-0.000 000 000 742 147 676 646 658 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 658(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
-0.000 000 000 742 147 676 646 658(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:

|-0.000 000 000 742 147 676 646 658| = 0.000 000 000 742 147 676 646 658


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: 0.000 000 000 742 147 676 646 658.

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) 0.000 000 000 742 147 676 646 658 × 2 = 0 + 0.000 000 001 484 295 353 293 316;
  • 2) 0.000 000 001 484 295 353 293 316 × 2 = 0 + 0.000 000 002 968 590 706 586 632;
  • 3) 0.000 000 002 968 590 706 586 632 × 2 = 0 + 0.000 000 005 937 181 413 173 264;
  • 4) 0.000 000 005 937 181 413 173 264 × 2 = 0 + 0.000 000 011 874 362 826 346 528;
  • 5) 0.000 000 011 874 362 826 346 528 × 2 = 0 + 0.000 000 023 748 725 652 693 056;
  • 6) 0.000 000 023 748 725 652 693 056 × 2 = 0 + 0.000 000 047 497 451 305 386 112;
  • 7) 0.000 000 047 497 451 305 386 112 × 2 = 0 + 0.000 000 094 994 902 610 772 224;
  • 8) 0.000 000 094 994 902 610 772 224 × 2 = 0 + 0.000 000 189 989 805 221 544 448;
  • 9) 0.000 000 189 989 805 221 544 448 × 2 = 0 + 0.000 000 379 979 610 443 088 896;
  • 10) 0.000 000 379 979 610 443 088 896 × 2 = 0 + 0.000 000 759 959 220 886 177 792;
  • 11) 0.000 000 759 959 220 886 177 792 × 2 = 0 + 0.000 001 519 918 441 772 355 584;
  • 12) 0.000 001 519 918 441 772 355 584 × 2 = 0 + 0.000 003 039 836 883 544 711 168;
  • 13) 0.000 003 039 836 883 544 711 168 × 2 = 0 + 0.000 006 079 673 767 089 422 336;
  • 14) 0.000 006 079 673 767 089 422 336 × 2 = 0 + 0.000 012 159 347 534 178 844 672;
  • 15) 0.000 012 159 347 534 178 844 672 × 2 = 0 + 0.000 024 318 695 068 357 689 344;
  • 16) 0.000 024 318 695 068 357 689 344 × 2 = 0 + 0.000 048 637 390 136 715 378 688;
  • 17) 0.000 048 637 390 136 715 378 688 × 2 = 0 + 0.000 097 274 780 273 430 757 376;
  • 18) 0.000 097 274 780 273 430 757 376 × 2 = 0 + 0.000 194 549 560 546 861 514 752;
  • 19) 0.000 194 549 560 546 861 514 752 × 2 = 0 + 0.000 389 099 121 093 723 029 504;
  • 20) 0.000 389 099 121 093 723 029 504 × 2 = 0 + 0.000 778 198 242 187 446 059 008;
  • 21) 0.000 778 198 242 187 446 059 008 × 2 = 0 + 0.001 556 396 484 374 892 118 016;
  • 22) 0.001 556 396 484 374 892 118 016 × 2 = 0 + 0.003 112 792 968 749 784 236 032;
  • 23) 0.003 112 792 968 749 784 236 032 × 2 = 0 + 0.006 225 585 937 499 568 472 064;
  • 24) 0.006 225 585 937 499 568 472 064 × 2 = 0 + 0.012 451 171 874 999 136 944 128;
  • 25) 0.012 451 171 874 999 136 944 128 × 2 = 0 + 0.024 902 343 749 998 273 888 256;
  • 26) 0.024 902 343 749 998 273 888 256 × 2 = 0 + 0.049 804 687 499 996 547 776 512;
  • 27) 0.049 804 687 499 996 547 776 512 × 2 = 0 + 0.099 609 374 999 993 095 553 024;
  • 28) 0.099 609 374 999 993 095 553 024 × 2 = 0 + 0.199 218 749 999 986 191 106 048;
  • 29) 0.199 218 749 999 986 191 106 048 × 2 = 0 + 0.398 437 499 999 972 382 212 096;
  • 30) 0.398 437 499 999 972 382 212 096 × 2 = 0 + 0.796 874 999 999 944 764 424 192;
  • 31) 0.796 874 999 999 944 764 424 192 × 2 = 1 + 0.593 749 999 999 889 528 848 384;
  • 32) 0.593 749 999 999 889 528 848 384 × 2 = 1 + 0.187 499 999 999 779 057 696 768;
  • 33) 0.187 499 999 999 779 057 696 768 × 2 = 0 + 0.374 999 999 999 558 115 393 536;
  • 34) 0.374 999 999 999 558 115 393 536 × 2 = 0 + 0.749 999 999 999 116 230 787 072;
  • 35) 0.749 999 999 999 116 230 787 072 × 2 = 1 + 0.499 999 999 998 232 461 574 144;
  • 36) 0.499 999 999 998 232 461 574 144 × 2 = 0 + 0.999 999 999 996 464 923 148 288;
  • 37) 0.999 999 999 996 464 923 148 288 × 2 = 1 + 0.999 999 999 992 929 846 296 576;
  • 38) 0.999 999 999 992 929 846 296 576 × 2 = 1 + 0.999 999 999 985 859 692 593 152;
  • 39) 0.999 999 999 985 859 692 593 152 × 2 = 1 + 0.999 999 999 971 719 385 186 304;
  • 40) 0.999 999 999 971 719 385 186 304 × 2 = 1 + 0.999 999 999 943 438 770 372 608;
  • 41) 0.999 999 999 943 438 770 372 608 × 2 = 1 + 0.999 999 999 886 877 540 745 216;
  • 42) 0.999 999 999 886 877 540 745 216 × 2 = 1 + 0.999 999 999 773 755 081 490 432;
  • 43) 0.999 999 999 773 755 081 490 432 × 2 = 1 + 0.999 999 999 547 510 162 980 864;
  • 44) 0.999 999 999 547 510 162 980 864 × 2 = 1 + 0.999 999 999 095 020 325 961 728;
  • 45) 0.999 999 999 095 020 325 961 728 × 2 = 1 + 0.999 999 998 190 040 651 923 456;
  • 46) 0.999 999 998 190 040 651 923 456 × 2 = 1 + 0.999 999 996 380 081 303 846 912;
  • 47) 0.999 999 996 380 081 303 846 912 × 2 = 1 + 0.999 999 992 760 162 607 693 824;
  • 48) 0.999 999 992 760 162 607 693 824 × 2 = 1 + 0.999 999 985 520 325 215 387 648;
  • 49) 0.999 999 985 520 325 215 387 648 × 2 = 1 + 0.999 999 971 040 650 430 775 296;
  • 50) 0.999 999 971 040 650 430 775 296 × 2 = 1 + 0.999 999 942 081 300 861 550 592;
  • 51) 0.999 999 942 081 300 861 550 592 × 2 = 1 + 0.999 999 884 162 601 723 101 184;
  • 52) 0.999 999 884 162 601 723 101 184 × 2 = 1 + 0.999 999 768 325 203 446 202 368;
  • 53) 0.999 999 768 325 203 446 202 368 × 2 = 1 + 0.999 999 536 650 406 892 404 736;
  • 54) 0.999 999 536 650 406 892 404 736 × 2 = 1 + 0.999 999 073 300 813 784 809 472;

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:


0.000 000 000 742 147 676 646 658(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 658(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:


0.000 000 000 742 147 676 646 658(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 -0.000 000 000 742 147 676 646 658 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 698 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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