-0.000 000 000 742 147 672 88 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 672 88(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
-0.000 000 000 742 147 672 88(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:

|-0.000 000 000 742 147 672 88| = 0.000 000 000 742 147 672 88


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: 0.000 000 000 742 147 672 88.

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) 0.000 000 000 742 147 672 88 × 2 = 0 + 0.000 000 001 484 295 345 76;
  • 2) 0.000 000 001 484 295 345 76 × 2 = 0 + 0.000 000 002 968 590 691 52;
  • 3) 0.000 000 002 968 590 691 52 × 2 = 0 + 0.000 000 005 937 181 383 04;
  • 4) 0.000 000 005 937 181 383 04 × 2 = 0 + 0.000 000 011 874 362 766 08;
  • 5) 0.000 000 011 874 362 766 08 × 2 = 0 + 0.000 000 023 748 725 532 16;
  • 6) 0.000 000 023 748 725 532 16 × 2 = 0 + 0.000 000 047 497 451 064 32;
  • 7) 0.000 000 047 497 451 064 32 × 2 = 0 + 0.000 000 094 994 902 128 64;
  • 8) 0.000 000 094 994 902 128 64 × 2 = 0 + 0.000 000 189 989 804 257 28;
  • 9) 0.000 000 189 989 804 257 28 × 2 = 0 + 0.000 000 379 979 608 514 56;
  • 10) 0.000 000 379 979 608 514 56 × 2 = 0 + 0.000 000 759 959 217 029 12;
  • 11) 0.000 000 759 959 217 029 12 × 2 = 0 + 0.000 001 519 918 434 058 24;
  • 12) 0.000 001 519 918 434 058 24 × 2 = 0 + 0.000 003 039 836 868 116 48;
  • 13) 0.000 003 039 836 868 116 48 × 2 = 0 + 0.000 006 079 673 736 232 96;
  • 14) 0.000 006 079 673 736 232 96 × 2 = 0 + 0.000 012 159 347 472 465 92;
  • 15) 0.000 012 159 347 472 465 92 × 2 = 0 + 0.000 024 318 694 944 931 84;
  • 16) 0.000 024 318 694 944 931 84 × 2 = 0 + 0.000 048 637 389 889 863 68;
  • 17) 0.000 048 637 389 889 863 68 × 2 = 0 + 0.000 097 274 779 779 727 36;
  • 18) 0.000 097 274 779 779 727 36 × 2 = 0 + 0.000 194 549 559 559 454 72;
  • 19) 0.000 194 549 559 559 454 72 × 2 = 0 + 0.000 389 099 119 118 909 44;
  • 20) 0.000 389 099 119 118 909 44 × 2 = 0 + 0.000 778 198 238 237 818 88;
  • 21) 0.000 778 198 238 237 818 88 × 2 = 0 + 0.001 556 396 476 475 637 76;
  • 22) 0.001 556 396 476 475 637 76 × 2 = 0 + 0.003 112 792 952 951 275 52;
  • 23) 0.003 112 792 952 951 275 52 × 2 = 0 + 0.006 225 585 905 902 551 04;
  • 24) 0.006 225 585 905 902 551 04 × 2 = 0 + 0.012 451 171 811 805 102 08;
  • 25) 0.012 451 171 811 805 102 08 × 2 = 0 + 0.024 902 343 623 610 204 16;
  • 26) 0.024 902 343 623 610 204 16 × 2 = 0 + 0.049 804 687 247 220 408 32;
  • 27) 0.049 804 687 247 220 408 32 × 2 = 0 + 0.099 609 374 494 440 816 64;
  • 28) 0.099 609 374 494 440 816 64 × 2 = 0 + 0.199 218 748 988 881 633 28;
  • 29) 0.199 218 748 988 881 633 28 × 2 = 0 + 0.398 437 497 977 763 266 56;
  • 30) 0.398 437 497 977 763 266 56 × 2 = 0 + 0.796 874 995 955 526 533 12;
  • 31) 0.796 874 995 955 526 533 12 × 2 = 1 + 0.593 749 991 911 053 066 24;
  • 32) 0.593 749 991 911 053 066 24 × 2 = 1 + 0.187 499 983 822 106 132 48;
  • 33) 0.187 499 983 822 106 132 48 × 2 = 0 + 0.374 999 967 644 212 264 96;
  • 34) 0.374 999 967 644 212 264 96 × 2 = 0 + 0.749 999 935 288 424 529 92;
  • 35) 0.749 999 935 288 424 529 92 × 2 = 1 + 0.499 999 870 576 849 059 84;
  • 36) 0.499 999 870 576 849 059 84 × 2 = 0 + 0.999 999 741 153 698 119 68;
  • 37) 0.999 999 741 153 698 119 68 × 2 = 1 + 0.999 999 482 307 396 239 36;
  • 38) 0.999 999 482 307 396 239 36 × 2 = 1 + 0.999 998 964 614 792 478 72;
  • 39) 0.999 998 964 614 792 478 72 × 2 = 1 + 0.999 997 929 229 584 957 44;
  • 40) 0.999 997 929 229 584 957 44 × 2 = 1 + 0.999 995 858 459 169 914 88;
  • 41) 0.999 995 858 459 169 914 88 × 2 = 1 + 0.999 991 716 918 339 829 76;
  • 42) 0.999 991 716 918 339 829 76 × 2 = 1 + 0.999 983 433 836 679 659 52;
  • 43) 0.999 983 433 836 679 659 52 × 2 = 1 + 0.999 966 867 673 359 319 04;
  • 44) 0.999 966 867 673 359 319 04 × 2 = 1 + 0.999 933 735 346 718 638 08;
  • 45) 0.999 933 735 346 718 638 08 × 2 = 1 + 0.999 867 470 693 437 276 16;
  • 46) 0.999 867 470 693 437 276 16 × 2 = 1 + 0.999 734 941 386 874 552 32;
  • 47) 0.999 734 941 386 874 552 32 × 2 = 1 + 0.999 469 882 773 749 104 64;
  • 48) 0.999 469 882 773 749 104 64 × 2 = 1 + 0.998 939 765 547 498 209 28;
  • 49) 0.998 939 765 547 498 209 28 × 2 = 1 + 0.997 879 531 094 996 418 56;
  • 50) 0.997 879 531 094 996 418 56 × 2 = 1 + 0.995 759 062 189 992 837 12;
  • 51) 0.995 759 062 189 992 837 12 × 2 = 1 + 0.991 518 124 379 985 674 24;
  • 52) 0.991 518 124 379 985 674 24 × 2 = 1 + 0.983 036 248 759 971 348 48;
  • 53) 0.983 036 248 759 971 348 48 × 2 = 1 + 0.966 072 497 519 942 696 96;
  • 54) 0.966 072 497 519 942 696 96 × 2 = 1 + 0.932 144 995 039 885 393 92;

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:


0.000 000 000 742 147 672 88(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 672 88(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:


0.000 000 000 742 147 672 88(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 -0.000 000 000 742 147 672 88 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 673 81 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 minutes

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