0.002 083 332 339 922 638 887 230 732 507 305 214 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.002 083 332 339 922 638 887 230 732 507 305 214 4(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
0.002 083 332 339 922 638 887 230 732 507 305 214 4(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.002 083 332 339 922 638 887 230 732 507 305 214 4.

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) 0.002 083 332 339 922 638 887 230 732 507 305 214 4 × 2 = 0 + 0.004 166 664 679 845 277 774 461 465 014 610 428 8;
  • 2) 0.004 166 664 679 845 277 774 461 465 014 610 428 8 × 2 = 0 + 0.008 333 329 359 690 555 548 922 930 029 220 857 6;
  • 3) 0.008 333 329 359 690 555 548 922 930 029 220 857 6 × 2 = 0 + 0.016 666 658 719 381 111 097 845 860 058 441 715 2;
  • 4) 0.016 666 658 719 381 111 097 845 860 058 441 715 2 × 2 = 0 + 0.033 333 317 438 762 222 195 691 720 116 883 430 4;
  • 5) 0.033 333 317 438 762 222 195 691 720 116 883 430 4 × 2 = 0 + 0.066 666 634 877 524 444 391 383 440 233 766 860 8;
  • 6) 0.066 666 634 877 524 444 391 383 440 233 766 860 8 × 2 = 0 + 0.133 333 269 755 048 888 782 766 880 467 533 721 6;
  • 7) 0.133 333 269 755 048 888 782 766 880 467 533 721 6 × 2 = 0 + 0.266 666 539 510 097 777 565 533 760 935 067 443 2;
  • 8) 0.266 666 539 510 097 777 565 533 760 935 067 443 2 × 2 = 0 + 0.533 333 079 020 195 555 131 067 521 870 134 886 4;
  • 9) 0.533 333 079 020 195 555 131 067 521 870 134 886 4 × 2 = 1 + 0.066 666 158 040 391 110 262 135 043 740 269 772 8;
  • 10) 0.066 666 158 040 391 110 262 135 043 740 269 772 8 × 2 = 0 + 0.133 332 316 080 782 220 524 270 087 480 539 545 6;
  • 11) 0.133 332 316 080 782 220 524 270 087 480 539 545 6 × 2 = 0 + 0.266 664 632 161 564 441 048 540 174 961 079 091 2;
  • 12) 0.266 664 632 161 564 441 048 540 174 961 079 091 2 × 2 = 0 + 0.533 329 264 323 128 882 097 080 349 922 158 182 4;
  • 13) 0.533 329 264 323 128 882 097 080 349 922 158 182 4 × 2 = 1 + 0.066 658 528 646 257 764 194 160 699 844 316 364 8;
  • 14) 0.066 658 528 646 257 764 194 160 699 844 316 364 8 × 2 = 0 + 0.133 317 057 292 515 528 388 321 399 688 632 729 6;
  • 15) 0.133 317 057 292 515 528 388 321 399 688 632 729 6 × 2 = 0 + 0.266 634 114 585 031 056 776 642 799 377 265 459 2;
  • 16) 0.266 634 114 585 031 056 776 642 799 377 265 459 2 × 2 = 0 + 0.533 268 229 170 062 113 553 285 598 754 530 918 4;
  • 17) 0.533 268 229 170 062 113 553 285 598 754 530 918 4 × 2 = 1 + 0.066 536 458 340 124 227 106 571 197 509 061 836 8;
  • 18) 0.066 536 458 340 124 227 106 571 197 509 061 836 8 × 2 = 0 + 0.133 072 916 680 248 454 213 142 395 018 123 673 6;
  • 19) 0.133 072 916 680 248 454 213 142 395 018 123 673 6 × 2 = 0 + 0.266 145 833 360 496 908 426 284 790 036 247 347 2;
  • 20) 0.266 145 833 360 496 908 426 284 790 036 247 347 2 × 2 = 0 + 0.532 291 666 720 993 816 852 569 580 072 494 694 4;
  • 21) 0.532 291 666 720 993 816 852 569 580 072 494 694 4 × 2 = 1 + 0.064 583 333 441 987 633 705 139 160 144 989 388 8;
  • 22) 0.064 583 333 441 987 633 705 139 160 144 989 388 8 × 2 = 0 + 0.129 166 666 883 975 267 410 278 320 289 978 777 6;
  • 23) 0.129 166 666 883 975 267 410 278 320 289 978 777 6 × 2 = 0 + 0.258 333 333 767 950 534 820 556 640 579 957 555 2;
  • 24) 0.258 333 333 767 950 534 820 556 640 579 957 555 2 × 2 = 0 + 0.516 666 667 535 901 069 641 113 281 159 915 110 4;
  • 25) 0.516 666 667 535 901 069 641 113 281 159 915 110 4 × 2 = 1 + 0.033 333 335 071 802 139 282 226 562 319 830 220 8;
  • 26) 0.033 333 335 071 802 139 282 226 562 319 830 220 8 × 2 = 0 + 0.066 666 670 143 604 278 564 453 124 639 660 441 6;
  • 27) 0.066 666 670 143 604 278 564 453 124 639 660 441 6 × 2 = 0 + 0.133 333 340 287 208 557 128 906 249 279 320 883 2;
  • 28) 0.133 333 340 287 208 557 128 906 249 279 320 883 2 × 2 = 0 + 0.266 666 680 574 417 114 257 812 498 558 641 766 4;
  • 29) 0.266 666 680 574 417 114 257 812 498 558 641 766 4 × 2 = 0 + 0.533 333 361 148 834 228 515 624 997 117 283 532 8;
  • 30) 0.533 333 361 148 834 228 515 624 997 117 283 532 8 × 2 = 1 + 0.066 666 722 297 668 457 031 249 994 234 567 065 6;
  • 31) 0.066 666 722 297 668 457 031 249 994 234 567 065 6 × 2 = 0 + 0.133 333 444 595 336 914 062 499 988 469 134 131 2;
  • 32) 0.133 333 444 595 336 914 062 499 988 469 134 131 2 × 2 = 0 + 0.266 666 889 190 673 828 124 999 976 938 268 262 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).


4. 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:


0.002 083 332 339 922 638 887 230 732 507 305 214 4(10) =

0.0000 0000 1000 1000 1000 1000 1000 0100(2)

5. Positive number before normalization:

0.002 083 332 339 922 638 887 230 732 507 305 214 4(10) =

0.0000 0000 1000 1000 1000 1000 1000 0100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the right, so that only one non zero digit remains to the left of it:


0.002 083 332 339 922 638 887 230 732 507 305 214 4(10) =

0.0000 0000 1000 1000 1000 1000 1000 0100(2) =

0.0000 0000 1000 1000 1000 1000 1000 0100(2) × 20 =

1.0001 0001 0001 0001 0000 100(2) × 2-9


7. 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 0 (a positive number)

Exponent (unadjusted): -9

Mantissa (not normalized):
1.0001 0001 0001 0001 0000 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-9 + 2(8-1) - 1 =

(-9 + 127)(10) =

118(10)


9. 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;
  • 118 ÷ 2 = 59 + 0;
  • 59 ÷ 2 = 29 + 1;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

118(10) =

0111 0110(2)


11. 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. 000 1000 1000 1000 1000 0100 =

000 1000 1000 1000 1000 0100


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0111 0110

Mantissa (23 bits) =
000 1000 1000 1000 1000 0100


Decimal number 0.002 083 332 339 922 638 887 230 732 507 305 214 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 0110 - 000 1000 1000 1000 1000 0100

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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