-0.000 000 000 742 147 858 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 858(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
-0.000 000 000 742 147 858(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:

|-0.000 000 000 742 147 858| = 0.000 000 000 742 147 858


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: 0.000 000 000 742 147 858.

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) 0.000 000 000 742 147 858 × 2 = 0 + 0.000 000 001 484 295 716;
  • 2) 0.000 000 001 484 295 716 × 2 = 0 + 0.000 000 002 968 591 432;
  • 3) 0.000 000 002 968 591 432 × 2 = 0 + 0.000 000 005 937 182 864;
  • 4) 0.000 000 005 937 182 864 × 2 = 0 + 0.000 000 011 874 365 728;
  • 5) 0.000 000 011 874 365 728 × 2 = 0 + 0.000 000 023 748 731 456;
  • 6) 0.000 000 023 748 731 456 × 2 = 0 + 0.000 000 047 497 462 912;
  • 7) 0.000 000 047 497 462 912 × 2 = 0 + 0.000 000 094 994 925 824;
  • 8) 0.000 000 094 994 925 824 × 2 = 0 + 0.000 000 189 989 851 648;
  • 9) 0.000 000 189 989 851 648 × 2 = 0 + 0.000 000 379 979 703 296;
  • 10) 0.000 000 379 979 703 296 × 2 = 0 + 0.000 000 759 959 406 592;
  • 11) 0.000 000 759 959 406 592 × 2 = 0 + 0.000 001 519 918 813 184;
  • 12) 0.000 001 519 918 813 184 × 2 = 0 + 0.000 003 039 837 626 368;
  • 13) 0.000 003 039 837 626 368 × 2 = 0 + 0.000 006 079 675 252 736;
  • 14) 0.000 006 079 675 252 736 × 2 = 0 + 0.000 012 159 350 505 472;
  • 15) 0.000 012 159 350 505 472 × 2 = 0 + 0.000 024 318 701 010 944;
  • 16) 0.000 024 318 701 010 944 × 2 = 0 + 0.000 048 637 402 021 888;
  • 17) 0.000 048 637 402 021 888 × 2 = 0 + 0.000 097 274 804 043 776;
  • 18) 0.000 097 274 804 043 776 × 2 = 0 + 0.000 194 549 608 087 552;
  • 19) 0.000 194 549 608 087 552 × 2 = 0 + 0.000 389 099 216 175 104;
  • 20) 0.000 389 099 216 175 104 × 2 = 0 + 0.000 778 198 432 350 208;
  • 21) 0.000 778 198 432 350 208 × 2 = 0 + 0.001 556 396 864 700 416;
  • 22) 0.001 556 396 864 700 416 × 2 = 0 + 0.003 112 793 729 400 832;
  • 23) 0.003 112 793 729 400 832 × 2 = 0 + 0.006 225 587 458 801 664;
  • 24) 0.006 225 587 458 801 664 × 2 = 0 + 0.012 451 174 917 603 328;
  • 25) 0.012 451 174 917 603 328 × 2 = 0 + 0.024 902 349 835 206 656;
  • 26) 0.024 902 349 835 206 656 × 2 = 0 + 0.049 804 699 670 413 312;
  • 27) 0.049 804 699 670 413 312 × 2 = 0 + 0.099 609 399 340 826 624;
  • 28) 0.099 609 399 340 826 624 × 2 = 0 + 0.199 218 798 681 653 248;
  • 29) 0.199 218 798 681 653 248 × 2 = 0 + 0.398 437 597 363 306 496;
  • 30) 0.398 437 597 363 306 496 × 2 = 0 + 0.796 875 194 726 612 992;
  • 31) 0.796 875 194 726 612 992 × 2 = 1 + 0.593 750 389 453 225 984;
  • 32) 0.593 750 389 453 225 984 × 2 = 1 + 0.187 500 778 906 451 968;
  • 33) 0.187 500 778 906 451 968 × 2 = 0 + 0.375 001 557 812 903 936;
  • 34) 0.375 001 557 812 903 936 × 2 = 0 + 0.750 003 115 625 807 872;
  • 35) 0.750 003 115 625 807 872 × 2 = 1 + 0.500 006 231 251 615 744;
  • 36) 0.500 006 231 251 615 744 × 2 = 1 + 0.000 012 462 503 231 488;
  • 37) 0.000 012 462 503 231 488 × 2 = 0 + 0.000 024 925 006 462 976;
  • 38) 0.000 024 925 006 462 976 × 2 = 0 + 0.000 049 850 012 925 952;
  • 39) 0.000 049 850 012 925 952 × 2 = 0 + 0.000 099 700 025 851 904;
  • 40) 0.000 099 700 025 851 904 × 2 = 0 + 0.000 199 400 051 703 808;
  • 41) 0.000 199 400 051 703 808 × 2 = 0 + 0.000 398 800 103 407 616;
  • 42) 0.000 398 800 103 407 616 × 2 = 0 + 0.000 797 600 206 815 232;
  • 43) 0.000 797 600 206 815 232 × 2 = 0 + 0.001 595 200 413 630 464;
  • 44) 0.001 595 200 413 630 464 × 2 = 0 + 0.003 190 400 827 260 928;
  • 45) 0.003 190 400 827 260 928 × 2 = 0 + 0.006 380 801 654 521 856;
  • 46) 0.006 380 801 654 521 856 × 2 = 0 + 0.012 761 603 309 043 712;
  • 47) 0.012 761 603 309 043 712 × 2 = 0 + 0.025 523 206 618 087 424;
  • 48) 0.025 523 206 618 087 424 × 2 = 0 + 0.051 046 413 236 174 848;
  • 49) 0.051 046 413 236 174 848 × 2 = 0 + 0.102 092 826 472 349 696;
  • 50) 0.102 092 826 472 349 696 × 2 = 0 + 0.204 185 652 944 699 392;
  • 51) 0.204 185 652 944 699 392 × 2 = 0 + 0.408 371 305 889 398 784;
  • 52) 0.408 371 305 889 398 784 × 2 = 0 + 0.816 742 611 778 797 568;
  • 53) 0.816 742 611 778 797 568 × 2 = 1 + 0.633 485 223 557 595 136;
  • 54) 0.633 485 223 557 595 136 × 2 = 1 + 0.266 970 447 115 190 272;

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:


0.000 000 000 742 147 858(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 858(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 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:


0.000 000 000 742 147 858(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 11(2) × 20 =

1.1001 1000 0000 0000 0000 011(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 011


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 0011 =

100 1100 0000 0000 0000 0011


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 0011


Decimal number -0.000 000 000 742 147 858 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0011

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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