-0.000 000 000 742 147 676 682 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 682(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
-0.000 000 000 742 147 676 682(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:

|-0.000 000 000 742 147 676 682| = 0.000 000 000 742 147 676 682


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: 0.000 000 000 742 147 676 682.

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) 0.000 000 000 742 147 676 682 × 2 = 0 + 0.000 000 001 484 295 353 364;
  • 2) 0.000 000 001 484 295 353 364 × 2 = 0 + 0.000 000 002 968 590 706 728;
  • 3) 0.000 000 002 968 590 706 728 × 2 = 0 + 0.000 000 005 937 181 413 456;
  • 4) 0.000 000 005 937 181 413 456 × 2 = 0 + 0.000 000 011 874 362 826 912;
  • 5) 0.000 000 011 874 362 826 912 × 2 = 0 + 0.000 000 023 748 725 653 824;
  • 6) 0.000 000 023 748 725 653 824 × 2 = 0 + 0.000 000 047 497 451 307 648;
  • 7) 0.000 000 047 497 451 307 648 × 2 = 0 + 0.000 000 094 994 902 615 296;
  • 8) 0.000 000 094 994 902 615 296 × 2 = 0 + 0.000 000 189 989 805 230 592;
  • 9) 0.000 000 189 989 805 230 592 × 2 = 0 + 0.000 000 379 979 610 461 184;
  • 10) 0.000 000 379 979 610 461 184 × 2 = 0 + 0.000 000 759 959 220 922 368;
  • 11) 0.000 000 759 959 220 922 368 × 2 = 0 + 0.000 001 519 918 441 844 736;
  • 12) 0.000 001 519 918 441 844 736 × 2 = 0 + 0.000 003 039 836 883 689 472;
  • 13) 0.000 003 039 836 883 689 472 × 2 = 0 + 0.000 006 079 673 767 378 944;
  • 14) 0.000 006 079 673 767 378 944 × 2 = 0 + 0.000 012 159 347 534 757 888;
  • 15) 0.000 012 159 347 534 757 888 × 2 = 0 + 0.000 024 318 695 069 515 776;
  • 16) 0.000 024 318 695 069 515 776 × 2 = 0 + 0.000 048 637 390 139 031 552;
  • 17) 0.000 048 637 390 139 031 552 × 2 = 0 + 0.000 097 274 780 278 063 104;
  • 18) 0.000 097 274 780 278 063 104 × 2 = 0 + 0.000 194 549 560 556 126 208;
  • 19) 0.000 194 549 560 556 126 208 × 2 = 0 + 0.000 389 099 121 112 252 416;
  • 20) 0.000 389 099 121 112 252 416 × 2 = 0 + 0.000 778 198 242 224 504 832;
  • 21) 0.000 778 198 242 224 504 832 × 2 = 0 + 0.001 556 396 484 449 009 664;
  • 22) 0.001 556 396 484 449 009 664 × 2 = 0 + 0.003 112 792 968 898 019 328;
  • 23) 0.003 112 792 968 898 019 328 × 2 = 0 + 0.006 225 585 937 796 038 656;
  • 24) 0.006 225 585 937 796 038 656 × 2 = 0 + 0.012 451 171 875 592 077 312;
  • 25) 0.012 451 171 875 592 077 312 × 2 = 0 + 0.024 902 343 751 184 154 624;
  • 26) 0.024 902 343 751 184 154 624 × 2 = 0 + 0.049 804 687 502 368 309 248;
  • 27) 0.049 804 687 502 368 309 248 × 2 = 0 + 0.099 609 375 004 736 618 496;
  • 28) 0.099 609 375 004 736 618 496 × 2 = 0 + 0.199 218 750 009 473 236 992;
  • 29) 0.199 218 750 009 473 236 992 × 2 = 0 + 0.398 437 500 018 946 473 984;
  • 30) 0.398 437 500 018 946 473 984 × 2 = 0 + 0.796 875 000 037 892 947 968;
  • 31) 0.796 875 000 037 892 947 968 × 2 = 1 + 0.593 750 000 075 785 895 936;
  • 32) 0.593 750 000 075 785 895 936 × 2 = 1 + 0.187 500 000 151 571 791 872;
  • 33) 0.187 500 000 151 571 791 872 × 2 = 0 + 0.375 000 000 303 143 583 744;
  • 34) 0.375 000 000 303 143 583 744 × 2 = 0 + 0.750 000 000 606 287 167 488;
  • 35) 0.750 000 000 606 287 167 488 × 2 = 1 + 0.500 000 001 212 574 334 976;
  • 36) 0.500 000 001 212 574 334 976 × 2 = 1 + 0.000 000 002 425 148 669 952;
  • 37) 0.000 000 002 425 148 669 952 × 2 = 0 + 0.000 000 004 850 297 339 904;
  • 38) 0.000 000 004 850 297 339 904 × 2 = 0 + 0.000 000 009 700 594 679 808;
  • 39) 0.000 000 009 700 594 679 808 × 2 = 0 + 0.000 000 019 401 189 359 616;
  • 40) 0.000 000 019 401 189 359 616 × 2 = 0 + 0.000 000 038 802 378 719 232;
  • 41) 0.000 000 038 802 378 719 232 × 2 = 0 + 0.000 000 077 604 757 438 464;
  • 42) 0.000 000 077 604 757 438 464 × 2 = 0 + 0.000 000 155 209 514 876 928;
  • 43) 0.000 000 155 209 514 876 928 × 2 = 0 + 0.000 000 310 419 029 753 856;
  • 44) 0.000 000 310 419 029 753 856 × 2 = 0 + 0.000 000 620 838 059 507 712;
  • 45) 0.000 000 620 838 059 507 712 × 2 = 0 + 0.000 001 241 676 119 015 424;
  • 46) 0.000 001 241 676 119 015 424 × 2 = 0 + 0.000 002 483 352 238 030 848;
  • 47) 0.000 002 483 352 238 030 848 × 2 = 0 + 0.000 004 966 704 476 061 696;
  • 48) 0.000 004 966 704 476 061 696 × 2 = 0 + 0.000 009 933 408 952 123 392;
  • 49) 0.000 009 933 408 952 123 392 × 2 = 0 + 0.000 019 866 817 904 246 784;
  • 50) 0.000 019 866 817 904 246 784 × 2 = 0 + 0.000 039 733 635 808 493 568;
  • 51) 0.000 039 733 635 808 493 568 × 2 = 0 + 0.000 079 467 271 616 987 136;
  • 52) 0.000 079 467 271 616 987 136 × 2 = 0 + 0.000 158 934 543 233 974 272;
  • 53) 0.000 158 934 543 233 974 272 × 2 = 0 + 0.000 317 869 086 467 948 544;
  • 54) 0.000 317 869 086 467 948 544 × 2 = 0 + 0.000 635 738 172 935 897 088;

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:


0.000 000 000 742 147 676 682(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 682(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:


0.000 000 000 742 147 676 682(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 -0.000 000 000 742 147 676 682 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