-0.000 000 000 742 147 676 646 727 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 727(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
-0.000 000 000 742 147 676 646 727(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:

|-0.000 000 000 742 147 676 646 727| = 0.000 000 000 742 147 676 646 727


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: 0.000 000 000 742 147 676 646 727.

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) 0.000 000 000 742 147 676 646 727 × 2 = 0 + 0.000 000 001 484 295 353 293 454;
  • 2) 0.000 000 001 484 295 353 293 454 × 2 = 0 + 0.000 000 002 968 590 706 586 908;
  • 3) 0.000 000 002 968 590 706 586 908 × 2 = 0 + 0.000 000 005 937 181 413 173 816;
  • 4) 0.000 000 005 937 181 413 173 816 × 2 = 0 + 0.000 000 011 874 362 826 347 632;
  • 5) 0.000 000 011 874 362 826 347 632 × 2 = 0 + 0.000 000 023 748 725 652 695 264;
  • 6) 0.000 000 023 748 725 652 695 264 × 2 = 0 + 0.000 000 047 497 451 305 390 528;
  • 7) 0.000 000 047 497 451 305 390 528 × 2 = 0 + 0.000 000 094 994 902 610 781 056;
  • 8) 0.000 000 094 994 902 610 781 056 × 2 = 0 + 0.000 000 189 989 805 221 562 112;
  • 9) 0.000 000 189 989 805 221 562 112 × 2 = 0 + 0.000 000 379 979 610 443 124 224;
  • 10) 0.000 000 379 979 610 443 124 224 × 2 = 0 + 0.000 000 759 959 220 886 248 448;
  • 11) 0.000 000 759 959 220 886 248 448 × 2 = 0 + 0.000 001 519 918 441 772 496 896;
  • 12) 0.000 001 519 918 441 772 496 896 × 2 = 0 + 0.000 003 039 836 883 544 993 792;
  • 13) 0.000 003 039 836 883 544 993 792 × 2 = 0 + 0.000 006 079 673 767 089 987 584;
  • 14) 0.000 006 079 673 767 089 987 584 × 2 = 0 + 0.000 012 159 347 534 179 975 168;
  • 15) 0.000 012 159 347 534 179 975 168 × 2 = 0 + 0.000 024 318 695 068 359 950 336;
  • 16) 0.000 024 318 695 068 359 950 336 × 2 = 0 + 0.000 048 637 390 136 719 900 672;
  • 17) 0.000 048 637 390 136 719 900 672 × 2 = 0 + 0.000 097 274 780 273 439 801 344;
  • 18) 0.000 097 274 780 273 439 801 344 × 2 = 0 + 0.000 194 549 560 546 879 602 688;
  • 19) 0.000 194 549 560 546 879 602 688 × 2 = 0 + 0.000 389 099 121 093 759 205 376;
  • 20) 0.000 389 099 121 093 759 205 376 × 2 = 0 + 0.000 778 198 242 187 518 410 752;
  • 21) 0.000 778 198 242 187 518 410 752 × 2 = 0 + 0.001 556 396 484 375 036 821 504;
  • 22) 0.001 556 396 484 375 036 821 504 × 2 = 0 + 0.003 112 792 968 750 073 643 008;
  • 23) 0.003 112 792 968 750 073 643 008 × 2 = 0 + 0.006 225 585 937 500 147 286 016;
  • 24) 0.006 225 585 937 500 147 286 016 × 2 = 0 + 0.012 451 171 875 000 294 572 032;
  • 25) 0.012 451 171 875 000 294 572 032 × 2 = 0 + 0.024 902 343 750 000 589 144 064;
  • 26) 0.024 902 343 750 000 589 144 064 × 2 = 0 + 0.049 804 687 500 001 178 288 128;
  • 27) 0.049 804 687 500 001 178 288 128 × 2 = 0 + 0.099 609 375 000 002 356 576 256;
  • 28) 0.099 609 375 000 002 356 576 256 × 2 = 0 + 0.199 218 750 000 004 713 152 512;
  • 29) 0.199 218 750 000 004 713 152 512 × 2 = 0 + 0.398 437 500 000 009 426 305 024;
  • 30) 0.398 437 500 000 009 426 305 024 × 2 = 0 + 0.796 875 000 000 018 852 610 048;
  • 31) 0.796 875 000 000 018 852 610 048 × 2 = 1 + 0.593 750 000 000 037 705 220 096;
  • 32) 0.593 750 000 000 037 705 220 096 × 2 = 1 + 0.187 500 000 000 075 410 440 192;
  • 33) 0.187 500 000 000 075 410 440 192 × 2 = 0 + 0.375 000 000 000 150 820 880 384;
  • 34) 0.375 000 000 000 150 820 880 384 × 2 = 0 + 0.750 000 000 000 301 641 760 768;
  • 35) 0.750 000 000 000 301 641 760 768 × 2 = 1 + 0.500 000 000 000 603 283 521 536;
  • 36) 0.500 000 000 000 603 283 521 536 × 2 = 1 + 0.000 000 000 001 206 567 043 072;
  • 37) 0.000 000 000 001 206 567 043 072 × 2 = 0 + 0.000 000 000 002 413 134 086 144;
  • 38) 0.000 000 000 002 413 134 086 144 × 2 = 0 + 0.000 000 000 004 826 268 172 288;
  • 39) 0.000 000 000 004 826 268 172 288 × 2 = 0 + 0.000 000 000 009 652 536 344 576;
  • 40) 0.000 000 000 009 652 536 344 576 × 2 = 0 + 0.000 000 000 019 305 072 689 152;
  • 41) 0.000 000 000 019 305 072 689 152 × 2 = 0 + 0.000 000 000 038 610 145 378 304;
  • 42) 0.000 000 000 038 610 145 378 304 × 2 = 0 + 0.000 000 000 077 220 290 756 608;
  • 43) 0.000 000 000 077 220 290 756 608 × 2 = 0 + 0.000 000 000 154 440 581 513 216;
  • 44) 0.000 000 000 154 440 581 513 216 × 2 = 0 + 0.000 000 000 308 881 163 026 432;
  • 45) 0.000 000 000 308 881 163 026 432 × 2 = 0 + 0.000 000 000 617 762 326 052 864;
  • 46) 0.000 000 000 617 762 326 052 864 × 2 = 0 + 0.000 000 001 235 524 652 105 728;
  • 47) 0.000 000 001 235 524 652 105 728 × 2 = 0 + 0.000 000 002 471 049 304 211 456;
  • 48) 0.000 000 002 471 049 304 211 456 × 2 = 0 + 0.000 000 004 942 098 608 422 912;
  • 49) 0.000 000 004 942 098 608 422 912 × 2 = 0 + 0.000 000 009 884 197 216 845 824;
  • 50) 0.000 000 009 884 197 216 845 824 × 2 = 0 + 0.000 000 019 768 394 433 691 648;
  • 51) 0.000 000 019 768 394 433 691 648 × 2 = 0 + 0.000 000 039 536 788 867 383 296;
  • 52) 0.000 000 039 536 788 867 383 296 × 2 = 0 + 0.000 000 079 073 577 734 766 592;
  • 53) 0.000 000 079 073 577 734 766 592 × 2 = 0 + 0.000 000 158 147 155 469 533 184;
  • 54) 0.000 000 158 147 155 469 533 184 × 2 = 0 + 0.000 000 316 294 310 939 066 368;

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:


0.000 000 000 742 147 676 646 727(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 727(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:


0.000 000 000 742 147 676 646 727(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 -0.000 000 000 742 147 676 646 727 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