-0.000 000 000 742 147 676 599 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 599 1(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
-0.000 000 000 742 147 676 599 1(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:

|-0.000 000 000 742 147 676 599 1| = 0.000 000 000 742 147 676 599 1


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: 0.000 000 000 742 147 676 599 1.

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) 0.000 000 000 742 147 676 599 1 × 2 = 0 + 0.000 000 001 484 295 353 198 2;
  • 2) 0.000 000 001 484 295 353 198 2 × 2 = 0 + 0.000 000 002 968 590 706 396 4;
  • 3) 0.000 000 002 968 590 706 396 4 × 2 = 0 + 0.000 000 005 937 181 412 792 8;
  • 4) 0.000 000 005 937 181 412 792 8 × 2 = 0 + 0.000 000 011 874 362 825 585 6;
  • 5) 0.000 000 011 874 362 825 585 6 × 2 = 0 + 0.000 000 023 748 725 651 171 2;
  • 6) 0.000 000 023 748 725 651 171 2 × 2 = 0 + 0.000 000 047 497 451 302 342 4;
  • 7) 0.000 000 047 497 451 302 342 4 × 2 = 0 + 0.000 000 094 994 902 604 684 8;
  • 8) 0.000 000 094 994 902 604 684 8 × 2 = 0 + 0.000 000 189 989 805 209 369 6;
  • 9) 0.000 000 189 989 805 209 369 6 × 2 = 0 + 0.000 000 379 979 610 418 739 2;
  • 10) 0.000 000 379 979 610 418 739 2 × 2 = 0 + 0.000 000 759 959 220 837 478 4;
  • 11) 0.000 000 759 959 220 837 478 4 × 2 = 0 + 0.000 001 519 918 441 674 956 8;
  • 12) 0.000 001 519 918 441 674 956 8 × 2 = 0 + 0.000 003 039 836 883 349 913 6;
  • 13) 0.000 003 039 836 883 349 913 6 × 2 = 0 + 0.000 006 079 673 766 699 827 2;
  • 14) 0.000 006 079 673 766 699 827 2 × 2 = 0 + 0.000 012 159 347 533 399 654 4;
  • 15) 0.000 012 159 347 533 399 654 4 × 2 = 0 + 0.000 024 318 695 066 799 308 8;
  • 16) 0.000 024 318 695 066 799 308 8 × 2 = 0 + 0.000 048 637 390 133 598 617 6;
  • 17) 0.000 048 637 390 133 598 617 6 × 2 = 0 + 0.000 097 274 780 267 197 235 2;
  • 18) 0.000 097 274 780 267 197 235 2 × 2 = 0 + 0.000 194 549 560 534 394 470 4;
  • 19) 0.000 194 549 560 534 394 470 4 × 2 = 0 + 0.000 389 099 121 068 788 940 8;
  • 20) 0.000 389 099 121 068 788 940 8 × 2 = 0 + 0.000 778 198 242 137 577 881 6;
  • 21) 0.000 778 198 242 137 577 881 6 × 2 = 0 + 0.001 556 396 484 275 155 763 2;
  • 22) 0.001 556 396 484 275 155 763 2 × 2 = 0 + 0.003 112 792 968 550 311 526 4;
  • 23) 0.003 112 792 968 550 311 526 4 × 2 = 0 + 0.006 225 585 937 100 623 052 8;
  • 24) 0.006 225 585 937 100 623 052 8 × 2 = 0 + 0.012 451 171 874 201 246 105 6;
  • 25) 0.012 451 171 874 201 246 105 6 × 2 = 0 + 0.024 902 343 748 402 492 211 2;
  • 26) 0.024 902 343 748 402 492 211 2 × 2 = 0 + 0.049 804 687 496 804 984 422 4;
  • 27) 0.049 804 687 496 804 984 422 4 × 2 = 0 + 0.099 609 374 993 609 968 844 8;
  • 28) 0.099 609 374 993 609 968 844 8 × 2 = 0 + 0.199 218 749 987 219 937 689 6;
  • 29) 0.199 218 749 987 219 937 689 6 × 2 = 0 + 0.398 437 499 974 439 875 379 2;
  • 30) 0.398 437 499 974 439 875 379 2 × 2 = 0 + 0.796 874 999 948 879 750 758 4;
  • 31) 0.796 874 999 948 879 750 758 4 × 2 = 1 + 0.593 749 999 897 759 501 516 8;
  • 32) 0.593 749 999 897 759 501 516 8 × 2 = 1 + 0.187 499 999 795 519 003 033 6;
  • 33) 0.187 499 999 795 519 003 033 6 × 2 = 0 + 0.374 999 999 591 038 006 067 2;
  • 34) 0.374 999 999 591 038 006 067 2 × 2 = 0 + 0.749 999 999 182 076 012 134 4;
  • 35) 0.749 999 999 182 076 012 134 4 × 2 = 1 + 0.499 999 998 364 152 024 268 8;
  • 36) 0.499 999 998 364 152 024 268 8 × 2 = 0 + 0.999 999 996 728 304 048 537 6;
  • 37) 0.999 999 996 728 304 048 537 6 × 2 = 1 + 0.999 999 993 456 608 097 075 2;
  • 38) 0.999 999 993 456 608 097 075 2 × 2 = 1 + 0.999 999 986 913 216 194 150 4;
  • 39) 0.999 999 986 913 216 194 150 4 × 2 = 1 + 0.999 999 973 826 432 388 300 8;
  • 40) 0.999 999 973 826 432 388 300 8 × 2 = 1 + 0.999 999 947 652 864 776 601 6;
  • 41) 0.999 999 947 652 864 776 601 6 × 2 = 1 + 0.999 999 895 305 729 553 203 2;
  • 42) 0.999 999 895 305 729 553 203 2 × 2 = 1 + 0.999 999 790 611 459 106 406 4;
  • 43) 0.999 999 790 611 459 106 406 4 × 2 = 1 + 0.999 999 581 222 918 212 812 8;
  • 44) 0.999 999 581 222 918 212 812 8 × 2 = 1 + 0.999 999 162 445 836 425 625 6;
  • 45) 0.999 999 162 445 836 425 625 6 × 2 = 1 + 0.999 998 324 891 672 851 251 2;
  • 46) 0.999 998 324 891 672 851 251 2 × 2 = 1 + 0.999 996 649 783 345 702 502 4;
  • 47) 0.999 996 649 783 345 702 502 4 × 2 = 1 + 0.999 993 299 566 691 405 004 8;
  • 48) 0.999 993 299 566 691 405 004 8 × 2 = 1 + 0.999 986 599 133 382 810 009 6;
  • 49) 0.999 986 599 133 382 810 009 6 × 2 = 1 + 0.999 973 198 266 765 620 019 2;
  • 50) 0.999 973 198 266 765 620 019 2 × 2 = 1 + 0.999 946 396 533 531 240 038 4;
  • 51) 0.999 946 396 533 531 240 038 4 × 2 = 1 + 0.999 892 793 067 062 480 076 8;
  • 52) 0.999 892 793 067 062 480 076 8 × 2 = 1 + 0.999 785 586 134 124 960 153 6;
  • 53) 0.999 785 586 134 124 960 153 6 × 2 = 1 + 0.999 571 172 268 249 920 307 2;
  • 54) 0.999 571 172 268 249 920 307 2 × 2 = 1 + 0.999 142 344 536 499 840 614 4;

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:


0.000 000 000 742 147 676 599 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 599 1(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:


0.000 000 000 742 147 676 599 1(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 -0.000 000 000 742 147 676 599 1 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