-0.000 000 000 742 147 676 646 683 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 683(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
-0.000 000 000 742 147 676 646 683(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:

|-0.000 000 000 742 147 676 646 683| = 0.000 000 000 742 147 676 646 683


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: 0.000 000 000 742 147 676 646 683.

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) 0.000 000 000 742 147 676 646 683 × 2 = 0 + 0.000 000 001 484 295 353 293 366;
  • 2) 0.000 000 001 484 295 353 293 366 × 2 = 0 + 0.000 000 002 968 590 706 586 732;
  • 3) 0.000 000 002 968 590 706 586 732 × 2 = 0 + 0.000 000 005 937 181 413 173 464;
  • 4) 0.000 000 005 937 181 413 173 464 × 2 = 0 + 0.000 000 011 874 362 826 346 928;
  • 5) 0.000 000 011 874 362 826 346 928 × 2 = 0 + 0.000 000 023 748 725 652 693 856;
  • 6) 0.000 000 023 748 725 652 693 856 × 2 = 0 + 0.000 000 047 497 451 305 387 712;
  • 7) 0.000 000 047 497 451 305 387 712 × 2 = 0 + 0.000 000 094 994 902 610 775 424;
  • 8) 0.000 000 094 994 902 610 775 424 × 2 = 0 + 0.000 000 189 989 805 221 550 848;
  • 9) 0.000 000 189 989 805 221 550 848 × 2 = 0 + 0.000 000 379 979 610 443 101 696;
  • 10) 0.000 000 379 979 610 443 101 696 × 2 = 0 + 0.000 000 759 959 220 886 203 392;
  • 11) 0.000 000 759 959 220 886 203 392 × 2 = 0 + 0.000 001 519 918 441 772 406 784;
  • 12) 0.000 001 519 918 441 772 406 784 × 2 = 0 + 0.000 003 039 836 883 544 813 568;
  • 13) 0.000 003 039 836 883 544 813 568 × 2 = 0 + 0.000 006 079 673 767 089 627 136;
  • 14) 0.000 006 079 673 767 089 627 136 × 2 = 0 + 0.000 012 159 347 534 179 254 272;
  • 15) 0.000 012 159 347 534 179 254 272 × 2 = 0 + 0.000 024 318 695 068 358 508 544;
  • 16) 0.000 024 318 695 068 358 508 544 × 2 = 0 + 0.000 048 637 390 136 717 017 088;
  • 17) 0.000 048 637 390 136 717 017 088 × 2 = 0 + 0.000 097 274 780 273 434 034 176;
  • 18) 0.000 097 274 780 273 434 034 176 × 2 = 0 + 0.000 194 549 560 546 868 068 352;
  • 19) 0.000 194 549 560 546 868 068 352 × 2 = 0 + 0.000 389 099 121 093 736 136 704;
  • 20) 0.000 389 099 121 093 736 136 704 × 2 = 0 + 0.000 778 198 242 187 472 273 408;
  • 21) 0.000 778 198 242 187 472 273 408 × 2 = 0 + 0.001 556 396 484 374 944 546 816;
  • 22) 0.001 556 396 484 374 944 546 816 × 2 = 0 + 0.003 112 792 968 749 889 093 632;
  • 23) 0.003 112 792 968 749 889 093 632 × 2 = 0 + 0.006 225 585 937 499 778 187 264;
  • 24) 0.006 225 585 937 499 778 187 264 × 2 = 0 + 0.012 451 171 874 999 556 374 528;
  • 25) 0.012 451 171 874 999 556 374 528 × 2 = 0 + 0.024 902 343 749 999 112 749 056;
  • 26) 0.024 902 343 749 999 112 749 056 × 2 = 0 + 0.049 804 687 499 998 225 498 112;
  • 27) 0.049 804 687 499 998 225 498 112 × 2 = 0 + 0.099 609 374 999 996 450 996 224;
  • 28) 0.099 609 374 999 996 450 996 224 × 2 = 0 + 0.199 218 749 999 992 901 992 448;
  • 29) 0.199 218 749 999 992 901 992 448 × 2 = 0 + 0.398 437 499 999 985 803 984 896;
  • 30) 0.398 437 499 999 985 803 984 896 × 2 = 0 + 0.796 874 999 999 971 607 969 792;
  • 31) 0.796 874 999 999 971 607 969 792 × 2 = 1 + 0.593 749 999 999 943 215 939 584;
  • 32) 0.593 749 999 999 943 215 939 584 × 2 = 1 + 0.187 499 999 999 886 431 879 168;
  • 33) 0.187 499 999 999 886 431 879 168 × 2 = 0 + 0.374 999 999 999 772 863 758 336;
  • 34) 0.374 999 999 999 772 863 758 336 × 2 = 0 + 0.749 999 999 999 545 727 516 672;
  • 35) 0.749 999 999 999 545 727 516 672 × 2 = 1 + 0.499 999 999 999 091 455 033 344;
  • 36) 0.499 999 999 999 091 455 033 344 × 2 = 0 + 0.999 999 999 998 182 910 066 688;
  • 37) 0.999 999 999 998 182 910 066 688 × 2 = 1 + 0.999 999 999 996 365 820 133 376;
  • 38) 0.999 999 999 996 365 820 133 376 × 2 = 1 + 0.999 999 999 992 731 640 266 752;
  • 39) 0.999 999 999 992 731 640 266 752 × 2 = 1 + 0.999 999 999 985 463 280 533 504;
  • 40) 0.999 999 999 985 463 280 533 504 × 2 = 1 + 0.999 999 999 970 926 561 067 008;
  • 41) 0.999 999 999 970 926 561 067 008 × 2 = 1 + 0.999 999 999 941 853 122 134 016;
  • 42) 0.999 999 999 941 853 122 134 016 × 2 = 1 + 0.999 999 999 883 706 244 268 032;
  • 43) 0.999 999 999 883 706 244 268 032 × 2 = 1 + 0.999 999 999 767 412 488 536 064;
  • 44) 0.999 999 999 767 412 488 536 064 × 2 = 1 + 0.999 999 999 534 824 977 072 128;
  • 45) 0.999 999 999 534 824 977 072 128 × 2 = 1 + 0.999 999 999 069 649 954 144 256;
  • 46) 0.999 999 999 069 649 954 144 256 × 2 = 1 + 0.999 999 998 139 299 908 288 512;
  • 47) 0.999 999 998 139 299 908 288 512 × 2 = 1 + 0.999 999 996 278 599 816 577 024;
  • 48) 0.999 999 996 278 599 816 577 024 × 2 = 1 + 0.999 999 992 557 199 633 154 048;
  • 49) 0.999 999 992 557 199 633 154 048 × 2 = 1 + 0.999 999 985 114 399 266 308 096;
  • 50) 0.999 999 985 114 399 266 308 096 × 2 = 1 + 0.999 999 970 228 798 532 616 192;
  • 51) 0.999 999 970 228 798 532 616 192 × 2 = 1 + 0.999 999 940 457 597 065 232 384;
  • 52) 0.999 999 940 457 597 065 232 384 × 2 = 1 + 0.999 999 880 915 194 130 464 768;
  • 53) 0.999 999 880 915 194 130 464 768 × 2 = 1 + 0.999 999 761 830 388 260 929 536;
  • 54) 0.999 999 761 830 388 260 929 536 × 2 = 1 + 0.999 999 523 660 776 521 859 072;

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:


0.000 000 000 742 147 676 646 683(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 683(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:


0.000 000 000 742 147 676 646 683(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 -0.000 000 000 742 147 676 646 683 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 587 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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