-0.000 000 000 742 147 676 646 749 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 749(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
-0.000 000 000 742 147 676 646 749(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:

|-0.000 000 000 742 147 676 646 749| = 0.000 000 000 742 147 676 646 749


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: 0.000 000 000 742 147 676 646 749.

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) 0.000 000 000 742 147 676 646 749 × 2 = 0 + 0.000 000 001 484 295 353 293 498;
  • 2) 0.000 000 001 484 295 353 293 498 × 2 = 0 + 0.000 000 002 968 590 706 586 996;
  • 3) 0.000 000 002 968 590 706 586 996 × 2 = 0 + 0.000 000 005 937 181 413 173 992;
  • 4) 0.000 000 005 937 181 413 173 992 × 2 = 0 + 0.000 000 011 874 362 826 347 984;
  • 5) 0.000 000 011 874 362 826 347 984 × 2 = 0 + 0.000 000 023 748 725 652 695 968;
  • 6) 0.000 000 023 748 725 652 695 968 × 2 = 0 + 0.000 000 047 497 451 305 391 936;
  • 7) 0.000 000 047 497 451 305 391 936 × 2 = 0 + 0.000 000 094 994 902 610 783 872;
  • 8) 0.000 000 094 994 902 610 783 872 × 2 = 0 + 0.000 000 189 989 805 221 567 744;
  • 9) 0.000 000 189 989 805 221 567 744 × 2 = 0 + 0.000 000 379 979 610 443 135 488;
  • 10) 0.000 000 379 979 610 443 135 488 × 2 = 0 + 0.000 000 759 959 220 886 270 976;
  • 11) 0.000 000 759 959 220 886 270 976 × 2 = 0 + 0.000 001 519 918 441 772 541 952;
  • 12) 0.000 001 519 918 441 772 541 952 × 2 = 0 + 0.000 003 039 836 883 545 083 904;
  • 13) 0.000 003 039 836 883 545 083 904 × 2 = 0 + 0.000 006 079 673 767 090 167 808;
  • 14) 0.000 006 079 673 767 090 167 808 × 2 = 0 + 0.000 012 159 347 534 180 335 616;
  • 15) 0.000 012 159 347 534 180 335 616 × 2 = 0 + 0.000 024 318 695 068 360 671 232;
  • 16) 0.000 024 318 695 068 360 671 232 × 2 = 0 + 0.000 048 637 390 136 721 342 464;
  • 17) 0.000 048 637 390 136 721 342 464 × 2 = 0 + 0.000 097 274 780 273 442 684 928;
  • 18) 0.000 097 274 780 273 442 684 928 × 2 = 0 + 0.000 194 549 560 546 885 369 856;
  • 19) 0.000 194 549 560 546 885 369 856 × 2 = 0 + 0.000 389 099 121 093 770 739 712;
  • 20) 0.000 389 099 121 093 770 739 712 × 2 = 0 + 0.000 778 198 242 187 541 479 424;
  • 21) 0.000 778 198 242 187 541 479 424 × 2 = 0 + 0.001 556 396 484 375 082 958 848;
  • 22) 0.001 556 396 484 375 082 958 848 × 2 = 0 + 0.003 112 792 968 750 165 917 696;
  • 23) 0.003 112 792 968 750 165 917 696 × 2 = 0 + 0.006 225 585 937 500 331 835 392;
  • 24) 0.006 225 585 937 500 331 835 392 × 2 = 0 + 0.012 451 171 875 000 663 670 784;
  • 25) 0.012 451 171 875 000 663 670 784 × 2 = 0 + 0.024 902 343 750 001 327 341 568;
  • 26) 0.024 902 343 750 001 327 341 568 × 2 = 0 + 0.049 804 687 500 002 654 683 136;
  • 27) 0.049 804 687 500 002 654 683 136 × 2 = 0 + 0.099 609 375 000 005 309 366 272;
  • 28) 0.099 609 375 000 005 309 366 272 × 2 = 0 + 0.199 218 750 000 010 618 732 544;
  • 29) 0.199 218 750 000 010 618 732 544 × 2 = 0 + 0.398 437 500 000 021 237 465 088;
  • 30) 0.398 437 500 000 021 237 465 088 × 2 = 0 + 0.796 875 000 000 042 474 930 176;
  • 31) 0.796 875 000 000 042 474 930 176 × 2 = 1 + 0.593 750 000 000 084 949 860 352;
  • 32) 0.593 750 000 000 084 949 860 352 × 2 = 1 + 0.187 500 000 000 169 899 720 704;
  • 33) 0.187 500 000 000 169 899 720 704 × 2 = 0 + 0.375 000 000 000 339 799 441 408;
  • 34) 0.375 000 000 000 339 799 441 408 × 2 = 0 + 0.750 000 000 000 679 598 882 816;
  • 35) 0.750 000 000 000 679 598 882 816 × 2 = 1 + 0.500 000 000 001 359 197 765 632;
  • 36) 0.500 000 000 001 359 197 765 632 × 2 = 1 + 0.000 000 000 002 718 395 531 264;
  • 37) 0.000 000 000 002 718 395 531 264 × 2 = 0 + 0.000 000 000 005 436 791 062 528;
  • 38) 0.000 000 000 005 436 791 062 528 × 2 = 0 + 0.000 000 000 010 873 582 125 056;
  • 39) 0.000 000 000 010 873 582 125 056 × 2 = 0 + 0.000 000 000 021 747 164 250 112;
  • 40) 0.000 000 000 021 747 164 250 112 × 2 = 0 + 0.000 000 000 043 494 328 500 224;
  • 41) 0.000 000 000 043 494 328 500 224 × 2 = 0 + 0.000 000 000 086 988 657 000 448;
  • 42) 0.000 000 000 086 988 657 000 448 × 2 = 0 + 0.000 000 000 173 977 314 000 896;
  • 43) 0.000 000 000 173 977 314 000 896 × 2 = 0 + 0.000 000 000 347 954 628 001 792;
  • 44) 0.000 000 000 347 954 628 001 792 × 2 = 0 + 0.000 000 000 695 909 256 003 584;
  • 45) 0.000 000 000 695 909 256 003 584 × 2 = 0 + 0.000 000 001 391 818 512 007 168;
  • 46) 0.000 000 001 391 818 512 007 168 × 2 = 0 + 0.000 000 002 783 637 024 014 336;
  • 47) 0.000 000 002 783 637 024 014 336 × 2 = 0 + 0.000 000 005 567 274 048 028 672;
  • 48) 0.000 000 005 567 274 048 028 672 × 2 = 0 + 0.000 000 011 134 548 096 057 344;
  • 49) 0.000 000 011 134 548 096 057 344 × 2 = 0 + 0.000 000 022 269 096 192 114 688;
  • 50) 0.000 000 022 269 096 192 114 688 × 2 = 0 + 0.000 000 044 538 192 384 229 376;
  • 51) 0.000 000 044 538 192 384 229 376 × 2 = 0 + 0.000 000 089 076 384 768 458 752;
  • 52) 0.000 000 089 076 384 768 458 752 × 2 = 0 + 0.000 000 178 152 769 536 917 504;
  • 53) 0.000 000 178 152 769 536 917 504 × 2 = 0 + 0.000 000 356 305 539 073 835 008;
  • 54) 0.000 000 356 305 539 073 835 008 × 2 = 0 + 0.000 000 712 611 078 147 670 016;

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:


0.000 000 000 742 147 676 646 749(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 749(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:


0.000 000 000 742 147 676 646 749(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 -0.000 000 000 742 147 676 646 749 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