-0.000 000 000 742 147 676 565 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 565(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
-0.000 000 000 742 147 676 565(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:

|-0.000 000 000 742 147 676 565| = 0.000 000 000 742 147 676 565


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: 0.000 000 000 742 147 676 565.

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) 0.000 000 000 742 147 676 565 × 2 = 0 + 0.000 000 001 484 295 353 13;
  • 2) 0.000 000 001 484 295 353 13 × 2 = 0 + 0.000 000 002 968 590 706 26;
  • 3) 0.000 000 002 968 590 706 26 × 2 = 0 + 0.000 000 005 937 181 412 52;
  • 4) 0.000 000 005 937 181 412 52 × 2 = 0 + 0.000 000 011 874 362 825 04;
  • 5) 0.000 000 011 874 362 825 04 × 2 = 0 + 0.000 000 023 748 725 650 08;
  • 6) 0.000 000 023 748 725 650 08 × 2 = 0 + 0.000 000 047 497 451 300 16;
  • 7) 0.000 000 047 497 451 300 16 × 2 = 0 + 0.000 000 094 994 902 600 32;
  • 8) 0.000 000 094 994 902 600 32 × 2 = 0 + 0.000 000 189 989 805 200 64;
  • 9) 0.000 000 189 989 805 200 64 × 2 = 0 + 0.000 000 379 979 610 401 28;
  • 10) 0.000 000 379 979 610 401 28 × 2 = 0 + 0.000 000 759 959 220 802 56;
  • 11) 0.000 000 759 959 220 802 56 × 2 = 0 + 0.000 001 519 918 441 605 12;
  • 12) 0.000 001 519 918 441 605 12 × 2 = 0 + 0.000 003 039 836 883 210 24;
  • 13) 0.000 003 039 836 883 210 24 × 2 = 0 + 0.000 006 079 673 766 420 48;
  • 14) 0.000 006 079 673 766 420 48 × 2 = 0 + 0.000 012 159 347 532 840 96;
  • 15) 0.000 012 159 347 532 840 96 × 2 = 0 + 0.000 024 318 695 065 681 92;
  • 16) 0.000 024 318 695 065 681 92 × 2 = 0 + 0.000 048 637 390 131 363 84;
  • 17) 0.000 048 637 390 131 363 84 × 2 = 0 + 0.000 097 274 780 262 727 68;
  • 18) 0.000 097 274 780 262 727 68 × 2 = 0 + 0.000 194 549 560 525 455 36;
  • 19) 0.000 194 549 560 525 455 36 × 2 = 0 + 0.000 389 099 121 050 910 72;
  • 20) 0.000 389 099 121 050 910 72 × 2 = 0 + 0.000 778 198 242 101 821 44;
  • 21) 0.000 778 198 242 101 821 44 × 2 = 0 + 0.001 556 396 484 203 642 88;
  • 22) 0.001 556 396 484 203 642 88 × 2 = 0 + 0.003 112 792 968 407 285 76;
  • 23) 0.003 112 792 968 407 285 76 × 2 = 0 + 0.006 225 585 936 814 571 52;
  • 24) 0.006 225 585 936 814 571 52 × 2 = 0 + 0.012 451 171 873 629 143 04;
  • 25) 0.012 451 171 873 629 143 04 × 2 = 0 + 0.024 902 343 747 258 286 08;
  • 26) 0.024 902 343 747 258 286 08 × 2 = 0 + 0.049 804 687 494 516 572 16;
  • 27) 0.049 804 687 494 516 572 16 × 2 = 0 + 0.099 609 374 989 033 144 32;
  • 28) 0.099 609 374 989 033 144 32 × 2 = 0 + 0.199 218 749 978 066 288 64;
  • 29) 0.199 218 749 978 066 288 64 × 2 = 0 + 0.398 437 499 956 132 577 28;
  • 30) 0.398 437 499 956 132 577 28 × 2 = 0 + 0.796 874 999 912 265 154 56;
  • 31) 0.796 874 999 912 265 154 56 × 2 = 1 + 0.593 749 999 824 530 309 12;
  • 32) 0.593 749 999 824 530 309 12 × 2 = 1 + 0.187 499 999 649 060 618 24;
  • 33) 0.187 499 999 649 060 618 24 × 2 = 0 + 0.374 999 999 298 121 236 48;
  • 34) 0.374 999 999 298 121 236 48 × 2 = 0 + 0.749 999 998 596 242 472 96;
  • 35) 0.749 999 998 596 242 472 96 × 2 = 1 + 0.499 999 997 192 484 945 92;
  • 36) 0.499 999 997 192 484 945 92 × 2 = 0 + 0.999 999 994 384 969 891 84;
  • 37) 0.999 999 994 384 969 891 84 × 2 = 1 + 0.999 999 988 769 939 783 68;
  • 38) 0.999 999 988 769 939 783 68 × 2 = 1 + 0.999 999 977 539 879 567 36;
  • 39) 0.999 999 977 539 879 567 36 × 2 = 1 + 0.999 999 955 079 759 134 72;
  • 40) 0.999 999 955 079 759 134 72 × 2 = 1 + 0.999 999 910 159 518 269 44;
  • 41) 0.999 999 910 159 518 269 44 × 2 = 1 + 0.999 999 820 319 036 538 88;
  • 42) 0.999 999 820 319 036 538 88 × 2 = 1 + 0.999 999 640 638 073 077 76;
  • 43) 0.999 999 640 638 073 077 76 × 2 = 1 + 0.999 999 281 276 146 155 52;
  • 44) 0.999 999 281 276 146 155 52 × 2 = 1 + 0.999 998 562 552 292 311 04;
  • 45) 0.999 998 562 552 292 311 04 × 2 = 1 + 0.999 997 125 104 584 622 08;
  • 46) 0.999 997 125 104 584 622 08 × 2 = 1 + 0.999 994 250 209 169 244 16;
  • 47) 0.999 994 250 209 169 244 16 × 2 = 1 + 0.999 988 500 418 338 488 32;
  • 48) 0.999 988 500 418 338 488 32 × 2 = 1 + 0.999 977 000 836 676 976 64;
  • 49) 0.999 977 000 836 676 976 64 × 2 = 1 + 0.999 954 001 673 353 953 28;
  • 50) 0.999 954 001 673 353 953 28 × 2 = 1 + 0.999 908 003 346 707 906 56;
  • 51) 0.999 908 003 346 707 906 56 × 2 = 1 + 0.999 816 006 693 415 813 12;
  • 52) 0.999 816 006 693 415 813 12 × 2 = 1 + 0.999 632 013 386 831 626 24;
  • 53) 0.999 632 013 386 831 626 24 × 2 = 1 + 0.999 264 026 773 663 252 48;
  • 54) 0.999 264 026 773 663 252 48 × 2 = 1 + 0.998 528 053 547 326 504 96;

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:


0.000 000 000 742 147 676 565(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 565(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:


0.000 000 000 742 147 676 565(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 -0.000 000 000 742 147 676 565 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 493 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