-0.000 000 000 742 147 676 179 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 179(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
-0.000 000 000 742 147 676 179(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:

|-0.000 000 000 742 147 676 179| = 0.000 000 000 742 147 676 179


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: 0.000 000 000 742 147 676 179.

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) 0.000 000 000 742 147 676 179 × 2 = 0 + 0.000 000 001 484 295 352 358;
  • 2) 0.000 000 001 484 295 352 358 × 2 = 0 + 0.000 000 002 968 590 704 716;
  • 3) 0.000 000 002 968 590 704 716 × 2 = 0 + 0.000 000 005 937 181 409 432;
  • 4) 0.000 000 005 937 181 409 432 × 2 = 0 + 0.000 000 011 874 362 818 864;
  • 5) 0.000 000 011 874 362 818 864 × 2 = 0 + 0.000 000 023 748 725 637 728;
  • 6) 0.000 000 023 748 725 637 728 × 2 = 0 + 0.000 000 047 497 451 275 456;
  • 7) 0.000 000 047 497 451 275 456 × 2 = 0 + 0.000 000 094 994 902 550 912;
  • 8) 0.000 000 094 994 902 550 912 × 2 = 0 + 0.000 000 189 989 805 101 824;
  • 9) 0.000 000 189 989 805 101 824 × 2 = 0 + 0.000 000 379 979 610 203 648;
  • 10) 0.000 000 379 979 610 203 648 × 2 = 0 + 0.000 000 759 959 220 407 296;
  • 11) 0.000 000 759 959 220 407 296 × 2 = 0 + 0.000 001 519 918 440 814 592;
  • 12) 0.000 001 519 918 440 814 592 × 2 = 0 + 0.000 003 039 836 881 629 184;
  • 13) 0.000 003 039 836 881 629 184 × 2 = 0 + 0.000 006 079 673 763 258 368;
  • 14) 0.000 006 079 673 763 258 368 × 2 = 0 + 0.000 012 159 347 526 516 736;
  • 15) 0.000 012 159 347 526 516 736 × 2 = 0 + 0.000 024 318 695 053 033 472;
  • 16) 0.000 024 318 695 053 033 472 × 2 = 0 + 0.000 048 637 390 106 066 944;
  • 17) 0.000 048 637 390 106 066 944 × 2 = 0 + 0.000 097 274 780 212 133 888;
  • 18) 0.000 097 274 780 212 133 888 × 2 = 0 + 0.000 194 549 560 424 267 776;
  • 19) 0.000 194 549 560 424 267 776 × 2 = 0 + 0.000 389 099 120 848 535 552;
  • 20) 0.000 389 099 120 848 535 552 × 2 = 0 + 0.000 778 198 241 697 071 104;
  • 21) 0.000 778 198 241 697 071 104 × 2 = 0 + 0.001 556 396 483 394 142 208;
  • 22) 0.001 556 396 483 394 142 208 × 2 = 0 + 0.003 112 792 966 788 284 416;
  • 23) 0.003 112 792 966 788 284 416 × 2 = 0 + 0.006 225 585 933 576 568 832;
  • 24) 0.006 225 585 933 576 568 832 × 2 = 0 + 0.012 451 171 867 153 137 664;
  • 25) 0.012 451 171 867 153 137 664 × 2 = 0 + 0.024 902 343 734 306 275 328;
  • 26) 0.024 902 343 734 306 275 328 × 2 = 0 + 0.049 804 687 468 612 550 656;
  • 27) 0.049 804 687 468 612 550 656 × 2 = 0 + 0.099 609 374 937 225 101 312;
  • 28) 0.099 609 374 937 225 101 312 × 2 = 0 + 0.199 218 749 874 450 202 624;
  • 29) 0.199 218 749 874 450 202 624 × 2 = 0 + 0.398 437 499 748 900 405 248;
  • 30) 0.398 437 499 748 900 405 248 × 2 = 0 + 0.796 874 999 497 800 810 496;
  • 31) 0.796 874 999 497 800 810 496 × 2 = 1 + 0.593 749 998 995 601 620 992;
  • 32) 0.593 749 998 995 601 620 992 × 2 = 1 + 0.187 499 997 991 203 241 984;
  • 33) 0.187 499 997 991 203 241 984 × 2 = 0 + 0.374 999 995 982 406 483 968;
  • 34) 0.374 999 995 982 406 483 968 × 2 = 0 + 0.749 999 991 964 812 967 936;
  • 35) 0.749 999 991 964 812 967 936 × 2 = 1 + 0.499 999 983 929 625 935 872;
  • 36) 0.499 999 983 929 625 935 872 × 2 = 0 + 0.999 999 967 859 251 871 744;
  • 37) 0.999 999 967 859 251 871 744 × 2 = 1 + 0.999 999 935 718 503 743 488;
  • 38) 0.999 999 935 718 503 743 488 × 2 = 1 + 0.999 999 871 437 007 486 976;
  • 39) 0.999 999 871 437 007 486 976 × 2 = 1 + 0.999 999 742 874 014 973 952;
  • 40) 0.999 999 742 874 014 973 952 × 2 = 1 + 0.999 999 485 748 029 947 904;
  • 41) 0.999 999 485 748 029 947 904 × 2 = 1 + 0.999 998 971 496 059 895 808;
  • 42) 0.999 998 971 496 059 895 808 × 2 = 1 + 0.999 997 942 992 119 791 616;
  • 43) 0.999 997 942 992 119 791 616 × 2 = 1 + 0.999 995 885 984 239 583 232;
  • 44) 0.999 995 885 984 239 583 232 × 2 = 1 + 0.999 991 771 968 479 166 464;
  • 45) 0.999 991 771 968 479 166 464 × 2 = 1 + 0.999 983 543 936 958 332 928;
  • 46) 0.999 983 543 936 958 332 928 × 2 = 1 + 0.999 967 087 873 916 665 856;
  • 47) 0.999 967 087 873 916 665 856 × 2 = 1 + 0.999 934 175 747 833 331 712;
  • 48) 0.999 934 175 747 833 331 712 × 2 = 1 + 0.999 868 351 495 666 663 424;
  • 49) 0.999 868 351 495 666 663 424 × 2 = 1 + 0.999 736 702 991 333 326 848;
  • 50) 0.999 736 702 991 333 326 848 × 2 = 1 + 0.999 473 405 982 666 653 696;
  • 51) 0.999 473 405 982 666 653 696 × 2 = 1 + 0.998 946 811 965 333 307 392;
  • 52) 0.998 946 811 965 333 307 392 × 2 = 1 + 0.997 893 623 930 666 614 784;
  • 53) 0.997 893 623 930 666 614 784 × 2 = 1 + 0.995 787 247 861 333 229 568;
  • 54) 0.995 787 247 861 333 229 568 × 2 = 1 + 0.991 574 495 722 666 459 136;

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:


0.000 000 000 742 147 676 179(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 179(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:


0.000 000 000 742 147 676 179(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 -0.000 000 000 742 147 676 179 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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