-0.000 000 000 742 147 679 09 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 679 09(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
-0.000 000 000 742 147 679 09(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:

|-0.000 000 000 742 147 679 09| = 0.000 000 000 742 147 679 09


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: 0.000 000 000 742 147 679 09.

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) 0.000 000 000 742 147 679 09 × 2 = 0 + 0.000 000 001 484 295 358 18;
  • 2) 0.000 000 001 484 295 358 18 × 2 = 0 + 0.000 000 002 968 590 716 36;
  • 3) 0.000 000 002 968 590 716 36 × 2 = 0 + 0.000 000 005 937 181 432 72;
  • 4) 0.000 000 005 937 181 432 72 × 2 = 0 + 0.000 000 011 874 362 865 44;
  • 5) 0.000 000 011 874 362 865 44 × 2 = 0 + 0.000 000 023 748 725 730 88;
  • 6) 0.000 000 023 748 725 730 88 × 2 = 0 + 0.000 000 047 497 451 461 76;
  • 7) 0.000 000 047 497 451 461 76 × 2 = 0 + 0.000 000 094 994 902 923 52;
  • 8) 0.000 000 094 994 902 923 52 × 2 = 0 + 0.000 000 189 989 805 847 04;
  • 9) 0.000 000 189 989 805 847 04 × 2 = 0 + 0.000 000 379 979 611 694 08;
  • 10) 0.000 000 379 979 611 694 08 × 2 = 0 + 0.000 000 759 959 223 388 16;
  • 11) 0.000 000 759 959 223 388 16 × 2 = 0 + 0.000 001 519 918 446 776 32;
  • 12) 0.000 001 519 918 446 776 32 × 2 = 0 + 0.000 003 039 836 893 552 64;
  • 13) 0.000 003 039 836 893 552 64 × 2 = 0 + 0.000 006 079 673 787 105 28;
  • 14) 0.000 006 079 673 787 105 28 × 2 = 0 + 0.000 012 159 347 574 210 56;
  • 15) 0.000 012 159 347 574 210 56 × 2 = 0 + 0.000 024 318 695 148 421 12;
  • 16) 0.000 024 318 695 148 421 12 × 2 = 0 + 0.000 048 637 390 296 842 24;
  • 17) 0.000 048 637 390 296 842 24 × 2 = 0 + 0.000 097 274 780 593 684 48;
  • 18) 0.000 097 274 780 593 684 48 × 2 = 0 + 0.000 194 549 561 187 368 96;
  • 19) 0.000 194 549 561 187 368 96 × 2 = 0 + 0.000 389 099 122 374 737 92;
  • 20) 0.000 389 099 122 374 737 92 × 2 = 0 + 0.000 778 198 244 749 475 84;
  • 21) 0.000 778 198 244 749 475 84 × 2 = 0 + 0.001 556 396 489 498 951 68;
  • 22) 0.001 556 396 489 498 951 68 × 2 = 0 + 0.003 112 792 978 997 903 36;
  • 23) 0.003 112 792 978 997 903 36 × 2 = 0 + 0.006 225 585 957 995 806 72;
  • 24) 0.006 225 585 957 995 806 72 × 2 = 0 + 0.012 451 171 915 991 613 44;
  • 25) 0.012 451 171 915 991 613 44 × 2 = 0 + 0.024 902 343 831 983 226 88;
  • 26) 0.024 902 343 831 983 226 88 × 2 = 0 + 0.049 804 687 663 966 453 76;
  • 27) 0.049 804 687 663 966 453 76 × 2 = 0 + 0.099 609 375 327 932 907 52;
  • 28) 0.099 609 375 327 932 907 52 × 2 = 0 + 0.199 218 750 655 865 815 04;
  • 29) 0.199 218 750 655 865 815 04 × 2 = 0 + 0.398 437 501 311 731 630 08;
  • 30) 0.398 437 501 311 731 630 08 × 2 = 0 + 0.796 875 002 623 463 260 16;
  • 31) 0.796 875 002 623 463 260 16 × 2 = 1 + 0.593 750 005 246 926 520 32;
  • 32) 0.593 750 005 246 926 520 32 × 2 = 1 + 0.187 500 010 493 853 040 64;
  • 33) 0.187 500 010 493 853 040 64 × 2 = 0 + 0.375 000 020 987 706 081 28;
  • 34) 0.375 000 020 987 706 081 28 × 2 = 0 + 0.750 000 041 975 412 162 56;
  • 35) 0.750 000 041 975 412 162 56 × 2 = 1 + 0.500 000 083 950 824 325 12;
  • 36) 0.500 000 083 950 824 325 12 × 2 = 1 + 0.000 000 167 901 648 650 24;
  • 37) 0.000 000 167 901 648 650 24 × 2 = 0 + 0.000 000 335 803 297 300 48;
  • 38) 0.000 000 335 803 297 300 48 × 2 = 0 + 0.000 000 671 606 594 600 96;
  • 39) 0.000 000 671 606 594 600 96 × 2 = 0 + 0.000 001 343 213 189 201 92;
  • 40) 0.000 001 343 213 189 201 92 × 2 = 0 + 0.000 002 686 426 378 403 84;
  • 41) 0.000 002 686 426 378 403 84 × 2 = 0 + 0.000 005 372 852 756 807 68;
  • 42) 0.000 005 372 852 756 807 68 × 2 = 0 + 0.000 010 745 705 513 615 36;
  • 43) 0.000 010 745 705 513 615 36 × 2 = 0 + 0.000 021 491 411 027 230 72;
  • 44) 0.000 021 491 411 027 230 72 × 2 = 0 + 0.000 042 982 822 054 461 44;
  • 45) 0.000 042 982 822 054 461 44 × 2 = 0 + 0.000 085 965 644 108 922 88;
  • 46) 0.000 085 965 644 108 922 88 × 2 = 0 + 0.000 171 931 288 217 845 76;
  • 47) 0.000 171 931 288 217 845 76 × 2 = 0 + 0.000 343 862 576 435 691 52;
  • 48) 0.000 343 862 576 435 691 52 × 2 = 0 + 0.000 687 725 152 871 383 04;
  • 49) 0.000 687 725 152 871 383 04 × 2 = 0 + 0.001 375 450 305 742 766 08;
  • 50) 0.001 375 450 305 742 766 08 × 2 = 0 + 0.002 750 900 611 485 532 16;
  • 51) 0.002 750 900 611 485 532 16 × 2 = 0 + 0.005 501 801 222 971 064 32;
  • 52) 0.005 501 801 222 971 064 32 × 2 = 0 + 0.011 003 602 445 942 128 64;
  • 53) 0.011 003 602 445 942 128 64 × 2 = 0 + 0.022 007 204 891 884 257 28;
  • 54) 0.022 007 204 891 884 257 28 × 2 = 0 + 0.044 014 409 783 768 514 56;

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:


0.000 000 000 742 147 679 09(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 679 09(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:


0.000 000 000 742 147 679 09(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 -0.000 000 000 742 147 679 09 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