-0.000 000 000 742 147 676 269 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 269(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
-0.000 000 000 742 147 676 269(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:

|-0.000 000 000 742 147 676 269| = 0.000 000 000 742 147 676 269


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: 0.000 000 000 742 147 676 269.

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) 0.000 000 000 742 147 676 269 × 2 = 0 + 0.000 000 001 484 295 352 538;
  • 2) 0.000 000 001 484 295 352 538 × 2 = 0 + 0.000 000 002 968 590 705 076;
  • 3) 0.000 000 002 968 590 705 076 × 2 = 0 + 0.000 000 005 937 181 410 152;
  • 4) 0.000 000 005 937 181 410 152 × 2 = 0 + 0.000 000 011 874 362 820 304;
  • 5) 0.000 000 011 874 362 820 304 × 2 = 0 + 0.000 000 023 748 725 640 608;
  • 6) 0.000 000 023 748 725 640 608 × 2 = 0 + 0.000 000 047 497 451 281 216;
  • 7) 0.000 000 047 497 451 281 216 × 2 = 0 + 0.000 000 094 994 902 562 432;
  • 8) 0.000 000 094 994 902 562 432 × 2 = 0 + 0.000 000 189 989 805 124 864;
  • 9) 0.000 000 189 989 805 124 864 × 2 = 0 + 0.000 000 379 979 610 249 728;
  • 10) 0.000 000 379 979 610 249 728 × 2 = 0 + 0.000 000 759 959 220 499 456;
  • 11) 0.000 000 759 959 220 499 456 × 2 = 0 + 0.000 001 519 918 440 998 912;
  • 12) 0.000 001 519 918 440 998 912 × 2 = 0 + 0.000 003 039 836 881 997 824;
  • 13) 0.000 003 039 836 881 997 824 × 2 = 0 + 0.000 006 079 673 763 995 648;
  • 14) 0.000 006 079 673 763 995 648 × 2 = 0 + 0.000 012 159 347 527 991 296;
  • 15) 0.000 012 159 347 527 991 296 × 2 = 0 + 0.000 024 318 695 055 982 592;
  • 16) 0.000 024 318 695 055 982 592 × 2 = 0 + 0.000 048 637 390 111 965 184;
  • 17) 0.000 048 637 390 111 965 184 × 2 = 0 + 0.000 097 274 780 223 930 368;
  • 18) 0.000 097 274 780 223 930 368 × 2 = 0 + 0.000 194 549 560 447 860 736;
  • 19) 0.000 194 549 560 447 860 736 × 2 = 0 + 0.000 389 099 120 895 721 472;
  • 20) 0.000 389 099 120 895 721 472 × 2 = 0 + 0.000 778 198 241 791 442 944;
  • 21) 0.000 778 198 241 791 442 944 × 2 = 0 + 0.001 556 396 483 582 885 888;
  • 22) 0.001 556 396 483 582 885 888 × 2 = 0 + 0.003 112 792 967 165 771 776;
  • 23) 0.003 112 792 967 165 771 776 × 2 = 0 + 0.006 225 585 934 331 543 552;
  • 24) 0.006 225 585 934 331 543 552 × 2 = 0 + 0.012 451 171 868 663 087 104;
  • 25) 0.012 451 171 868 663 087 104 × 2 = 0 + 0.024 902 343 737 326 174 208;
  • 26) 0.024 902 343 737 326 174 208 × 2 = 0 + 0.049 804 687 474 652 348 416;
  • 27) 0.049 804 687 474 652 348 416 × 2 = 0 + 0.099 609 374 949 304 696 832;
  • 28) 0.099 609 374 949 304 696 832 × 2 = 0 + 0.199 218 749 898 609 393 664;
  • 29) 0.199 218 749 898 609 393 664 × 2 = 0 + 0.398 437 499 797 218 787 328;
  • 30) 0.398 437 499 797 218 787 328 × 2 = 0 + 0.796 874 999 594 437 574 656;
  • 31) 0.796 874 999 594 437 574 656 × 2 = 1 + 0.593 749 999 188 875 149 312;
  • 32) 0.593 749 999 188 875 149 312 × 2 = 1 + 0.187 499 998 377 750 298 624;
  • 33) 0.187 499 998 377 750 298 624 × 2 = 0 + 0.374 999 996 755 500 597 248;
  • 34) 0.374 999 996 755 500 597 248 × 2 = 0 + 0.749 999 993 511 001 194 496;
  • 35) 0.749 999 993 511 001 194 496 × 2 = 1 + 0.499 999 987 022 002 388 992;
  • 36) 0.499 999 987 022 002 388 992 × 2 = 0 + 0.999 999 974 044 004 777 984;
  • 37) 0.999 999 974 044 004 777 984 × 2 = 1 + 0.999 999 948 088 009 555 968;
  • 38) 0.999 999 948 088 009 555 968 × 2 = 1 + 0.999 999 896 176 019 111 936;
  • 39) 0.999 999 896 176 019 111 936 × 2 = 1 + 0.999 999 792 352 038 223 872;
  • 40) 0.999 999 792 352 038 223 872 × 2 = 1 + 0.999 999 584 704 076 447 744;
  • 41) 0.999 999 584 704 076 447 744 × 2 = 1 + 0.999 999 169 408 152 895 488;
  • 42) 0.999 999 169 408 152 895 488 × 2 = 1 + 0.999 998 338 816 305 790 976;
  • 43) 0.999 998 338 816 305 790 976 × 2 = 1 + 0.999 996 677 632 611 581 952;
  • 44) 0.999 996 677 632 611 581 952 × 2 = 1 + 0.999 993 355 265 223 163 904;
  • 45) 0.999 993 355 265 223 163 904 × 2 = 1 + 0.999 986 710 530 446 327 808;
  • 46) 0.999 986 710 530 446 327 808 × 2 = 1 + 0.999 973 421 060 892 655 616;
  • 47) 0.999 973 421 060 892 655 616 × 2 = 1 + 0.999 946 842 121 785 311 232;
  • 48) 0.999 946 842 121 785 311 232 × 2 = 1 + 0.999 893 684 243 570 622 464;
  • 49) 0.999 893 684 243 570 622 464 × 2 = 1 + 0.999 787 368 487 141 244 928;
  • 50) 0.999 787 368 487 141 244 928 × 2 = 1 + 0.999 574 736 974 282 489 856;
  • 51) 0.999 574 736 974 282 489 856 × 2 = 1 + 0.999 149 473 948 564 979 712;
  • 52) 0.999 149 473 948 564 979 712 × 2 = 1 + 0.998 298 947 897 129 959 424;
  • 53) 0.998 298 947 897 129 959 424 × 2 = 1 + 0.996 597 895 794 259 918 848;
  • 54) 0.996 597 895 794 259 918 848 × 2 = 1 + 0.993 195 791 588 519 837 696;

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:


0.000 000 000 742 147 676 269(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 269(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:


0.000 000 000 742 147 676 269(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 -0.000 000 000 742 147 676 269 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 357 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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