-0.000 000 000 742 147 675 951 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 951(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
-0.000 000 000 742 147 675 951(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:

|-0.000 000 000 742 147 675 951| = 0.000 000 000 742 147 675 951


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: 0.000 000 000 742 147 675 951.

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) 0.000 000 000 742 147 675 951 × 2 = 0 + 0.000 000 001 484 295 351 902;
  • 2) 0.000 000 001 484 295 351 902 × 2 = 0 + 0.000 000 002 968 590 703 804;
  • 3) 0.000 000 002 968 590 703 804 × 2 = 0 + 0.000 000 005 937 181 407 608;
  • 4) 0.000 000 005 937 181 407 608 × 2 = 0 + 0.000 000 011 874 362 815 216;
  • 5) 0.000 000 011 874 362 815 216 × 2 = 0 + 0.000 000 023 748 725 630 432;
  • 6) 0.000 000 023 748 725 630 432 × 2 = 0 + 0.000 000 047 497 451 260 864;
  • 7) 0.000 000 047 497 451 260 864 × 2 = 0 + 0.000 000 094 994 902 521 728;
  • 8) 0.000 000 094 994 902 521 728 × 2 = 0 + 0.000 000 189 989 805 043 456;
  • 9) 0.000 000 189 989 805 043 456 × 2 = 0 + 0.000 000 379 979 610 086 912;
  • 10) 0.000 000 379 979 610 086 912 × 2 = 0 + 0.000 000 759 959 220 173 824;
  • 11) 0.000 000 759 959 220 173 824 × 2 = 0 + 0.000 001 519 918 440 347 648;
  • 12) 0.000 001 519 918 440 347 648 × 2 = 0 + 0.000 003 039 836 880 695 296;
  • 13) 0.000 003 039 836 880 695 296 × 2 = 0 + 0.000 006 079 673 761 390 592;
  • 14) 0.000 006 079 673 761 390 592 × 2 = 0 + 0.000 012 159 347 522 781 184;
  • 15) 0.000 012 159 347 522 781 184 × 2 = 0 + 0.000 024 318 695 045 562 368;
  • 16) 0.000 024 318 695 045 562 368 × 2 = 0 + 0.000 048 637 390 091 124 736;
  • 17) 0.000 048 637 390 091 124 736 × 2 = 0 + 0.000 097 274 780 182 249 472;
  • 18) 0.000 097 274 780 182 249 472 × 2 = 0 + 0.000 194 549 560 364 498 944;
  • 19) 0.000 194 549 560 364 498 944 × 2 = 0 + 0.000 389 099 120 728 997 888;
  • 20) 0.000 389 099 120 728 997 888 × 2 = 0 + 0.000 778 198 241 457 995 776;
  • 21) 0.000 778 198 241 457 995 776 × 2 = 0 + 0.001 556 396 482 915 991 552;
  • 22) 0.001 556 396 482 915 991 552 × 2 = 0 + 0.003 112 792 965 831 983 104;
  • 23) 0.003 112 792 965 831 983 104 × 2 = 0 + 0.006 225 585 931 663 966 208;
  • 24) 0.006 225 585 931 663 966 208 × 2 = 0 + 0.012 451 171 863 327 932 416;
  • 25) 0.012 451 171 863 327 932 416 × 2 = 0 + 0.024 902 343 726 655 864 832;
  • 26) 0.024 902 343 726 655 864 832 × 2 = 0 + 0.049 804 687 453 311 729 664;
  • 27) 0.049 804 687 453 311 729 664 × 2 = 0 + 0.099 609 374 906 623 459 328;
  • 28) 0.099 609 374 906 623 459 328 × 2 = 0 + 0.199 218 749 813 246 918 656;
  • 29) 0.199 218 749 813 246 918 656 × 2 = 0 + 0.398 437 499 626 493 837 312;
  • 30) 0.398 437 499 626 493 837 312 × 2 = 0 + 0.796 874 999 252 987 674 624;
  • 31) 0.796 874 999 252 987 674 624 × 2 = 1 + 0.593 749 998 505 975 349 248;
  • 32) 0.593 749 998 505 975 349 248 × 2 = 1 + 0.187 499 997 011 950 698 496;
  • 33) 0.187 499 997 011 950 698 496 × 2 = 0 + 0.374 999 994 023 901 396 992;
  • 34) 0.374 999 994 023 901 396 992 × 2 = 0 + 0.749 999 988 047 802 793 984;
  • 35) 0.749 999 988 047 802 793 984 × 2 = 1 + 0.499 999 976 095 605 587 968;
  • 36) 0.499 999 976 095 605 587 968 × 2 = 0 + 0.999 999 952 191 211 175 936;
  • 37) 0.999 999 952 191 211 175 936 × 2 = 1 + 0.999 999 904 382 422 351 872;
  • 38) 0.999 999 904 382 422 351 872 × 2 = 1 + 0.999 999 808 764 844 703 744;
  • 39) 0.999 999 808 764 844 703 744 × 2 = 1 + 0.999 999 617 529 689 407 488;
  • 40) 0.999 999 617 529 689 407 488 × 2 = 1 + 0.999 999 235 059 378 814 976;
  • 41) 0.999 999 235 059 378 814 976 × 2 = 1 + 0.999 998 470 118 757 629 952;
  • 42) 0.999 998 470 118 757 629 952 × 2 = 1 + 0.999 996 940 237 515 259 904;
  • 43) 0.999 996 940 237 515 259 904 × 2 = 1 + 0.999 993 880 475 030 519 808;
  • 44) 0.999 993 880 475 030 519 808 × 2 = 1 + 0.999 987 760 950 061 039 616;
  • 45) 0.999 987 760 950 061 039 616 × 2 = 1 + 0.999 975 521 900 122 079 232;
  • 46) 0.999 975 521 900 122 079 232 × 2 = 1 + 0.999 951 043 800 244 158 464;
  • 47) 0.999 951 043 800 244 158 464 × 2 = 1 + 0.999 902 087 600 488 316 928;
  • 48) 0.999 902 087 600 488 316 928 × 2 = 1 + 0.999 804 175 200 976 633 856;
  • 49) 0.999 804 175 200 976 633 856 × 2 = 1 + 0.999 608 350 401 953 267 712;
  • 50) 0.999 608 350 401 953 267 712 × 2 = 1 + 0.999 216 700 803 906 535 424;
  • 51) 0.999 216 700 803 906 535 424 × 2 = 1 + 0.998 433 401 607 813 070 848;
  • 52) 0.998 433 401 607 813 070 848 × 2 = 1 + 0.996 866 803 215 626 141 696;
  • 53) 0.996 866 803 215 626 141 696 × 2 = 1 + 0.993 733 606 431 252 283 392;
  • 54) 0.993 733 606 431 252 283 392 × 2 = 1 + 0.987 467 212 862 504 566 784;

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:


0.000 000 000 742 147 675 951(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 951(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:


0.000 000 000 742 147 675 951(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 -0.000 000 000 742 147 675 951 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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