-0.000 000 000 742 147 695 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 695 3(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
-0.000 000 000 742 147 695 3(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:

|-0.000 000 000 742 147 695 3| = 0.000 000 000 742 147 695 3


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: 0.000 000 000 742 147 695 3.

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) 0.000 000 000 742 147 695 3 × 2 = 0 + 0.000 000 001 484 295 390 6;
  • 2) 0.000 000 001 484 295 390 6 × 2 = 0 + 0.000 000 002 968 590 781 2;
  • 3) 0.000 000 002 968 590 781 2 × 2 = 0 + 0.000 000 005 937 181 562 4;
  • 4) 0.000 000 005 937 181 562 4 × 2 = 0 + 0.000 000 011 874 363 124 8;
  • 5) 0.000 000 011 874 363 124 8 × 2 = 0 + 0.000 000 023 748 726 249 6;
  • 6) 0.000 000 023 748 726 249 6 × 2 = 0 + 0.000 000 047 497 452 499 2;
  • 7) 0.000 000 047 497 452 499 2 × 2 = 0 + 0.000 000 094 994 904 998 4;
  • 8) 0.000 000 094 994 904 998 4 × 2 = 0 + 0.000 000 189 989 809 996 8;
  • 9) 0.000 000 189 989 809 996 8 × 2 = 0 + 0.000 000 379 979 619 993 6;
  • 10) 0.000 000 379 979 619 993 6 × 2 = 0 + 0.000 000 759 959 239 987 2;
  • 11) 0.000 000 759 959 239 987 2 × 2 = 0 + 0.000 001 519 918 479 974 4;
  • 12) 0.000 001 519 918 479 974 4 × 2 = 0 + 0.000 003 039 836 959 948 8;
  • 13) 0.000 003 039 836 959 948 8 × 2 = 0 + 0.000 006 079 673 919 897 6;
  • 14) 0.000 006 079 673 919 897 6 × 2 = 0 + 0.000 012 159 347 839 795 2;
  • 15) 0.000 012 159 347 839 795 2 × 2 = 0 + 0.000 024 318 695 679 590 4;
  • 16) 0.000 024 318 695 679 590 4 × 2 = 0 + 0.000 048 637 391 359 180 8;
  • 17) 0.000 048 637 391 359 180 8 × 2 = 0 + 0.000 097 274 782 718 361 6;
  • 18) 0.000 097 274 782 718 361 6 × 2 = 0 + 0.000 194 549 565 436 723 2;
  • 19) 0.000 194 549 565 436 723 2 × 2 = 0 + 0.000 389 099 130 873 446 4;
  • 20) 0.000 389 099 130 873 446 4 × 2 = 0 + 0.000 778 198 261 746 892 8;
  • 21) 0.000 778 198 261 746 892 8 × 2 = 0 + 0.001 556 396 523 493 785 6;
  • 22) 0.001 556 396 523 493 785 6 × 2 = 0 + 0.003 112 793 046 987 571 2;
  • 23) 0.003 112 793 046 987 571 2 × 2 = 0 + 0.006 225 586 093 975 142 4;
  • 24) 0.006 225 586 093 975 142 4 × 2 = 0 + 0.012 451 172 187 950 284 8;
  • 25) 0.012 451 172 187 950 284 8 × 2 = 0 + 0.024 902 344 375 900 569 6;
  • 26) 0.024 902 344 375 900 569 6 × 2 = 0 + 0.049 804 688 751 801 139 2;
  • 27) 0.049 804 688 751 801 139 2 × 2 = 0 + 0.099 609 377 503 602 278 4;
  • 28) 0.099 609 377 503 602 278 4 × 2 = 0 + 0.199 218 755 007 204 556 8;
  • 29) 0.199 218 755 007 204 556 8 × 2 = 0 + 0.398 437 510 014 409 113 6;
  • 30) 0.398 437 510 014 409 113 6 × 2 = 0 + 0.796 875 020 028 818 227 2;
  • 31) 0.796 875 020 028 818 227 2 × 2 = 1 + 0.593 750 040 057 636 454 4;
  • 32) 0.593 750 040 057 636 454 4 × 2 = 1 + 0.187 500 080 115 272 908 8;
  • 33) 0.187 500 080 115 272 908 8 × 2 = 0 + 0.375 000 160 230 545 817 6;
  • 34) 0.375 000 160 230 545 817 6 × 2 = 0 + 0.750 000 320 461 091 635 2;
  • 35) 0.750 000 320 461 091 635 2 × 2 = 1 + 0.500 000 640 922 183 270 4;
  • 36) 0.500 000 640 922 183 270 4 × 2 = 1 + 0.000 001 281 844 366 540 8;
  • 37) 0.000 001 281 844 366 540 8 × 2 = 0 + 0.000 002 563 688 733 081 6;
  • 38) 0.000 002 563 688 733 081 6 × 2 = 0 + 0.000 005 127 377 466 163 2;
  • 39) 0.000 005 127 377 466 163 2 × 2 = 0 + 0.000 010 254 754 932 326 4;
  • 40) 0.000 010 254 754 932 326 4 × 2 = 0 + 0.000 020 509 509 864 652 8;
  • 41) 0.000 020 509 509 864 652 8 × 2 = 0 + 0.000 041 019 019 729 305 6;
  • 42) 0.000 041 019 019 729 305 6 × 2 = 0 + 0.000 082 038 039 458 611 2;
  • 43) 0.000 082 038 039 458 611 2 × 2 = 0 + 0.000 164 076 078 917 222 4;
  • 44) 0.000 164 076 078 917 222 4 × 2 = 0 + 0.000 328 152 157 834 444 8;
  • 45) 0.000 328 152 157 834 444 8 × 2 = 0 + 0.000 656 304 315 668 889 6;
  • 46) 0.000 656 304 315 668 889 6 × 2 = 0 + 0.001 312 608 631 337 779 2;
  • 47) 0.001 312 608 631 337 779 2 × 2 = 0 + 0.002 625 217 262 675 558 4;
  • 48) 0.002 625 217 262 675 558 4 × 2 = 0 + 0.005 250 434 525 351 116 8;
  • 49) 0.005 250 434 525 351 116 8 × 2 = 0 + 0.010 500 869 050 702 233 6;
  • 50) 0.010 500 869 050 702 233 6 × 2 = 0 + 0.021 001 738 101 404 467 2;
  • 51) 0.021 001 738 101 404 467 2 × 2 = 0 + 0.042 003 476 202 808 934 4;
  • 52) 0.042 003 476 202 808 934 4 × 2 = 0 + 0.084 006 952 405 617 868 8;
  • 53) 0.084 006 952 405 617 868 8 × 2 = 0 + 0.168 013 904 811 235 737 6;
  • 54) 0.168 013 904 811 235 737 6 × 2 = 0 + 0.336 027 809 622 471 475 2;

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:


0.000 000 000 742 147 695 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 695 3(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:


0.000 000 000 742 147 695 3(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 -0.000 000 000 742 147 695 3 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