-0.000 000 000 742 147 676 646 689 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 689(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
-0.000 000 000 742 147 676 646 689(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:

|-0.000 000 000 742 147 676 646 689| = 0.000 000 000 742 147 676 646 689


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: 0.000 000 000 742 147 676 646 689.

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) 0.000 000 000 742 147 676 646 689 × 2 = 0 + 0.000 000 001 484 295 353 293 378;
  • 2) 0.000 000 001 484 295 353 293 378 × 2 = 0 + 0.000 000 002 968 590 706 586 756;
  • 3) 0.000 000 002 968 590 706 586 756 × 2 = 0 + 0.000 000 005 937 181 413 173 512;
  • 4) 0.000 000 005 937 181 413 173 512 × 2 = 0 + 0.000 000 011 874 362 826 347 024;
  • 5) 0.000 000 011 874 362 826 347 024 × 2 = 0 + 0.000 000 023 748 725 652 694 048;
  • 6) 0.000 000 023 748 725 652 694 048 × 2 = 0 + 0.000 000 047 497 451 305 388 096;
  • 7) 0.000 000 047 497 451 305 388 096 × 2 = 0 + 0.000 000 094 994 902 610 776 192;
  • 8) 0.000 000 094 994 902 610 776 192 × 2 = 0 + 0.000 000 189 989 805 221 552 384;
  • 9) 0.000 000 189 989 805 221 552 384 × 2 = 0 + 0.000 000 379 979 610 443 104 768;
  • 10) 0.000 000 379 979 610 443 104 768 × 2 = 0 + 0.000 000 759 959 220 886 209 536;
  • 11) 0.000 000 759 959 220 886 209 536 × 2 = 0 + 0.000 001 519 918 441 772 419 072;
  • 12) 0.000 001 519 918 441 772 419 072 × 2 = 0 + 0.000 003 039 836 883 544 838 144;
  • 13) 0.000 003 039 836 883 544 838 144 × 2 = 0 + 0.000 006 079 673 767 089 676 288;
  • 14) 0.000 006 079 673 767 089 676 288 × 2 = 0 + 0.000 012 159 347 534 179 352 576;
  • 15) 0.000 012 159 347 534 179 352 576 × 2 = 0 + 0.000 024 318 695 068 358 705 152;
  • 16) 0.000 024 318 695 068 358 705 152 × 2 = 0 + 0.000 048 637 390 136 717 410 304;
  • 17) 0.000 048 637 390 136 717 410 304 × 2 = 0 + 0.000 097 274 780 273 434 820 608;
  • 18) 0.000 097 274 780 273 434 820 608 × 2 = 0 + 0.000 194 549 560 546 869 641 216;
  • 19) 0.000 194 549 560 546 869 641 216 × 2 = 0 + 0.000 389 099 121 093 739 282 432;
  • 20) 0.000 389 099 121 093 739 282 432 × 2 = 0 + 0.000 778 198 242 187 478 564 864;
  • 21) 0.000 778 198 242 187 478 564 864 × 2 = 0 + 0.001 556 396 484 374 957 129 728;
  • 22) 0.001 556 396 484 374 957 129 728 × 2 = 0 + 0.003 112 792 968 749 914 259 456;
  • 23) 0.003 112 792 968 749 914 259 456 × 2 = 0 + 0.006 225 585 937 499 828 518 912;
  • 24) 0.006 225 585 937 499 828 518 912 × 2 = 0 + 0.012 451 171 874 999 657 037 824;
  • 25) 0.012 451 171 874 999 657 037 824 × 2 = 0 + 0.024 902 343 749 999 314 075 648;
  • 26) 0.024 902 343 749 999 314 075 648 × 2 = 0 + 0.049 804 687 499 998 628 151 296;
  • 27) 0.049 804 687 499 998 628 151 296 × 2 = 0 + 0.099 609 374 999 997 256 302 592;
  • 28) 0.099 609 374 999 997 256 302 592 × 2 = 0 + 0.199 218 749 999 994 512 605 184;
  • 29) 0.199 218 749 999 994 512 605 184 × 2 = 0 + 0.398 437 499 999 989 025 210 368;
  • 30) 0.398 437 499 999 989 025 210 368 × 2 = 0 + 0.796 874 999 999 978 050 420 736;
  • 31) 0.796 874 999 999 978 050 420 736 × 2 = 1 + 0.593 749 999 999 956 100 841 472;
  • 32) 0.593 749 999 999 956 100 841 472 × 2 = 1 + 0.187 499 999 999 912 201 682 944;
  • 33) 0.187 499 999 999 912 201 682 944 × 2 = 0 + 0.374 999 999 999 824 403 365 888;
  • 34) 0.374 999 999 999 824 403 365 888 × 2 = 0 + 0.749 999 999 999 648 806 731 776;
  • 35) 0.749 999 999 999 648 806 731 776 × 2 = 1 + 0.499 999 999 999 297 613 463 552;
  • 36) 0.499 999 999 999 297 613 463 552 × 2 = 0 + 0.999 999 999 998 595 226 927 104;
  • 37) 0.999 999 999 998 595 226 927 104 × 2 = 1 + 0.999 999 999 997 190 453 854 208;
  • 38) 0.999 999 999 997 190 453 854 208 × 2 = 1 + 0.999 999 999 994 380 907 708 416;
  • 39) 0.999 999 999 994 380 907 708 416 × 2 = 1 + 0.999 999 999 988 761 815 416 832;
  • 40) 0.999 999 999 988 761 815 416 832 × 2 = 1 + 0.999 999 999 977 523 630 833 664;
  • 41) 0.999 999 999 977 523 630 833 664 × 2 = 1 + 0.999 999 999 955 047 261 667 328;
  • 42) 0.999 999 999 955 047 261 667 328 × 2 = 1 + 0.999 999 999 910 094 523 334 656;
  • 43) 0.999 999 999 910 094 523 334 656 × 2 = 1 + 0.999 999 999 820 189 046 669 312;
  • 44) 0.999 999 999 820 189 046 669 312 × 2 = 1 + 0.999 999 999 640 378 093 338 624;
  • 45) 0.999 999 999 640 378 093 338 624 × 2 = 1 + 0.999 999 999 280 756 186 677 248;
  • 46) 0.999 999 999 280 756 186 677 248 × 2 = 1 + 0.999 999 998 561 512 373 354 496;
  • 47) 0.999 999 998 561 512 373 354 496 × 2 = 1 + 0.999 999 997 123 024 746 708 992;
  • 48) 0.999 999 997 123 024 746 708 992 × 2 = 1 + 0.999 999 994 246 049 493 417 984;
  • 49) 0.999 999 994 246 049 493 417 984 × 2 = 1 + 0.999 999 988 492 098 986 835 968;
  • 50) 0.999 999 988 492 098 986 835 968 × 2 = 1 + 0.999 999 976 984 197 973 671 936;
  • 51) 0.999 999 976 984 197 973 671 936 × 2 = 1 + 0.999 999 953 968 395 947 343 872;
  • 52) 0.999 999 953 968 395 947 343 872 × 2 = 1 + 0.999 999 907 936 791 894 687 744;
  • 53) 0.999 999 907 936 791 894 687 744 × 2 = 1 + 0.999 999 815 873 583 789 375 488;
  • 54) 0.999 999 815 873 583 789 375 488 × 2 = 1 + 0.999 999 631 747 167 578 750 976;

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:


0.000 000 000 742 147 676 646 689(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 689(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:


0.000 000 000 742 147 676 646 689(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 -0.000 000 000 742 147 676 646 689 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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