-0.000 000 000 742 147 676 023 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 023(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
-0.000 000 000 742 147 676 023(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:

|-0.000 000 000 742 147 676 023| = 0.000 000 000 742 147 676 023


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: 0.000 000 000 742 147 676 023.

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) 0.000 000 000 742 147 676 023 × 2 = 0 + 0.000 000 001 484 295 352 046;
  • 2) 0.000 000 001 484 295 352 046 × 2 = 0 + 0.000 000 002 968 590 704 092;
  • 3) 0.000 000 002 968 590 704 092 × 2 = 0 + 0.000 000 005 937 181 408 184;
  • 4) 0.000 000 005 937 181 408 184 × 2 = 0 + 0.000 000 011 874 362 816 368;
  • 5) 0.000 000 011 874 362 816 368 × 2 = 0 + 0.000 000 023 748 725 632 736;
  • 6) 0.000 000 023 748 725 632 736 × 2 = 0 + 0.000 000 047 497 451 265 472;
  • 7) 0.000 000 047 497 451 265 472 × 2 = 0 + 0.000 000 094 994 902 530 944;
  • 8) 0.000 000 094 994 902 530 944 × 2 = 0 + 0.000 000 189 989 805 061 888;
  • 9) 0.000 000 189 989 805 061 888 × 2 = 0 + 0.000 000 379 979 610 123 776;
  • 10) 0.000 000 379 979 610 123 776 × 2 = 0 + 0.000 000 759 959 220 247 552;
  • 11) 0.000 000 759 959 220 247 552 × 2 = 0 + 0.000 001 519 918 440 495 104;
  • 12) 0.000 001 519 918 440 495 104 × 2 = 0 + 0.000 003 039 836 880 990 208;
  • 13) 0.000 003 039 836 880 990 208 × 2 = 0 + 0.000 006 079 673 761 980 416;
  • 14) 0.000 006 079 673 761 980 416 × 2 = 0 + 0.000 012 159 347 523 960 832;
  • 15) 0.000 012 159 347 523 960 832 × 2 = 0 + 0.000 024 318 695 047 921 664;
  • 16) 0.000 024 318 695 047 921 664 × 2 = 0 + 0.000 048 637 390 095 843 328;
  • 17) 0.000 048 637 390 095 843 328 × 2 = 0 + 0.000 097 274 780 191 686 656;
  • 18) 0.000 097 274 780 191 686 656 × 2 = 0 + 0.000 194 549 560 383 373 312;
  • 19) 0.000 194 549 560 383 373 312 × 2 = 0 + 0.000 389 099 120 766 746 624;
  • 20) 0.000 389 099 120 766 746 624 × 2 = 0 + 0.000 778 198 241 533 493 248;
  • 21) 0.000 778 198 241 533 493 248 × 2 = 0 + 0.001 556 396 483 066 986 496;
  • 22) 0.001 556 396 483 066 986 496 × 2 = 0 + 0.003 112 792 966 133 972 992;
  • 23) 0.003 112 792 966 133 972 992 × 2 = 0 + 0.006 225 585 932 267 945 984;
  • 24) 0.006 225 585 932 267 945 984 × 2 = 0 + 0.012 451 171 864 535 891 968;
  • 25) 0.012 451 171 864 535 891 968 × 2 = 0 + 0.024 902 343 729 071 783 936;
  • 26) 0.024 902 343 729 071 783 936 × 2 = 0 + 0.049 804 687 458 143 567 872;
  • 27) 0.049 804 687 458 143 567 872 × 2 = 0 + 0.099 609 374 916 287 135 744;
  • 28) 0.099 609 374 916 287 135 744 × 2 = 0 + 0.199 218 749 832 574 271 488;
  • 29) 0.199 218 749 832 574 271 488 × 2 = 0 + 0.398 437 499 665 148 542 976;
  • 30) 0.398 437 499 665 148 542 976 × 2 = 0 + 0.796 874 999 330 297 085 952;
  • 31) 0.796 874 999 330 297 085 952 × 2 = 1 + 0.593 749 998 660 594 171 904;
  • 32) 0.593 749 998 660 594 171 904 × 2 = 1 + 0.187 499 997 321 188 343 808;
  • 33) 0.187 499 997 321 188 343 808 × 2 = 0 + 0.374 999 994 642 376 687 616;
  • 34) 0.374 999 994 642 376 687 616 × 2 = 0 + 0.749 999 989 284 753 375 232;
  • 35) 0.749 999 989 284 753 375 232 × 2 = 1 + 0.499 999 978 569 506 750 464;
  • 36) 0.499 999 978 569 506 750 464 × 2 = 0 + 0.999 999 957 139 013 500 928;
  • 37) 0.999 999 957 139 013 500 928 × 2 = 1 + 0.999 999 914 278 027 001 856;
  • 38) 0.999 999 914 278 027 001 856 × 2 = 1 + 0.999 999 828 556 054 003 712;
  • 39) 0.999 999 828 556 054 003 712 × 2 = 1 + 0.999 999 657 112 108 007 424;
  • 40) 0.999 999 657 112 108 007 424 × 2 = 1 + 0.999 999 314 224 216 014 848;
  • 41) 0.999 999 314 224 216 014 848 × 2 = 1 + 0.999 998 628 448 432 029 696;
  • 42) 0.999 998 628 448 432 029 696 × 2 = 1 + 0.999 997 256 896 864 059 392;
  • 43) 0.999 997 256 896 864 059 392 × 2 = 1 + 0.999 994 513 793 728 118 784;
  • 44) 0.999 994 513 793 728 118 784 × 2 = 1 + 0.999 989 027 587 456 237 568;
  • 45) 0.999 989 027 587 456 237 568 × 2 = 1 + 0.999 978 055 174 912 475 136;
  • 46) 0.999 978 055 174 912 475 136 × 2 = 1 + 0.999 956 110 349 824 950 272;
  • 47) 0.999 956 110 349 824 950 272 × 2 = 1 + 0.999 912 220 699 649 900 544;
  • 48) 0.999 912 220 699 649 900 544 × 2 = 1 + 0.999 824 441 399 299 801 088;
  • 49) 0.999 824 441 399 299 801 088 × 2 = 1 + 0.999 648 882 798 599 602 176;
  • 50) 0.999 648 882 798 599 602 176 × 2 = 1 + 0.999 297 765 597 199 204 352;
  • 51) 0.999 297 765 597 199 204 352 × 2 = 1 + 0.998 595 531 194 398 408 704;
  • 52) 0.998 595 531 194 398 408 704 × 2 = 1 + 0.997 191 062 388 796 817 408;
  • 53) 0.997 191 062 388 796 817 408 × 2 = 1 + 0.994 382 124 777 593 634 816;
  • 54) 0.994 382 124 777 593 634 816 × 2 = 1 + 0.988 764 249 555 187 269 632;

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:


0.000 000 000 742 147 676 023(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 023(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:


0.000 000 000 742 147 676 023(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 -0.000 000 000 742 147 676 023 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 675 983 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