-0.000 000 000 742 147 676 935 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 935(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
-0.000 000 000 742 147 676 935(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:

|-0.000 000 000 742 147 676 935| = 0.000 000 000 742 147 676 935


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: 0.000 000 000 742 147 676 935.

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) 0.000 000 000 742 147 676 935 × 2 = 0 + 0.000 000 001 484 295 353 87;
  • 2) 0.000 000 001 484 295 353 87 × 2 = 0 + 0.000 000 002 968 590 707 74;
  • 3) 0.000 000 002 968 590 707 74 × 2 = 0 + 0.000 000 005 937 181 415 48;
  • 4) 0.000 000 005 937 181 415 48 × 2 = 0 + 0.000 000 011 874 362 830 96;
  • 5) 0.000 000 011 874 362 830 96 × 2 = 0 + 0.000 000 023 748 725 661 92;
  • 6) 0.000 000 023 748 725 661 92 × 2 = 0 + 0.000 000 047 497 451 323 84;
  • 7) 0.000 000 047 497 451 323 84 × 2 = 0 + 0.000 000 094 994 902 647 68;
  • 8) 0.000 000 094 994 902 647 68 × 2 = 0 + 0.000 000 189 989 805 295 36;
  • 9) 0.000 000 189 989 805 295 36 × 2 = 0 + 0.000 000 379 979 610 590 72;
  • 10) 0.000 000 379 979 610 590 72 × 2 = 0 + 0.000 000 759 959 221 181 44;
  • 11) 0.000 000 759 959 221 181 44 × 2 = 0 + 0.000 001 519 918 442 362 88;
  • 12) 0.000 001 519 918 442 362 88 × 2 = 0 + 0.000 003 039 836 884 725 76;
  • 13) 0.000 003 039 836 884 725 76 × 2 = 0 + 0.000 006 079 673 769 451 52;
  • 14) 0.000 006 079 673 769 451 52 × 2 = 0 + 0.000 012 159 347 538 903 04;
  • 15) 0.000 012 159 347 538 903 04 × 2 = 0 + 0.000 024 318 695 077 806 08;
  • 16) 0.000 024 318 695 077 806 08 × 2 = 0 + 0.000 048 637 390 155 612 16;
  • 17) 0.000 048 637 390 155 612 16 × 2 = 0 + 0.000 097 274 780 311 224 32;
  • 18) 0.000 097 274 780 311 224 32 × 2 = 0 + 0.000 194 549 560 622 448 64;
  • 19) 0.000 194 549 560 622 448 64 × 2 = 0 + 0.000 389 099 121 244 897 28;
  • 20) 0.000 389 099 121 244 897 28 × 2 = 0 + 0.000 778 198 242 489 794 56;
  • 21) 0.000 778 198 242 489 794 56 × 2 = 0 + 0.001 556 396 484 979 589 12;
  • 22) 0.001 556 396 484 979 589 12 × 2 = 0 + 0.003 112 792 969 959 178 24;
  • 23) 0.003 112 792 969 959 178 24 × 2 = 0 + 0.006 225 585 939 918 356 48;
  • 24) 0.006 225 585 939 918 356 48 × 2 = 0 + 0.012 451 171 879 836 712 96;
  • 25) 0.012 451 171 879 836 712 96 × 2 = 0 + 0.024 902 343 759 673 425 92;
  • 26) 0.024 902 343 759 673 425 92 × 2 = 0 + 0.049 804 687 519 346 851 84;
  • 27) 0.049 804 687 519 346 851 84 × 2 = 0 + 0.099 609 375 038 693 703 68;
  • 28) 0.099 609 375 038 693 703 68 × 2 = 0 + 0.199 218 750 077 387 407 36;
  • 29) 0.199 218 750 077 387 407 36 × 2 = 0 + 0.398 437 500 154 774 814 72;
  • 30) 0.398 437 500 154 774 814 72 × 2 = 0 + 0.796 875 000 309 549 629 44;
  • 31) 0.796 875 000 309 549 629 44 × 2 = 1 + 0.593 750 000 619 099 258 88;
  • 32) 0.593 750 000 619 099 258 88 × 2 = 1 + 0.187 500 001 238 198 517 76;
  • 33) 0.187 500 001 238 198 517 76 × 2 = 0 + 0.375 000 002 476 397 035 52;
  • 34) 0.375 000 002 476 397 035 52 × 2 = 0 + 0.750 000 004 952 794 071 04;
  • 35) 0.750 000 004 952 794 071 04 × 2 = 1 + 0.500 000 009 905 588 142 08;
  • 36) 0.500 000 009 905 588 142 08 × 2 = 1 + 0.000 000 019 811 176 284 16;
  • 37) 0.000 000 019 811 176 284 16 × 2 = 0 + 0.000 000 039 622 352 568 32;
  • 38) 0.000 000 039 622 352 568 32 × 2 = 0 + 0.000 000 079 244 705 136 64;
  • 39) 0.000 000 079 244 705 136 64 × 2 = 0 + 0.000 000 158 489 410 273 28;
  • 40) 0.000 000 158 489 410 273 28 × 2 = 0 + 0.000 000 316 978 820 546 56;
  • 41) 0.000 000 316 978 820 546 56 × 2 = 0 + 0.000 000 633 957 641 093 12;
  • 42) 0.000 000 633 957 641 093 12 × 2 = 0 + 0.000 001 267 915 282 186 24;
  • 43) 0.000 001 267 915 282 186 24 × 2 = 0 + 0.000 002 535 830 564 372 48;
  • 44) 0.000 002 535 830 564 372 48 × 2 = 0 + 0.000 005 071 661 128 744 96;
  • 45) 0.000 005 071 661 128 744 96 × 2 = 0 + 0.000 010 143 322 257 489 92;
  • 46) 0.000 010 143 322 257 489 92 × 2 = 0 + 0.000 020 286 644 514 979 84;
  • 47) 0.000 020 286 644 514 979 84 × 2 = 0 + 0.000 040 573 289 029 959 68;
  • 48) 0.000 040 573 289 029 959 68 × 2 = 0 + 0.000 081 146 578 059 919 36;
  • 49) 0.000 081 146 578 059 919 36 × 2 = 0 + 0.000 162 293 156 119 838 72;
  • 50) 0.000 162 293 156 119 838 72 × 2 = 0 + 0.000 324 586 312 239 677 44;
  • 51) 0.000 324 586 312 239 677 44 × 2 = 0 + 0.000 649 172 624 479 354 88;
  • 52) 0.000 649 172 624 479 354 88 × 2 = 0 + 0.001 298 345 248 958 709 76;
  • 53) 0.001 298 345 248 958 709 76 × 2 = 0 + 0.002 596 690 497 917 419 52;
  • 54) 0.002 596 690 497 917 419 52 × 2 = 0 + 0.005 193 380 995 834 839 04;

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:


0.000 000 000 742 147 676 935(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 935(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:


0.000 000 000 742 147 676 935(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 -0.000 000 000 742 147 676 935 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