-0.000 000 000 742 147 676 646 653 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 653(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
-0.000 000 000 742 147 676 646 653(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:

|-0.000 000 000 742 147 676 646 653| = 0.000 000 000 742 147 676 646 653


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: 0.000 000 000 742 147 676 646 653.

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) 0.000 000 000 742 147 676 646 653 × 2 = 0 + 0.000 000 001 484 295 353 293 306;
  • 2) 0.000 000 001 484 295 353 293 306 × 2 = 0 + 0.000 000 002 968 590 706 586 612;
  • 3) 0.000 000 002 968 590 706 586 612 × 2 = 0 + 0.000 000 005 937 181 413 173 224;
  • 4) 0.000 000 005 937 181 413 173 224 × 2 = 0 + 0.000 000 011 874 362 826 346 448;
  • 5) 0.000 000 011 874 362 826 346 448 × 2 = 0 + 0.000 000 023 748 725 652 692 896;
  • 6) 0.000 000 023 748 725 652 692 896 × 2 = 0 + 0.000 000 047 497 451 305 385 792;
  • 7) 0.000 000 047 497 451 305 385 792 × 2 = 0 + 0.000 000 094 994 902 610 771 584;
  • 8) 0.000 000 094 994 902 610 771 584 × 2 = 0 + 0.000 000 189 989 805 221 543 168;
  • 9) 0.000 000 189 989 805 221 543 168 × 2 = 0 + 0.000 000 379 979 610 443 086 336;
  • 10) 0.000 000 379 979 610 443 086 336 × 2 = 0 + 0.000 000 759 959 220 886 172 672;
  • 11) 0.000 000 759 959 220 886 172 672 × 2 = 0 + 0.000 001 519 918 441 772 345 344;
  • 12) 0.000 001 519 918 441 772 345 344 × 2 = 0 + 0.000 003 039 836 883 544 690 688;
  • 13) 0.000 003 039 836 883 544 690 688 × 2 = 0 + 0.000 006 079 673 767 089 381 376;
  • 14) 0.000 006 079 673 767 089 381 376 × 2 = 0 + 0.000 012 159 347 534 178 762 752;
  • 15) 0.000 012 159 347 534 178 762 752 × 2 = 0 + 0.000 024 318 695 068 357 525 504;
  • 16) 0.000 024 318 695 068 357 525 504 × 2 = 0 + 0.000 048 637 390 136 715 051 008;
  • 17) 0.000 048 637 390 136 715 051 008 × 2 = 0 + 0.000 097 274 780 273 430 102 016;
  • 18) 0.000 097 274 780 273 430 102 016 × 2 = 0 + 0.000 194 549 560 546 860 204 032;
  • 19) 0.000 194 549 560 546 860 204 032 × 2 = 0 + 0.000 389 099 121 093 720 408 064;
  • 20) 0.000 389 099 121 093 720 408 064 × 2 = 0 + 0.000 778 198 242 187 440 816 128;
  • 21) 0.000 778 198 242 187 440 816 128 × 2 = 0 + 0.001 556 396 484 374 881 632 256;
  • 22) 0.001 556 396 484 374 881 632 256 × 2 = 0 + 0.003 112 792 968 749 763 264 512;
  • 23) 0.003 112 792 968 749 763 264 512 × 2 = 0 + 0.006 225 585 937 499 526 529 024;
  • 24) 0.006 225 585 937 499 526 529 024 × 2 = 0 + 0.012 451 171 874 999 053 058 048;
  • 25) 0.012 451 171 874 999 053 058 048 × 2 = 0 + 0.024 902 343 749 998 106 116 096;
  • 26) 0.024 902 343 749 998 106 116 096 × 2 = 0 + 0.049 804 687 499 996 212 232 192;
  • 27) 0.049 804 687 499 996 212 232 192 × 2 = 0 + 0.099 609 374 999 992 424 464 384;
  • 28) 0.099 609 374 999 992 424 464 384 × 2 = 0 + 0.199 218 749 999 984 848 928 768;
  • 29) 0.199 218 749 999 984 848 928 768 × 2 = 0 + 0.398 437 499 999 969 697 857 536;
  • 30) 0.398 437 499 999 969 697 857 536 × 2 = 0 + 0.796 874 999 999 939 395 715 072;
  • 31) 0.796 874 999 999 939 395 715 072 × 2 = 1 + 0.593 749 999 999 878 791 430 144;
  • 32) 0.593 749 999 999 878 791 430 144 × 2 = 1 + 0.187 499 999 999 757 582 860 288;
  • 33) 0.187 499 999 999 757 582 860 288 × 2 = 0 + 0.374 999 999 999 515 165 720 576;
  • 34) 0.374 999 999 999 515 165 720 576 × 2 = 0 + 0.749 999 999 999 030 331 441 152;
  • 35) 0.749 999 999 999 030 331 441 152 × 2 = 1 + 0.499 999 999 998 060 662 882 304;
  • 36) 0.499 999 999 998 060 662 882 304 × 2 = 0 + 0.999 999 999 996 121 325 764 608;
  • 37) 0.999 999 999 996 121 325 764 608 × 2 = 1 + 0.999 999 999 992 242 651 529 216;
  • 38) 0.999 999 999 992 242 651 529 216 × 2 = 1 + 0.999 999 999 984 485 303 058 432;
  • 39) 0.999 999 999 984 485 303 058 432 × 2 = 1 + 0.999 999 999 968 970 606 116 864;
  • 40) 0.999 999 999 968 970 606 116 864 × 2 = 1 + 0.999 999 999 937 941 212 233 728;
  • 41) 0.999 999 999 937 941 212 233 728 × 2 = 1 + 0.999 999 999 875 882 424 467 456;
  • 42) 0.999 999 999 875 882 424 467 456 × 2 = 1 + 0.999 999 999 751 764 848 934 912;
  • 43) 0.999 999 999 751 764 848 934 912 × 2 = 1 + 0.999 999 999 503 529 697 869 824;
  • 44) 0.999 999 999 503 529 697 869 824 × 2 = 1 + 0.999 999 999 007 059 395 739 648;
  • 45) 0.999 999 999 007 059 395 739 648 × 2 = 1 + 0.999 999 998 014 118 791 479 296;
  • 46) 0.999 999 998 014 118 791 479 296 × 2 = 1 + 0.999 999 996 028 237 582 958 592;
  • 47) 0.999 999 996 028 237 582 958 592 × 2 = 1 + 0.999 999 992 056 475 165 917 184;
  • 48) 0.999 999 992 056 475 165 917 184 × 2 = 1 + 0.999 999 984 112 950 331 834 368;
  • 49) 0.999 999 984 112 950 331 834 368 × 2 = 1 + 0.999 999 968 225 900 663 668 736;
  • 50) 0.999 999 968 225 900 663 668 736 × 2 = 1 + 0.999 999 936 451 801 327 337 472;
  • 51) 0.999 999 936 451 801 327 337 472 × 2 = 1 + 0.999 999 872 903 602 654 674 944;
  • 52) 0.999 999 872 903 602 654 674 944 × 2 = 1 + 0.999 999 745 807 205 309 349 888;
  • 53) 0.999 999 745 807 205 309 349 888 × 2 = 1 + 0.999 999 491 614 410 618 699 776;
  • 54) 0.999 999 491 614 410 618 699 776 × 2 = 1 + 0.999 998 983 228 821 237 399 552;

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:


0.000 000 000 742 147 676 646 653(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 653(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:


0.000 000 000 742 147 676 646 653(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 -0.000 000 000 742 147 676 646 653 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 555 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