-0.000 000 000 742 147 676 646 739 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 739 1(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
-0.000 000 000 742 147 676 646 739 1(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:

|-0.000 000 000 742 147 676 646 739 1| = 0.000 000 000 742 147 676 646 739 1


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: 0.000 000 000 742 147 676 646 739 1.

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) 0.000 000 000 742 147 676 646 739 1 × 2 = 0 + 0.000 000 001 484 295 353 293 478 2;
  • 2) 0.000 000 001 484 295 353 293 478 2 × 2 = 0 + 0.000 000 002 968 590 706 586 956 4;
  • 3) 0.000 000 002 968 590 706 586 956 4 × 2 = 0 + 0.000 000 005 937 181 413 173 912 8;
  • 4) 0.000 000 005 937 181 413 173 912 8 × 2 = 0 + 0.000 000 011 874 362 826 347 825 6;
  • 5) 0.000 000 011 874 362 826 347 825 6 × 2 = 0 + 0.000 000 023 748 725 652 695 651 2;
  • 6) 0.000 000 023 748 725 652 695 651 2 × 2 = 0 + 0.000 000 047 497 451 305 391 302 4;
  • 7) 0.000 000 047 497 451 305 391 302 4 × 2 = 0 + 0.000 000 094 994 902 610 782 604 8;
  • 8) 0.000 000 094 994 902 610 782 604 8 × 2 = 0 + 0.000 000 189 989 805 221 565 209 6;
  • 9) 0.000 000 189 989 805 221 565 209 6 × 2 = 0 + 0.000 000 379 979 610 443 130 419 2;
  • 10) 0.000 000 379 979 610 443 130 419 2 × 2 = 0 + 0.000 000 759 959 220 886 260 838 4;
  • 11) 0.000 000 759 959 220 886 260 838 4 × 2 = 0 + 0.000 001 519 918 441 772 521 676 8;
  • 12) 0.000 001 519 918 441 772 521 676 8 × 2 = 0 + 0.000 003 039 836 883 545 043 353 6;
  • 13) 0.000 003 039 836 883 545 043 353 6 × 2 = 0 + 0.000 006 079 673 767 090 086 707 2;
  • 14) 0.000 006 079 673 767 090 086 707 2 × 2 = 0 + 0.000 012 159 347 534 180 173 414 4;
  • 15) 0.000 012 159 347 534 180 173 414 4 × 2 = 0 + 0.000 024 318 695 068 360 346 828 8;
  • 16) 0.000 024 318 695 068 360 346 828 8 × 2 = 0 + 0.000 048 637 390 136 720 693 657 6;
  • 17) 0.000 048 637 390 136 720 693 657 6 × 2 = 0 + 0.000 097 274 780 273 441 387 315 2;
  • 18) 0.000 097 274 780 273 441 387 315 2 × 2 = 0 + 0.000 194 549 560 546 882 774 630 4;
  • 19) 0.000 194 549 560 546 882 774 630 4 × 2 = 0 + 0.000 389 099 121 093 765 549 260 8;
  • 20) 0.000 389 099 121 093 765 549 260 8 × 2 = 0 + 0.000 778 198 242 187 531 098 521 6;
  • 21) 0.000 778 198 242 187 531 098 521 6 × 2 = 0 + 0.001 556 396 484 375 062 197 043 2;
  • 22) 0.001 556 396 484 375 062 197 043 2 × 2 = 0 + 0.003 112 792 968 750 124 394 086 4;
  • 23) 0.003 112 792 968 750 124 394 086 4 × 2 = 0 + 0.006 225 585 937 500 248 788 172 8;
  • 24) 0.006 225 585 937 500 248 788 172 8 × 2 = 0 + 0.012 451 171 875 000 497 576 345 6;
  • 25) 0.012 451 171 875 000 497 576 345 6 × 2 = 0 + 0.024 902 343 750 000 995 152 691 2;
  • 26) 0.024 902 343 750 000 995 152 691 2 × 2 = 0 + 0.049 804 687 500 001 990 305 382 4;
  • 27) 0.049 804 687 500 001 990 305 382 4 × 2 = 0 + 0.099 609 375 000 003 980 610 764 8;
  • 28) 0.099 609 375 000 003 980 610 764 8 × 2 = 0 + 0.199 218 750 000 007 961 221 529 6;
  • 29) 0.199 218 750 000 007 961 221 529 6 × 2 = 0 + 0.398 437 500 000 015 922 443 059 2;
  • 30) 0.398 437 500 000 015 922 443 059 2 × 2 = 0 + 0.796 875 000 000 031 844 886 118 4;
  • 31) 0.796 875 000 000 031 844 886 118 4 × 2 = 1 + 0.593 750 000 000 063 689 772 236 8;
  • 32) 0.593 750 000 000 063 689 772 236 8 × 2 = 1 + 0.187 500 000 000 127 379 544 473 6;
  • 33) 0.187 500 000 000 127 379 544 473 6 × 2 = 0 + 0.375 000 000 000 254 759 088 947 2;
  • 34) 0.375 000 000 000 254 759 088 947 2 × 2 = 0 + 0.750 000 000 000 509 518 177 894 4;
  • 35) 0.750 000 000 000 509 518 177 894 4 × 2 = 1 + 0.500 000 000 001 019 036 355 788 8;
  • 36) 0.500 000 000 001 019 036 355 788 8 × 2 = 1 + 0.000 000 000 002 038 072 711 577 6;
  • 37) 0.000 000 000 002 038 072 711 577 6 × 2 = 0 + 0.000 000 000 004 076 145 423 155 2;
  • 38) 0.000 000 000 004 076 145 423 155 2 × 2 = 0 + 0.000 000 000 008 152 290 846 310 4;
  • 39) 0.000 000 000 008 152 290 846 310 4 × 2 = 0 + 0.000 000 000 016 304 581 692 620 8;
  • 40) 0.000 000 000 016 304 581 692 620 8 × 2 = 0 + 0.000 000 000 032 609 163 385 241 6;
  • 41) 0.000 000 000 032 609 163 385 241 6 × 2 = 0 + 0.000 000 000 065 218 326 770 483 2;
  • 42) 0.000 000 000 065 218 326 770 483 2 × 2 = 0 + 0.000 000 000 130 436 653 540 966 4;
  • 43) 0.000 000 000 130 436 653 540 966 4 × 2 = 0 + 0.000 000 000 260 873 307 081 932 8;
  • 44) 0.000 000 000 260 873 307 081 932 8 × 2 = 0 + 0.000 000 000 521 746 614 163 865 6;
  • 45) 0.000 000 000 521 746 614 163 865 6 × 2 = 0 + 0.000 000 001 043 493 228 327 731 2;
  • 46) 0.000 000 001 043 493 228 327 731 2 × 2 = 0 + 0.000 000 002 086 986 456 655 462 4;
  • 47) 0.000 000 002 086 986 456 655 462 4 × 2 = 0 + 0.000 000 004 173 972 913 310 924 8;
  • 48) 0.000 000 004 173 972 913 310 924 8 × 2 = 0 + 0.000 000 008 347 945 826 621 849 6;
  • 49) 0.000 000 008 347 945 826 621 849 6 × 2 = 0 + 0.000 000 016 695 891 653 243 699 2;
  • 50) 0.000 000 016 695 891 653 243 699 2 × 2 = 0 + 0.000 000 033 391 783 306 487 398 4;
  • 51) 0.000 000 033 391 783 306 487 398 4 × 2 = 0 + 0.000 000 066 783 566 612 974 796 8;
  • 52) 0.000 000 066 783 566 612 974 796 8 × 2 = 0 + 0.000 000 133 567 133 225 949 593 6;
  • 53) 0.000 000 133 567 133 225 949 593 6 × 2 = 0 + 0.000 000 267 134 266 451 899 187 2;
  • 54) 0.000 000 267 134 266 451 899 187 2 × 2 = 0 + 0.000 000 534 268 532 903 798 374 4;

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:


0.000 000 000 742 147 676 646 739 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 739 1(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:


0.000 000 000 742 147 676 646 739 1(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 -0.000 000 000 742 147 676 646 739 1 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