-0.000 000 000 742 147 807 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 807(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
-0.000 000 000 742 147 807(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:

|-0.000 000 000 742 147 807| = 0.000 000 000 742 147 807


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: 0.000 000 000 742 147 807.

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) 0.000 000 000 742 147 807 × 2 = 0 + 0.000 000 001 484 295 614;
  • 2) 0.000 000 001 484 295 614 × 2 = 0 + 0.000 000 002 968 591 228;
  • 3) 0.000 000 002 968 591 228 × 2 = 0 + 0.000 000 005 937 182 456;
  • 4) 0.000 000 005 937 182 456 × 2 = 0 + 0.000 000 011 874 364 912;
  • 5) 0.000 000 011 874 364 912 × 2 = 0 + 0.000 000 023 748 729 824;
  • 6) 0.000 000 023 748 729 824 × 2 = 0 + 0.000 000 047 497 459 648;
  • 7) 0.000 000 047 497 459 648 × 2 = 0 + 0.000 000 094 994 919 296;
  • 8) 0.000 000 094 994 919 296 × 2 = 0 + 0.000 000 189 989 838 592;
  • 9) 0.000 000 189 989 838 592 × 2 = 0 + 0.000 000 379 979 677 184;
  • 10) 0.000 000 379 979 677 184 × 2 = 0 + 0.000 000 759 959 354 368;
  • 11) 0.000 000 759 959 354 368 × 2 = 0 + 0.000 001 519 918 708 736;
  • 12) 0.000 001 519 918 708 736 × 2 = 0 + 0.000 003 039 837 417 472;
  • 13) 0.000 003 039 837 417 472 × 2 = 0 + 0.000 006 079 674 834 944;
  • 14) 0.000 006 079 674 834 944 × 2 = 0 + 0.000 012 159 349 669 888;
  • 15) 0.000 012 159 349 669 888 × 2 = 0 + 0.000 024 318 699 339 776;
  • 16) 0.000 024 318 699 339 776 × 2 = 0 + 0.000 048 637 398 679 552;
  • 17) 0.000 048 637 398 679 552 × 2 = 0 + 0.000 097 274 797 359 104;
  • 18) 0.000 097 274 797 359 104 × 2 = 0 + 0.000 194 549 594 718 208;
  • 19) 0.000 194 549 594 718 208 × 2 = 0 + 0.000 389 099 189 436 416;
  • 20) 0.000 389 099 189 436 416 × 2 = 0 + 0.000 778 198 378 872 832;
  • 21) 0.000 778 198 378 872 832 × 2 = 0 + 0.001 556 396 757 745 664;
  • 22) 0.001 556 396 757 745 664 × 2 = 0 + 0.003 112 793 515 491 328;
  • 23) 0.003 112 793 515 491 328 × 2 = 0 + 0.006 225 587 030 982 656;
  • 24) 0.006 225 587 030 982 656 × 2 = 0 + 0.012 451 174 061 965 312;
  • 25) 0.012 451 174 061 965 312 × 2 = 0 + 0.024 902 348 123 930 624;
  • 26) 0.024 902 348 123 930 624 × 2 = 0 + 0.049 804 696 247 861 248;
  • 27) 0.049 804 696 247 861 248 × 2 = 0 + 0.099 609 392 495 722 496;
  • 28) 0.099 609 392 495 722 496 × 2 = 0 + 0.199 218 784 991 444 992;
  • 29) 0.199 218 784 991 444 992 × 2 = 0 + 0.398 437 569 982 889 984;
  • 30) 0.398 437 569 982 889 984 × 2 = 0 + 0.796 875 139 965 779 968;
  • 31) 0.796 875 139 965 779 968 × 2 = 1 + 0.593 750 279 931 559 936;
  • 32) 0.593 750 279 931 559 936 × 2 = 1 + 0.187 500 559 863 119 872;
  • 33) 0.187 500 559 863 119 872 × 2 = 0 + 0.375 001 119 726 239 744;
  • 34) 0.375 001 119 726 239 744 × 2 = 0 + 0.750 002 239 452 479 488;
  • 35) 0.750 002 239 452 479 488 × 2 = 1 + 0.500 004 478 904 958 976;
  • 36) 0.500 004 478 904 958 976 × 2 = 1 + 0.000 008 957 809 917 952;
  • 37) 0.000 008 957 809 917 952 × 2 = 0 + 0.000 017 915 619 835 904;
  • 38) 0.000 017 915 619 835 904 × 2 = 0 + 0.000 035 831 239 671 808;
  • 39) 0.000 035 831 239 671 808 × 2 = 0 + 0.000 071 662 479 343 616;
  • 40) 0.000 071 662 479 343 616 × 2 = 0 + 0.000 143 324 958 687 232;
  • 41) 0.000 143 324 958 687 232 × 2 = 0 + 0.000 286 649 917 374 464;
  • 42) 0.000 286 649 917 374 464 × 2 = 0 + 0.000 573 299 834 748 928;
  • 43) 0.000 573 299 834 748 928 × 2 = 0 + 0.001 146 599 669 497 856;
  • 44) 0.001 146 599 669 497 856 × 2 = 0 + 0.002 293 199 338 995 712;
  • 45) 0.002 293 199 338 995 712 × 2 = 0 + 0.004 586 398 677 991 424;
  • 46) 0.004 586 398 677 991 424 × 2 = 0 + 0.009 172 797 355 982 848;
  • 47) 0.009 172 797 355 982 848 × 2 = 0 + 0.018 345 594 711 965 696;
  • 48) 0.018 345 594 711 965 696 × 2 = 0 + 0.036 691 189 423 931 392;
  • 49) 0.036 691 189 423 931 392 × 2 = 0 + 0.073 382 378 847 862 784;
  • 50) 0.073 382 378 847 862 784 × 2 = 0 + 0.146 764 757 695 725 568;
  • 51) 0.146 764 757 695 725 568 × 2 = 0 + 0.293 529 515 391 451 136;
  • 52) 0.293 529 515 391 451 136 × 2 = 0 + 0.587 059 030 782 902 272;
  • 53) 0.587 059 030 782 902 272 × 2 = 1 + 0.174 118 061 565 804 544;
  • 54) 0.174 118 061 565 804 544 × 2 = 0 + 0.348 236 123 131 609 088;

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:


0.000 000 000 742 147 807(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 807(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(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:


0.000 000 000 742 147 807(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) × 20 =

1.1001 1000 0000 0000 0000 010(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 010


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 0010 =

100 1100 0000 0000 0000 0010


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 0010


Decimal number -0.000 000 000 742 147 807 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0010

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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