473 28 × 2 = 0 + 0.000 017 825 996 010 946 56;
  • 45) 0.000 017 825 996 010 946 56 × 2 = 0 + 0.000 035 651 992 021 893 12;
  • 46) 0.000 035 651 992 021 893 12 × 2 = 0 + 0.000 071 303 984 043 786 24;
  • 47) 0.000 071 303 984 043 786 24 × 2 = 0 + 0.000 142 607 968 087 572 48;
  • 48) 0.000 142 607 968 087 572 48 × 2 = 0 + 0.000 285 215 936 175 144 96;
  • 49) 0.000 285 215 936 175 144 96 × 2 = 0 + 0.000 570 431 872 350 289 92;
  • 50) 0.000 570 431 872 350 289 92 × 2 = 0 + 0.001 140 863 744 700 579 84;
  • 51) 0.001 140 863 744 700 579 84 × 2 = 0 + 0.002 281 727 489 401 159 68;
  • 52) 0.002 281 727 489 401 159 68 × 2 = 0 + 0.004 563 454 978 802 319 36;
  • 53) 0.004 563 454 978 802 319 36 × 2 = 0 + 0.009 126 909 957 604 638 72;
  • 54) 0.009 126 909 957 604 638 72 × 2 = 0 + 0.018 253 819 915 209 277 44;

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:


0.000 000 000 742 147 677 66(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 677 66(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:


0.000 000 000 742 147 677 66(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 -0.000 000 000 742 147 677 66 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