-0.000 000 000 742 147 676 177 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 177(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
-0.000 000 000 742 147 676 177(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:

|-0.000 000 000 742 147 676 177| = 0.000 000 000 742 147 676 177


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: 0.000 000 000 742 147 676 177.

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) 0.000 000 000 742 147 676 177 × 2 = 0 + 0.000 000 001 484 295 352 354;
  • 2) 0.000 000 001 484 295 352 354 × 2 = 0 + 0.000 000 002 968 590 704 708;
  • 3) 0.000 000 002 968 590 704 708 × 2 = 0 + 0.000 000 005 937 181 409 416;
  • 4) 0.000 000 005 937 181 409 416 × 2 = 0 + 0.000 000 011 874 362 818 832;
  • 5) 0.000 000 011 874 362 818 832 × 2 = 0 + 0.000 000 023 748 725 637 664;
  • 6) 0.000 000 023 748 725 637 664 × 2 = 0 + 0.000 000 047 497 451 275 328;
  • 7) 0.000 000 047 497 451 275 328 × 2 = 0 + 0.000 000 094 994 902 550 656;
  • 8) 0.000 000 094 994 902 550 656 × 2 = 0 + 0.000 000 189 989 805 101 312;
  • 9) 0.000 000 189 989 805 101 312 × 2 = 0 + 0.000 000 379 979 610 202 624;
  • 10) 0.000 000 379 979 610 202 624 × 2 = 0 + 0.000 000 759 959 220 405 248;
  • 11) 0.000 000 759 959 220 405 248 × 2 = 0 + 0.000 001 519 918 440 810 496;
  • 12) 0.000 001 519 918 440 810 496 × 2 = 0 + 0.000 003 039 836 881 620 992;
  • 13) 0.000 003 039 836 881 620 992 × 2 = 0 + 0.000 006 079 673 763 241 984;
  • 14) 0.000 006 079 673 763 241 984 × 2 = 0 + 0.000 012 159 347 526 483 968;
  • 15) 0.000 012 159 347 526 483 968 × 2 = 0 + 0.000 024 318 695 052 967 936;
  • 16) 0.000 024 318 695 052 967 936 × 2 = 0 + 0.000 048 637 390 105 935 872;
  • 17) 0.000 048 637 390 105 935 872 × 2 = 0 + 0.000 097 274 780 211 871 744;
  • 18) 0.000 097 274 780 211 871 744 × 2 = 0 + 0.000 194 549 560 423 743 488;
  • 19) 0.000 194 549 560 423 743 488 × 2 = 0 + 0.000 389 099 120 847 486 976;
  • 20) 0.000 389 099 120 847 486 976 × 2 = 0 + 0.000 778 198 241 694 973 952;
  • 21) 0.000 778 198 241 694 973 952 × 2 = 0 + 0.001 556 396 483 389 947 904;
  • 22) 0.001 556 396 483 389 947 904 × 2 = 0 + 0.003 112 792 966 779 895 808;
  • 23) 0.003 112 792 966 779 895 808 × 2 = 0 + 0.006 225 585 933 559 791 616;
  • 24) 0.006 225 585 933 559 791 616 × 2 = 0 + 0.012 451 171 867 119 583 232;
  • 25) 0.012 451 171 867 119 583 232 × 2 = 0 + 0.024 902 343 734 239 166 464;
  • 26) 0.024 902 343 734 239 166 464 × 2 = 0 + 0.049 804 687 468 478 332 928;
  • 27) 0.049 804 687 468 478 332 928 × 2 = 0 + 0.099 609 374 936 956 665 856;
  • 28) 0.099 609 374 936 956 665 856 × 2 = 0 + 0.199 218 749 873 913 331 712;
  • 29) 0.199 218 749 873 913 331 712 × 2 = 0 + 0.398 437 499 747 826 663 424;
  • 30) 0.398 437 499 747 826 663 424 × 2 = 0 + 0.796 874 999 495 653 326 848;
  • 31) 0.796 874 999 495 653 326 848 × 2 = 1 + 0.593 749 998 991 306 653 696;
  • 32) 0.593 749 998 991 306 653 696 × 2 = 1 + 0.187 499 997 982 613 307 392;
  • 33) 0.187 499 997 982 613 307 392 × 2 = 0 + 0.374 999 995 965 226 614 784;
  • 34) 0.374 999 995 965 226 614 784 × 2 = 0 + 0.749 999 991 930 453 229 568;
  • 35) 0.749 999 991 930 453 229 568 × 2 = 1 + 0.499 999 983 860 906 459 136;
  • 36) 0.499 999 983 860 906 459 136 × 2 = 0 + 0.999 999 967 721 812 918 272;
  • 37) 0.999 999 967 721 812 918 272 × 2 = 1 + 0.999 999 935 443 625 836 544;
  • 38) 0.999 999 935 443 625 836 544 × 2 = 1 + 0.999 999 870 887 251 673 088;
  • 39) 0.999 999 870 887 251 673 088 × 2 = 1 + 0.999 999 741 774 503 346 176;
  • 40) 0.999 999 741 774 503 346 176 × 2 = 1 + 0.999 999 483 549 006 692 352;
  • 41) 0.999 999 483 549 006 692 352 × 2 = 1 + 0.999 998 967 098 013 384 704;
  • 42) 0.999 998 967 098 013 384 704 × 2 = 1 + 0.999 997 934 196 026 769 408;
  • 43) 0.999 997 934 196 026 769 408 × 2 = 1 + 0.999 995 868 392 053 538 816;
  • 44) 0.999 995 868 392 053 538 816 × 2 = 1 + 0.999 991 736 784 107 077 632;
  • 45) 0.999 991 736 784 107 077 632 × 2 = 1 + 0.999 983 473 568 214 155 264;
  • 46) 0.999 983 473 568 214 155 264 × 2 = 1 + 0.999 966 947 136 428 310 528;
  • 47) 0.999 966 947 136 428 310 528 × 2 = 1 + 0.999 933 894 272 856 621 056;
  • 48) 0.999 933 894 272 856 621 056 × 2 = 1 + 0.999 867 788 545 713 242 112;
  • 49) 0.999 867 788 545 713 242 112 × 2 = 1 + 0.999 735 577 091 426 484 224;
  • 50) 0.999 735 577 091 426 484 224 × 2 = 1 + 0.999 471 154 182 852 968 448;
  • 51) 0.999 471 154 182 852 968 448 × 2 = 1 + 0.998 942 308 365 705 936 896;
  • 52) 0.998 942 308 365 705 936 896 × 2 = 1 + 0.997 884 616 731 411 873 792;
  • 53) 0.997 884 616 731 411 873 792 × 2 = 1 + 0.995 769 233 462 823 747 584;
  • 54) 0.995 769 233 462 823 747 584 × 2 = 1 + 0.991 538 466 925 647 495 168;

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:


0.000 000 000 742 147 676 177(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 177(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:


0.000 000 000 742 147 676 177(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 -0.000 000 000 742 147 676 177 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