-0.000 000 000 742 147 676 646 704 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 704 2(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
-0.000 000 000 742 147 676 646 704 2(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:

|-0.000 000 000 742 147 676 646 704 2| = 0.000 000 000 742 147 676 646 704 2


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: 0.000 000 000 742 147 676 646 704 2.

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) 0.000 000 000 742 147 676 646 704 2 × 2 = 0 + 0.000 000 001 484 295 353 293 408 4;
  • 2) 0.000 000 001 484 295 353 293 408 4 × 2 = 0 + 0.000 000 002 968 590 706 586 816 8;
  • 3) 0.000 000 002 968 590 706 586 816 8 × 2 = 0 + 0.000 000 005 937 181 413 173 633 6;
  • 4) 0.000 000 005 937 181 413 173 633 6 × 2 = 0 + 0.000 000 011 874 362 826 347 267 2;
  • 5) 0.000 000 011 874 362 826 347 267 2 × 2 = 0 + 0.000 000 023 748 725 652 694 534 4;
  • 6) 0.000 000 023 748 725 652 694 534 4 × 2 = 0 + 0.000 000 047 497 451 305 389 068 8;
  • 7) 0.000 000 047 497 451 305 389 068 8 × 2 = 0 + 0.000 000 094 994 902 610 778 137 6;
  • 8) 0.000 000 094 994 902 610 778 137 6 × 2 = 0 + 0.000 000 189 989 805 221 556 275 2;
  • 9) 0.000 000 189 989 805 221 556 275 2 × 2 = 0 + 0.000 000 379 979 610 443 112 550 4;
  • 10) 0.000 000 379 979 610 443 112 550 4 × 2 = 0 + 0.000 000 759 959 220 886 225 100 8;
  • 11) 0.000 000 759 959 220 886 225 100 8 × 2 = 0 + 0.000 001 519 918 441 772 450 201 6;
  • 12) 0.000 001 519 918 441 772 450 201 6 × 2 = 0 + 0.000 003 039 836 883 544 900 403 2;
  • 13) 0.000 003 039 836 883 544 900 403 2 × 2 = 0 + 0.000 006 079 673 767 089 800 806 4;
  • 14) 0.000 006 079 673 767 089 800 806 4 × 2 = 0 + 0.000 012 159 347 534 179 601 612 8;
  • 15) 0.000 012 159 347 534 179 601 612 8 × 2 = 0 + 0.000 024 318 695 068 359 203 225 6;
  • 16) 0.000 024 318 695 068 359 203 225 6 × 2 = 0 + 0.000 048 637 390 136 718 406 451 2;
  • 17) 0.000 048 637 390 136 718 406 451 2 × 2 = 0 + 0.000 097 274 780 273 436 812 902 4;
  • 18) 0.000 097 274 780 273 436 812 902 4 × 2 = 0 + 0.000 194 549 560 546 873 625 804 8;
  • 19) 0.000 194 549 560 546 873 625 804 8 × 2 = 0 + 0.000 389 099 121 093 747 251 609 6;
  • 20) 0.000 389 099 121 093 747 251 609 6 × 2 = 0 + 0.000 778 198 242 187 494 503 219 2;
  • 21) 0.000 778 198 242 187 494 503 219 2 × 2 = 0 + 0.001 556 396 484 374 989 006 438 4;
  • 22) 0.001 556 396 484 374 989 006 438 4 × 2 = 0 + 0.003 112 792 968 749 978 012 876 8;
  • 23) 0.003 112 792 968 749 978 012 876 8 × 2 = 0 + 0.006 225 585 937 499 956 025 753 6;
  • 24) 0.006 225 585 937 499 956 025 753 6 × 2 = 0 + 0.012 451 171 874 999 912 051 507 2;
  • 25) 0.012 451 171 874 999 912 051 507 2 × 2 = 0 + 0.024 902 343 749 999 824 103 014 4;
  • 26) 0.024 902 343 749 999 824 103 014 4 × 2 = 0 + 0.049 804 687 499 999 648 206 028 8;
  • 27) 0.049 804 687 499 999 648 206 028 8 × 2 = 0 + 0.099 609 374 999 999 296 412 057 6;
  • 28) 0.099 609 374 999 999 296 412 057 6 × 2 = 0 + 0.199 218 749 999 998 592 824 115 2;
  • 29) 0.199 218 749 999 998 592 824 115 2 × 2 = 0 + 0.398 437 499 999 997 185 648 230 4;
  • 30) 0.398 437 499 999 997 185 648 230 4 × 2 = 0 + 0.796 874 999 999 994 371 296 460 8;
  • 31) 0.796 874 999 999 994 371 296 460 8 × 2 = 1 + 0.593 749 999 999 988 742 592 921 6;
  • 32) 0.593 749 999 999 988 742 592 921 6 × 2 = 1 + 0.187 499 999 999 977 485 185 843 2;
  • 33) 0.187 499 999 999 977 485 185 843 2 × 2 = 0 + 0.374 999 999 999 954 970 371 686 4;
  • 34) 0.374 999 999 999 954 970 371 686 4 × 2 = 0 + 0.749 999 999 999 909 940 743 372 8;
  • 35) 0.749 999 999 999 909 940 743 372 8 × 2 = 1 + 0.499 999 999 999 819 881 486 745 6;
  • 36) 0.499 999 999 999 819 881 486 745 6 × 2 = 0 + 0.999 999 999 999 639 762 973 491 2;
  • 37) 0.999 999 999 999 639 762 973 491 2 × 2 = 1 + 0.999 999 999 999 279 525 946 982 4;
  • 38) 0.999 999 999 999 279 525 946 982 4 × 2 = 1 + 0.999 999 999 998 559 051 893 964 8;
  • 39) 0.999 999 999 998 559 051 893 964 8 × 2 = 1 + 0.999 999 999 997 118 103 787 929 6;
  • 40) 0.999 999 999 997 118 103 787 929 6 × 2 = 1 + 0.999 999 999 994 236 207 575 859 2;
  • 41) 0.999 999 999 994 236 207 575 859 2 × 2 = 1 + 0.999 999 999 988 472 415 151 718 4;
  • 42) 0.999 999 999 988 472 415 151 718 4 × 2 = 1 + 0.999 999 999 976 944 830 303 436 8;
  • 43) 0.999 999 999 976 944 830 303 436 8 × 2 = 1 + 0.999 999 999 953 889 660 606 873 6;
  • 44) 0.999 999 999 953 889 660 606 873 6 × 2 = 1 + 0.999 999 999 907 779 321 213 747 2;
  • 45) 0.999 999 999 907 779 321 213 747 2 × 2 = 1 + 0.999 999 999 815 558 642 427 494 4;
  • 46) 0.999 999 999 815 558 642 427 494 4 × 2 = 1 + 0.999 999 999 631 117 284 854 988 8;
  • 47) 0.999 999 999 631 117 284 854 988 8 × 2 = 1 + 0.999 999 999 262 234 569 709 977 6;
  • 48) 0.999 999 999 262 234 569 709 977 6 × 2 = 1 + 0.999 999 998 524 469 139 419 955 2;
  • 49) 0.999 999 998 524 469 139 419 955 2 × 2 = 1 + 0.999 999 997 048 938 278 839 910 4;
  • 50) 0.999 999 997 048 938 278 839 910 4 × 2 = 1 + 0.999 999 994 097 876 557 679 820 8;
  • 51) 0.999 999 994 097 876 557 679 820 8 × 2 = 1 + 0.999 999 988 195 753 115 359 641 6;
  • 52) 0.999 999 988 195 753 115 359 641 6 × 2 = 1 + 0.999 999 976 391 506 230 719 283 2;
  • 53) 0.999 999 976 391 506 230 719 283 2 × 2 = 1 + 0.999 999 952 783 012 461 438 566 4;
  • 54) 0.999 999 952 783 012 461 438 566 4 × 2 = 1 + 0.999 999 905 566 024 922 877 132 8;

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:


0.000 000 000 742 147 676 646 704 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 704 2(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:


0.000 000 000 742 147 676 646 704 2(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 -0.000 000 000 742 147 676 646 704 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 708 3 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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