-0.000 000 000 742 147 676 727 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 727(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
-0.000 000 000 742 147 676 727(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:

|-0.000 000 000 742 147 676 727| = 0.000 000 000 742 147 676 727


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: 0.000 000 000 742 147 676 727.

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) 0.000 000 000 742 147 676 727 × 2 = 0 + 0.000 000 001 484 295 353 454;
  • 2) 0.000 000 001 484 295 353 454 × 2 = 0 + 0.000 000 002 968 590 706 908;
  • 3) 0.000 000 002 968 590 706 908 × 2 = 0 + 0.000 000 005 937 181 413 816;
  • 4) 0.000 000 005 937 181 413 816 × 2 = 0 + 0.000 000 011 874 362 827 632;
  • 5) 0.000 000 011 874 362 827 632 × 2 = 0 + 0.000 000 023 748 725 655 264;
  • 6) 0.000 000 023 748 725 655 264 × 2 = 0 + 0.000 000 047 497 451 310 528;
  • 7) 0.000 000 047 497 451 310 528 × 2 = 0 + 0.000 000 094 994 902 621 056;
  • 8) 0.000 000 094 994 902 621 056 × 2 = 0 + 0.000 000 189 989 805 242 112;
  • 9) 0.000 000 189 989 805 242 112 × 2 = 0 + 0.000 000 379 979 610 484 224;
  • 10) 0.000 000 379 979 610 484 224 × 2 = 0 + 0.000 000 759 959 220 968 448;
  • 11) 0.000 000 759 959 220 968 448 × 2 = 0 + 0.000 001 519 918 441 936 896;
  • 12) 0.000 001 519 918 441 936 896 × 2 = 0 + 0.000 003 039 836 883 873 792;
  • 13) 0.000 003 039 836 883 873 792 × 2 = 0 + 0.000 006 079 673 767 747 584;
  • 14) 0.000 006 079 673 767 747 584 × 2 = 0 + 0.000 012 159 347 535 495 168;
  • 15) 0.000 012 159 347 535 495 168 × 2 = 0 + 0.000 024 318 695 070 990 336;
  • 16) 0.000 024 318 695 070 990 336 × 2 = 0 + 0.000 048 637 390 141 980 672;
  • 17) 0.000 048 637 390 141 980 672 × 2 = 0 + 0.000 097 274 780 283 961 344;
  • 18) 0.000 097 274 780 283 961 344 × 2 = 0 + 0.000 194 549 560 567 922 688;
  • 19) 0.000 194 549 560 567 922 688 × 2 = 0 + 0.000 389 099 121 135 845 376;
  • 20) 0.000 389 099 121 135 845 376 × 2 = 0 + 0.000 778 198 242 271 690 752;
  • 21) 0.000 778 198 242 271 690 752 × 2 = 0 + 0.001 556 396 484 543 381 504;
  • 22) 0.001 556 396 484 543 381 504 × 2 = 0 + 0.003 112 792 969 086 763 008;
  • 23) 0.003 112 792 969 086 763 008 × 2 = 0 + 0.006 225 585 938 173 526 016;
  • 24) 0.006 225 585 938 173 526 016 × 2 = 0 + 0.012 451 171 876 347 052 032;
  • 25) 0.012 451 171 876 347 052 032 × 2 = 0 + 0.024 902 343 752 694 104 064;
  • 26) 0.024 902 343 752 694 104 064 × 2 = 0 + 0.049 804 687 505 388 208 128;
  • 27) 0.049 804 687 505 388 208 128 × 2 = 0 + 0.099 609 375 010 776 416 256;
  • 28) 0.099 609 375 010 776 416 256 × 2 = 0 + 0.199 218 750 021 552 832 512;
  • 29) 0.199 218 750 021 552 832 512 × 2 = 0 + 0.398 437 500 043 105 665 024;
  • 30) 0.398 437 500 043 105 665 024 × 2 = 0 + 0.796 875 000 086 211 330 048;
  • 31) 0.796 875 000 086 211 330 048 × 2 = 1 + 0.593 750 000 172 422 660 096;
  • 32) 0.593 750 000 172 422 660 096 × 2 = 1 + 0.187 500 000 344 845 320 192;
  • 33) 0.187 500 000 344 845 320 192 × 2 = 0 + 0.375 000 000 689 690 640 384;
  • 34) 0.375 000 000 689 690 640 384 × 2 = 0 + 0.750 000 001 379 381 280 768;
  • 35) 0.750 000 001 379 381 280 768 × 2 = 1 + 0.500 000 002 758 762 561 536;
  • 36) 0.500 000 002 758 762 561 536 × 2 = 1 + 0.000 000 005 517 525 123 072;
  • 37) 0.000 000 005 517 525 123 072 × 2 = 0 + 0.000 000 011 035 050 246 144;
  • 38) 0.000 000 011 035 050 246 144 × 2 = 0 + 0.000 000 022 070 100 492 288;
  • 39) 0.000 000 022 070 100 492 288 × 2 = 0 + 0.000 000 044 140 200 984 576;
  • 40) 0.000 000 044 140 200 984 576 × 2 = 0 + 0.000 000 088 280 401 969 152;
  • 41) 0.000 000 088 280 401 969 152 × 2 = 0 + 0.000 000 176 560 803 938 304;
  • 42) 0.000 000 176 560 803 938 304 × 2 = 0 + 0.000 000 353 121 607 876 608;
  • 43) 0.000 000 353 121 607 876 608 × 2 = 0 + 0.000 000 706 243 215 753 216;
  • 44) 0.000 000 706 243 215 753 216 × 2 = 0 + 0.000 001 412 486 431 506 432;
  • 45) 0.000 001 412 486 431 506 432 × 2 = 0 + 0.000 002 824 972 863 012 864;
  • 46) 0.000 002 824 972 863 012 864 × 2 = 0 + 0.000 005 649 945 726 025 728;
  • 47) 0.000 005 649 945 726 025 728 × 2 = 0 + 0.000 011 299 891 452 051 456;
  • 48) 0.000 011 299 891 452 051 456 × 2 = 0 + 0.000 022 599 782 904 102 912;
  • 49) 0.000 022 599 782 904 102 912 × 2 = 0 + 0.000 045 199 565 808 205 824;
  • 50) 0.000 045 199 565 808 205 824 × 2 = 0 + 0.000 090 399 131 616 411 648;
  • 51) 0.000 090 399 131 616 411 648 × 2 = 0 + 0.000 180 798 263 232 823 296;
  • 52) 0.000 180 798 263 232 823 296 × 2 = 0 + 0.000 361 596 526 465 646 592;
  • 53) 0.000 361 596 526 465 646 592 × 2 = 0 + 0.000 723 193 052 931 293 184;
  • 54) 0.000 723 193 052 931 293 184 × 2 = 0 + 0.001 446 386 105 862 586 368;

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:


0.000 000 000 742 147 676 727(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 727(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 727(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 727 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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