-0.000 000 000 742 147 676 646 710 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 710 8(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
-0.000 000 000 742 147 676 646 710 8(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:

|-0.000 000 000 742 147 676 646 710 8| = 0.000 000 000 742 147 676 646 710 8


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: 0.000 000 000 742 147 676 646 710 8.

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) 0.000 000 000 742 147 676 646 710 8 × 2 = 0 + 0.000 000 001 484 295 353 293 421 6;
  • 2) 0.000 000 001 484 295 353 293 421 6 × 2 = 0 + 0.000 000 002 968 590 706 586 843 2;
  • 3) 0.000 000 002 968 590 706 586 843 2 × 2 = 0 + 0.000 000 005 937 181 413 173 686 4;
  • 4) 0.000 000 005 937 181 413 173 686 4 × 2 = 0 + 0.000 000 011 874 362 826 347 372 8;
  • 5) 0.000 000 011 874 362 826 347 372 8 × 2 = 0 + 0.000 000 023 748 725 652 694 745 6;
  • 6) 0.000 000 023 748 725 652 694 745 6 × 2 = 0 + 0.000 000 047 497 451 305 389 491 2;
  • 7) 0.000 000 047 497 451 305 389 491 2 × 2 = 0 + 0.000 000 094 994 902 610 778 982 4;
  • 8) 0.000 000 094 994 902 610 778 982 4 × 2 = 0 + 0.000 000 189 989 805 221 557 964 8;
  • 9) 0.000 000 189 989 805 221 557 964 8 × 2 = 0 + 0.000 000 379 979 610 443 115 929 6;
  • 10) 0.000 000 379 979 610 443 115 929 6 × 2 = 0 + 0.000 000 759 959 220 886 231 859 2;
  • 11) 0.000 000 759 959 220 886 231 859 2 × 2 = 0 + 0.000 001 519 918 441 772 463 718 4;
  • 12) 0.000 001 519 918 441 772 463 718 4 × 2 = 0 + 0.000 003 039 836 883 544 927 436 8;
  • 13) 0.000 003 039 836 883 544 927 436 8 × 2 = 0 + 0.000 006 079 673 767 089 854 873 6;
  • 14) 0.000 006 079 673 767 089 854 873 6 × 2 = 0 + 0.000 012 159 347 534 179 709 747 2;
  • 15) 0.000 012 159 347 534 179 709 747 2 × 2 = 0 + 0.000 024 318 695 068 359 419 494 4;
  • 16) 0.000 024 318 695 068 359 419 494 4 × 2 = 0 + 0.000 048 637 390 136 718 838 988 8;
  • 17) 0.000 048 637 390 136 718 838 988 8 × 2 = 0 + 0.000 097 274 780 273 437 677 977 6;
  • 18) 0.000 097 274 780 273 437 677 977 6 × 2 = 0 + 0.000 194 549 560 546 875 355 955 2;
  • 19) 0.000 194 549 560 546 875 355 955 2 × 2 = 0 + 0.000 389 099 121 093 750 711 910 4;
  • 20) 0.000 389 099 121 093 750 711 910 4 × 2 = 0 + 0.000 778 198 242 187 501 423 820 8;
  • 21) 0.000 778 198 242 187 501 423 820 8 × 2 = 0 + 0.001 556 396 484 375 002 847 641 6;
  • 22) 0.001 556 396 484 375 002 847 641 6 × 2 = 0 + 0.003 112 792 968 750 005 695 283 2;
  • 23) 0.003 112 792 968 750 005 695 283 2 × 2 = 0 + 0.006 225 585 937 500 011 390 566 4;
  • 24) 0.006 225 585 937 500 011 390 566 4 × 2 = 0 + 0.012 451 171 875 000 022 781 132 8;
  • 25) 0.012 451 171 875 000 022 781 132 8 × 2 = 0 + 0.024 902 343 750 000 045 562 265 6;
  • 26) 0.024 902 343 750 000 045 562 265 6 × 2 = 0 + 0.049 804 687 500 000 091 124 531 2;
  • 27) 0.049 804 687 500 000 091 124 531 2 × 2 = 0 + 0.099 609 375 000 000 182 249 062 4;
  • 28) 0.099 609 375 000 000 182 249 062 4 × 2 = 0 + 0.199 218 750 000 000 364 498 124 8;
  • 29) 0.199 218 750 000 000 364 498 124 8 × 2 = 0 + 0.398 437 500 000 000 728 996 249 6;
  • 30) 0.398 437 500 000 000 728 996 249 6 × 2 = 0 + 0.796 875 000 000 001 457 992 499 2;
  • 31) 0.796 875 000 000 001 457 992 499 2 × 2 = 1 + 0.593 750 000 000 002 915 984 998 4;
  • 32) 0.593 750 000 000 002 915 984 998 4 × 2 = 1 + 0.187 500 000 000 005 831 969 996 8;
  • 33) 0.187 500 000 000 005 831 969 996 8 × 2 = 0 + 0.375 000 000 000 011 663 939 993 6;
  • 34) 0.375 000 000 000 011 663 939 993 6 × 2 = 0 + 0.750 000 000 000 023 327 879 987 2;
  • 35) 0.750 000 000 000 023 327 879 987 2 × 2 = 1 + 0.500 000 000 000 046 655 759 974 4;
  • 36) 0.500 000 000 000 046 655 759 974 4 × 2 = 1 + 0.000 000 000 000 093 311 519 948 8;
  • 37) 0.000 000 000 000 093 311 519 948 8 × 2 = 0 + 0.000 000 000 000 186 623 039 897 6;
  • 38) 0.000 000 000 000 186 623 039 897 6 × 2 = 0 + 0.000 000 000 000 373 246 079 795 2;
  • 39) 0.000 000 000 000 373 246 079 795 2 × 2 = 0 + 0.000 000 000 000 746 492 159 590 4;
  • 40) 0.000 000 000 000 746 492 159 590 4 × 2 = 0 + 0.000 000 000 001 492 984 319 180 8;
  • 41) 0.000 000 000 001 492 984 319 180 8 × 2 = 0 + 0.000 000 000 002 985 968 638 361 6;
  • 42) 0.000 000 000 002 985 968 638 361 6 × 2 = 0 + 0.000 000 000 005 971 937 276 723 2;
  • 43) 0.000 000 000 005 971 937 276 723 2 × 2 = 0 + 0.000 000 000 011 943 874 553 446 4;
  • 44) 0.000 000 000 011 943 874 553 446 4 × 2 = 0 + 0.000 000 000 023 887 749 106 892 8;
  • 45) 0.000 000 000 023 887 749 106 892 8 × 2 = 0 + 0.000 000 000 047 775 498 213 785 6;
  • 46) 0.000 000 000 047 775 498 213 785 6 × 2 = 0 + 0.000 000 000 095 550 996 427 571 2;
  • 47) 0.000 000 000 095 550 996 427 571 2 × 2 = 0 + 0.000 000 000 191 101 992 855 142 4;
  • 48) 0.000 000 000 191 101 992 855 142 4 × 2 = 0 + 0.000 000 000 382 203 985 710 284 8;
  • 49) 0.000 000 000 382 203 985 710 284 8 × 2 = 0 + 0.000 000 000 764 407 971 420 569 6;
  • 50) 0.000 000 000 764 407 971 420 569 6 × 2 = 0 + 0.000 000 001 528 815 942 841 139 2;
  • 51) 0.000 000 001 528 815 942 841 139 2 × 2 = 0 + 0.000 000 003 057 631 885 682 278 4;
  • 52) 0.000 000 003 057 631 885 682 278 4 × 2 = 0 + 0.000 000 006 115 263 771 364 556 8;
  • 53) 0.000 000 006 115 263 771 364 556 8 × 2 = 0 + 0.000 000 012 230 527 542 729 113 6;
  • 54) 0.000 000 012 230 527 542 729 113 6 × 2 = 0 + 0.000 000 024 461 055 085 458 227 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:


0.000 000 000 742 147 676 646 710 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 710 8(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:


0.000 000 000 742 147 676 646 710 8(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 -0.000 000 000 742 147 676 646 710 8 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