-0.000 000 000 742 147 676 620 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 620 8(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
-0.000 000 000 742 147 676 620 8(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:

|-0.000 000 000 742 147 676 620 8| = 0.000 000 000 742 147 676 620 8


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: 0.000 000 000 742 147 676 620 8.

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) 0.000 000 000 742 147 676 620 8 × 2 = 0 + 0.000 000 001 484 295 353 241 6;
  • 2) 0.000 000 001 484 295 353 241 6 × 2 = 0 + 0.000 000 002 968 590 706 483 2;
  • 3) 0.000 000 002 968 590 706 483 2 × 2 = 0 + 0.000 000 005 937 181 412 966 4;
  • 4) 0.000 000 005 937 181 412 966 4 × 2 = 0 + 0.000 000 011 874 362 825 932 8;
  • 5) 0.000 000 011 874 362 825 932 8 × 2 = 0 + 0.000 000 023 748 725 651 865 6;
  • 6) 0.000 000 023 748 725 651 865 6 × 2 = 0 + 0.000 000 047 497 451 303 731 2;
  • 7) 0.000 000 047 497 451 303 731 2 × 2 = 0 + 0.000 000 094 994 902 607 462 4;
  • 8) 0.000 000 094 994 902 607 462 4 × 2 = 0 + 0.000 000 189 989 805 214 924 8;
  • 9) 0.000 000 189 989 805 214 924 8 × 2 = 0 + 0.000 000 379 979 610 429 849 6;
  • 10) 0.000 000 379 979 610 429 849 6 × 2 = 0 + 0.000 000 759 959 220 859 699 2;
  • 11) 0.000 000 759 959 220 859 699 2 × 2 = 0 + 0.000 001 519 918 441 719 398 4;
  • 12) 0.000 001 519 918 441 719 398 4 × 2 = 0 + 0.000 003 039 836 883 438 796 8;
  • 13) 0.000 003 039 836 883 438 796 8 × 2 = 0 + 0.000 006 079 673 766 877 593 6;
  • 14) 0.000 006 079 673 766 877 593 6 × 2 = 0 + 0.000 012 159 347 533 755 187 2;
  • 15) 0.000 012 159 347 533 755 187 2 × 2 = 0 + 0.000 024 318 695 067 510 374 4;
  • 16) 0.000 024 318 695 067 510 374 4 × 2 = 0 + 0.000 048 637 390 135 020 748 8;
  • 17) 0.000 048 637 390 135 020 748 8 × 2 = 0 + 0.000 097 274 780 270 041 497 6;
  • 18) 0.000 097 274 780 270 041 497 6 × 2 = 0 + 0.000 194 549 560 540 082 995 2;
  • 19) 0.000 194 549 560 540 082 995 2 × 2 = 0 + 0.000 389 099 121 080 165 990 4;
  • 20) 0.000 389 099 121 080 165 990 4 × 2 = 0 + 0.000 778 198 242 160 331 980 8;
  • 21) 0.000 778 198 242 160 331 980 8 × 2 = 0 + 0.001 556 396 484 320 663 961 6;
  • 22) 0.001 556 396 484 320 663 961 6 × 2 = 0 + 0.003 112 792 968 641 327 923 2;
  • 23) 0.003 112 792 968 641 327 923 2 × 2 = 0 + 0.006 225 585 937 282 655 846 4;
  • 24) 0.006 225 585 937 282 655 846 4 × 2 = 0 + 0.012 451 171 874 565 311 692 8;
  • 25) 0.012 451 171 874 565 311 692 8 × 2 = 0 + 0.024 902 343 749 130 623 385 6;
  • 26) 0.024 902 343 749 130 623 385 6 × 2 = 0 + 0.049 804 687 498 261 246 771 2;
  • 27) 0.049 804 687 498 261 246 771 2 × 2 = 0 + 0.099 609 374 996 522 493 542 4;
  • 28) 0.099 609 374 996 522 493 542 4 × 2 = 0 + 0.199 218 749 993 044 987 084 8;
  • 29) 0.199 218 749 993 044 987 084 8 × 2 = 0 + 0.398 437 499 986 089 974 169 6;
  • 30) 0.398 437 499 986 089 974 169 6 × 2 = 0 + 0.796 874 999 972 179 948 339 2;
  • 31) 0.796 874 999 972 179 948 339 2 × 2 = 1 + 0.593 749 999 944 359 896 678 4;
  • 32) 0.593 749 999 944 359 896 678 4 × 2 = 1 + 0.187 499 999 888 719 793 356 8;
  • 33) 0.187 499 999 888 719 793 356 8 × 2 = 0 + 0.374 999 999 777 439 586 713 6;
  • 34) 0.374 999 999 777 439 586 713 6 × 2 = 0 + 0.749 999 999 554 879 173 427 2;
  • 35) 0.749 999 999 554 879 173 427 2 × 2 = 1 + 0.499 999 999 109 758 346 854 4;
  • 36) 0.499 999 999 109 758 346 854 4 × 2 = 0 + 0.999 999 998 219 516 693 708 8;
  • 37) 0.999 999 998 219 516 693 708 8 × 2 = 1 + 0.999 999 996 439 033 387 417 6;
  • 38) 0.999 999 996 439 033 387 417 6 × 2 = 1 + 0.999 999 992 878 066 774 835 2;
  • 39) 0.999 999 992 878 066 774 835 2 × 2 = 1 + 0.999 999 985 756 133 549 670 4;
  • 40) 0.999 999 985 756 133 549 670 4 × 2 = 1 + 0.999 999 971 512 267 099 340 8;
  • 41) 0.999 999 971 512 267 099 340 8 × 2 = 1 + 0.999 999 943 024 534 198 681 6;
  • 42) 0.999 999 943 024 534 198 681 6 × 2 = 1 + 0.999 999 886 049 068 397 363 2;
  • 43) 0.999 999 886 049 068 397 363 2 × 2 = 1 + 0.999 999 772 098 136 794 726 4;
  • 44) 0.999 999 772 098 136 794 726 4 × 2 = 1 + 0.999 999 544 196 273 589 452 8;
  • 45) 0.999 999 544 196 273 589 452 8 × 2 = 1 + 0.999 999 088 392 547 178 905 6;
  • 46) 0.999 999 088 392 547 178 905 6 × 2 = 1 + 0.999 998 176 785 094 357 811 2;
  • 47) 0.999 998 176 785 094 357 811 2 × 2 = 1 + 0.999 996 353 570 188 715 622 4;
  • 48) 0.999 996 353 570 188 715 622 4 × 2 = 1 + 0.999 992 707 140 377 431 244 8;
  • 49) 0.999 992 707 140 377 431 244 8 × 2 = 1 + 0.999 985 414 280 754 862 489 6;
  • 50) 0.999 985 414 280 754 862 489 6 × 2 = 1 + 0.999 970 828 561 509 724 979 2;
  • 51) 0.999 970 828 561 509 724 979 2 × 2 = 1 + 0.999 941 657 123 019 449 958 4;
  • 52) 0.999 941 657 123 019 449 958 4 × 2 = 1 + 0.999 883 314 246 038 899 916 8;
  • 53) 0.999 883 314 246 038 899 916 8 × 2 = 1 + 0.999 766 628 492 077 799 833 6;
  • 54) 0.999 766 628 492 077 799 833 6 × 2 = 1 + 0.999 533 256 984 155 599 667 2;

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:


0.000 000 000 742 147 676 620 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 620 8(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:


0.000 000 000 742 147 676 620 8(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 -0.000 000 000 742 147 676 620 8 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